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CHAPTER 19CHAPTER 19

3

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

SOLUTION

Simple harmonic motion. x x tm n= +sin( )w f

where w p pn

m

T

x

= = =

= = ¥ -

2 2

0 162 83

5 5 10 3

..

srad/s

mm m

Then x t= ¥ +-5 10 62 833 sin( . )f

Differentiate to obtain velocity and acceleration.

v x x t

a x x t

n m n

n m n

= = +

= = - +

&&&

w w f

w w f

cos( )

sin( )2

Maximum velocity. | | ( . )( )max&x xn m= = ¥ -w 62 83 5 10 3 &xmax .= 0 314 m/s �

Maximum acceleration. | | ( . ) ( )max&&x xn m= = ¥ -w2 2 362 83 5 10 &&x amax max .= = 19 74 m/s2 �

PROBLEM 19.1

Determine the maximum velocity and maximum acceleration of a particle which moves in simple harmonic motion with an amplitude of 5 mm and a period of 0.1 s.

4


SOLUTION


where w p p pn nf= = =2 2 40 80( )( ) rad/s

Differentiate to obtain velocity and acceleration.

v x x t

v x

a x x t

a x

n m n

m n m

n m n

m n m

= = +=

= = - +

=

&

&&

w w fw

w w f

w

cos( )

sin( )2

2

Use information given to obtain amplitude and maximum velocity.

xa

mm

n

= = =w p2 2

60

800 000950

m/sm

2

( ). xm = 0 950. mm �

v xm n m= = =w p( )( . ) .80 0 000950 2387 m/s vm = 239 mm/s �

PROBLEM 19.2

Determine the amplitude and maximum velocity of a particle which moves in simple harmonic motion with a maximum acceleration of 60 m/s2 and a frequency of 40 Hz.

5


PROBLEM 19.3

A particle moves in simple harmonic motion. Knowing that the amplitude is 300 mm and the maximum acceleration is 5 m/s2, determine the maximum velocity of the particle and the frequency of its motion.

SOLUTION

Simple harmonic motion. x x t x

x t

x

m n m

n

n

= + == +=

sin( ) .

( . ) sin( ) ( )

( . )( )c

w fw f

w

0 300

0 300

0 3

m

m

& oos( ) ( )

( . )( ) sin( )

( . )

w f

w w fn

n n

m

t

x t

a

+

= - +

=

m/s

(m/s)

| | m/s

&& 0 3

0 3

2

(( )wn ma2 5= m/s2

Natural frequency. w

ww

nm

n

n

a

f

2

0 3

5

0 316 667

4 082

= = =

=

=

| |

( . )

( )

( . ).

.

m

m/s

mrad/s

rad/s

22

nn

2p

fn = =( . )

( ).

4 082

20 6497

rad/s

rad/cycleHz

p fn = 0 650. Hz �

Maximum velocity. v xm m n= =w ( . )( . )0 3 4 082m rad/s vm = 1 225. m/s �

6


PROBLEM 19.4

A 5 kg block is supported by the spring shown. If the block is moved vertically downward from its equilibrium position and released, determine (a) the period and frequency of the resulting motion, (b) the maximum velocity and acceleration of the block if the amplitude of its motion is 50 mm.

SOLUTION

(a) Simple harmonic motion. x x tm n= +sin( )w f


w

w

t pw

n

n

n

nn

k

mk= =

=

= =

=

12000

12000

15

20 2 28 284

2

N/m

N/m

kg

rad/s

( )

.

t pn = =2

28 2840 22214

.. s t n = 0 222. s �

fnn

= = =1 1

0 222144 50

t .. Hz �

(b) x

x tm = =

= +50 0 05

0 05 28 284

mm m.

. sin ( . )f

Maximum velocity. v xm m n= =w ( . )( . )0 05 28 284 vm = 1 414. m/s �

Maximum acceleration. a xm m n= =w2 0 05( . )(28.284)2 am = 40 0. m/s2 �

7


PROBLEM 19.5

A 32-kg block is attached to a spring and can move without friction in a slot as shown. The block is in its equilibrium position when it is struck by a hammer, which imparts to the block an initial velocity of 250 mm/s. Determine (a) the period and frequency of the resulting motion, (b) the amplitude of the motion and the maximum acceleration of the block.

SOLUTION

(a) x x t

k

m

m n

n

n

nn

= +

= =¥

=

=

sin( )

.

w f

w

w

t pw

12 10

19 365

2

3 N/m

32 kg

rad/s

t p

t

n

n

=

=

2

19 3650 324

.. s fn

n

= = =1 1

0 3243 08

t .. Hz �

(b) At t x= =0 00, , &x v0 0 250= = mm/s

Thus, x xm n0 0 0= = +sin( ( ) )w f

and f = 0

&x v x x

v x

x

m n n m n

m

m

0 0

0

0 0

0 250 19 365

= = + == =

=

w w wcos( ( ) )

. ( . )m/s rad/s

(( . )

( . )

.

0 250

19 365

12 91 10 3

m/s

rad/s

mxm = ¥ - xm = 12 91. mm �

a xm m n= = ¥ -w2 312 91 10( . m)(19.365 rad/s)2 am = 4 84. m/s2 �

8


PROBLEM 19.6

A simple pendulum consisting of a bob attached to a cord oscillates in a vertical plane with a period of 1.3 s. Assuming simple harmonic motion and knowing that the maximum velocity of the bob is 400 mm/s, determine (a) the amplitude of the motion in degrees, (b) the maximum tangential acceleration of the bob.

SOLUTION

(a) Simple harmonic motion. q q w f

w pt

p

w

q q w w

= +

= =

=

=

m n

nn

n

m n n

tsin( )

( )

( . )

.

cos(

2 2

1 3

4 833

s

rad/s& tt

v l l

m m n

m m m n

+

=

= =

f

q q w

q q w

)&

&

qwm

m

n

v

l= (1)

For a simple pendulum, wng

l=

Thus, lg

ln

= =

=w2 2

9 81

4 833

0 4200

.

( . )

.

m/s

rad/s

m

2

Amplitude.

From Eq. (1), qwm

m

n

v

l= =

( . )

( . )( . / )

0 4

0 42 4 833

m/s

m rad s

qm = 0 19706. rad qm = ∞11 29. �

(b) Maximum tangential acceleration. a lt = &&q Maximum tangential acceleration occurs when &&q is maximum.

&&&&

q q w w f

q q w

q w

= - +

=

=

=

m n n

m n

t m n

t

t

a l

a

2

2

2

0 4

sin( )

( )

( ) ( .

max

max

max 2200 0 19706 4 833 2m rad rad/s)( . )( . ) ( ) .at m = 1 933 m/s2 �

9


PROBLEM 19.7

A simple pendulum consisting of a bob attached to a cord of length l = 800 mm oscillates in a vertical plane. Assuming simple harmonic motion and knowing that the bob is released from rest when q = ∞6 , determine (a) the frequency of oscillation, (b) the maximum velocity of the bob.

SOLUTION

(a) Frequency. wng

l= =

( . )

( . )

9 81

0 8

m/s

m

2

ww

p p

n

nnf

=

= =

3 502

2

3 502

2

.

( . )

rad/s

rad/s fn = 0 557. Hz �

(b) Simple harmonic motion. q q w f= +m ntsin( )

where qm = ∞ =6 0 10472. rad

Maximum velocity. &q q w w f= +m n ntcos( )

&&&

q q w

q q wm m n

m m m n

m

v l l

v

=

= = =

= ¥ -

( . )( . )( . )

.

0 8 0 10472 3 502

293 4 10 3

m

mm/s vm = 293 mm/s �

10


PROBLEM 19.8

An instrument package A is bolted to a shaker table as shown. The table moves vertically in simple harmonic motion at the same frequency as the variable-speed motor which drives it. The package is to be tested at a peak acceleration of 50 m/s2. Knowing that the amplitude of the shaker table is 60 mm, determine (a) the required speed of the motor in rpm, (b) the maximum velocity of the table.

SOLUTION

In simple harmonic motion,

a x n

n

n

n

max max

( . )

( . )

.

=

=

==

w

w

ww

2

2

2 2

50 0 06

833 33

28 8675

m/s m

rad/s

r

2

aad/s

Hz (cycles per second)

fnn=

=

=

wp

p

228 8675

24 5945

.

.

(a) Motor speed. ( . )( )4 5945 60rev/s s/min speed rpm= 276 �

(b) Maximum velocity. v x nmax max ( . )( . )= =w 0 06 28 8675m rad/s vmax .= 1 732 m/s �

11


PROBLEM 19.9

The motion of a particle is described by the equation x t t= +5 2 4 2sin cos , where x is expressed in meters and t in seconds. Determine (a) the period of the motion, (b) its amplitude, (c) its phase angle.

SOLUTION

For simple harmonic motion x x tm n= +sin( )w f

Double angle formula (trigonometry): sin( ) (sin )(cos ) (sin )(cos )A B A B B A+ = +

Let A t Bn= =w f,

Then x x t

x x t x t

x x

m n

m n m n

m

= += +=

sin( )

(sin )(cos ) (sin )(cos )

( cos

w fw f f wf))(sin ) ( sin )(cos )w f wn m nt x t+

Given x t t= +5 2 4 2sin cos

Comparing, w fn mx= =2 5cos (1)

xm sinf = 4 (2)

(a) t pw

p pnn

= = =2 2

2( )rad/ss t = 3 14. s �

(b) Squaring Eqs. (1) and (2) and adding,

x x

x x

m m

m m

2 2 2 2 2 2

2 2 2 2

4 5

41

cos sin

(cos sin )

f f

f f

+ = +

+ = = m2 xm = 6 40. m �

(c) Dividing Eq. (2) by Eq. (1), tanf = 4

5 f = ∞38 7. �

12


PROBLEM 19.10

An instrument package B is placed on the shaking table C as shown. The table is made to move horizontally in simple harmonic motion with a frequency of 3 Hz. Knowing that the coefficient of static friction is ms = 0 40. between the package and the table, determine the largest allowable amplitude of the motion if the package is not to slip on the table. Given answers in both SI and U.S. customary units.

SOLUTION

Maximum allowable acceleration of B.

ms = 0 40.

+ SF ma= : F ma

mg ma

a g a g

m m

s m

m s m

=== =

mm 0 40.

Simple harmonic motion. f

a x

g x

x

nn

n

m m n

m

m

= =

=

=

=

= ¥

32

6

0 40 6

1 1258 1

2

2

Hz

rad/s

rad/s

wp

w p

w

p. ( )

. 00 3- g

Largest allowable amplitude.

SI: xm = ¥ = ¥- -1 1258 10 9 81 11 044 103 3. ( . ) . m xm = 11 04. mm �

13


PROBLEM 19.11

A 32-kg block attached to a spring of constant k = 12 kN/m can move without friction in a slot as shown. The block is given an initial 300-mm displacement downward from its equilibrium position and released. Determine 1.5 s after the block has been released (a) the total distance traveled by the block, (b) the acceleration of the block.

SOLUTION

(a) x x t

k

m

m n

n

n

n

= +

= = ¥

=

=

sin( )

(

(

.

w f

w

w

t

12 10

32

19 365

2

3 N/m)

kg)

rad/s

ppw

p

tn

n

=

=

( )

( . )

.

2

19 365

0 3245 s

Initial conditions. x x

x

x x

x

m

m n

m

( ) . ( )

. sin( )

( ) cos( )

0 0 3 0 0

0 3 0

0 0 0

2

= == += = +

=

m, &

&f

w f

f p

==

= +ÊËÁ

ˆ¯

=

=

0 3

0 3 19 3652

0 3245

1 5 0

.

( ) ( . )sin . , .

( . ( .

x t t

x

np t s

s) 33 19 365 1 52

0 2147

1 5 0 3 19 36

)sin ( . )( . ) .

( . ) ( . )( .

+ÈÎÍ

˘˚

= -

=

p m

s&x 55 19 365 1 52

4 057)cos ( . )( . ) .+ÈÎÍ

˘˚

= Øp m/s

In one cycle, block travels ( )( .4 0 3 m) 1.2 m=

To travel 4 cycles, it takes (4 cyc)(0.3245 s/cyc) 1.2980 s=

at t = 1 5. s. Thus, total distance traveled is

4 1 2 0 6 0 3 0 2147 5 49( . ) . ( . . ) .+ + - = m �

(b) &&x( . ) ( . )( . ) sin ( . )( . ) .1 5 0 3 19 365 19 365 1 52

80 52= - +ÈÎÍ

˘˚

=p m/s22 Ø �

14


PROBLEM 19.12

A 2 kg block is supported as shown by a spring of constant k = 400 N/m, which can act in tension or compression. The block is in its equilibrium position when it is struck from below by a hammer, which imparts to the block an upward velocity of 2.5 m/s. Determine (a) the time required for the block to move 100 mm upward, (b) the corresponding velocity and acceleration of the block.

SOLUTION


Natural frequency. wnk

mk= =, 400 N/m

w

wf

f

n

n

mx x

x x

=

= == = +==

400

2

10 2 14 1421

0 0 0

0

0

N/m

rad/s.

( ) sin( )

( )& mm n

m m

x

x x

x

w cos( )

( ) .

. ( . ) .

( .

0 0

0 2 5

2 5 14 1421 0 17678

0 1

+== ==

& m/s

m

77678 14 1421)sin( . )(t m/s) (1)

(a) Time at mm ( m)x x= =100 0 1.

0 1 0 17678 14 1421

14 14210 04252

1 0 10 17678

. . sin( . )

sin

..

..

=

=( )

=-

t

t s

st = 0 046. �

Note: Since w is in rad/s, convert the argument of sin-1 to radians.

(b) Velocity and acceleration.

&&&

&

x x t

x x t

t

x

m n n

m n n

=

= -==

w w

w w

cos( )

sin

.

( . )( .

2

0 04252

0 17678 14 14221 14 1421 0 04252

2 0615

)cos[( . )( . )]

.&x = m/s v = ≠2 06. m/s �

&&x = -

= -

( . )( . ) sin[( . )( . )]0 1768 14 1421 14 1421 0 04252

20

2

m/s2 a = Ø20 0. m/s2 �

15


PROBLEM 19.13

In Problem 19.12, determine the position, velocity, and acceleration of the block 0.90 s after it has been struck by the hammer.

SOLUTION



w

w

n

n

n

k

mk= =

=

= =

,

.

400

400

2

10 2 14 1421

N/m

N/m

kg

rad/s

x x

x x x

x

m

m n

m

( ) sin( )

( ) cos( ) ( ) .

. (

0 0 0

0

0 0 0 0 2 5

2 5

= = +== + ==

ff

w& & m/s

114 1421 0 17678

0 17678 14 1421

. ) .

( . )sin ( . )(

x

x tm =

= m

m/s)

Simple harmonic motion. x x t

x x t

x x t

m n

m n n

m n n

= += +

= - +

sin( )

cos( )

sin( )

w fw w f

w w f

&&& 2

Note: arguments of sin and cos are in radians.

At 0.90 s: x = =( . )sin[( . )( . )] .0 17678 14 1421 0 90 0 02843 m x = ≠0 0284. m �

&x = = -( . )( . )cos[( . )( . )] .0 17678 14 1421 14 1421 0 90 2 4675 m/s v = ≠2 47. m/s �

&&x = - = -( . )( . ) sin[( . )( . )] .0 17678 14 1421 14 1421 0 90 5 68592 m/s2 a = Ø5 69. m/s2 �

16


PROBLEM 19.14

The bob of a simple pendulum of length l = 800 mm is released from rest when q = + ∞5 . Assuming simple harmonic motion, determine 1.6 s after release (a) the angle q, (b) the magnitudes of the velocity and acceleration of the bob.

SOLUTION

q q w f w

w

= + = =

=

m n n

n

tg

lsin( )

.

..

9 81

0 83 502

m/s

m rad/s

Initial conditions: q p

q

q p q f

q q

( )( )( )

( )

( ) sin( )

( )

0 55

180

0 0

05

1800

0 0

= ∞ =

=

= = +

= =

rad&

&m

mm n

m

t

w f

f p

q p

q p p

cos( )

sin .

0

25

1805

1803 502

2

+

=

=

= +ÊËÁ

ˆ¯

rad

(a) At t = 1 6. s: q p p= +ÈÎÍ

˘˚

5

1803 502 1 6

2sin ( . )( . ) q = =0 06786. rad 3.89° �

(b)

17


PROBLEM 19.14 (Continued)

&q q w w f p p= + = ÊËÁ

ˆ¯

+ÈÎÍ

˘m n ntcos( ) ( . )cos ( . )( . )

5

1803 502 3 502 1 6

2 ˚=&q( .1 6 s) 0.19223 rad/s

v l= = =&q ( .0 800 m)(0.19223 rad/s) 0.1538 m/s �

&&q q w w fp p

= - +

= -ÊËÁ

ˆ¯

+

m n nt2

25

1803 502 3 502 1 6

sin( )

( . ) sin ( . )( . )22

0 8319

0 8

2 2

ÈÎÍ

˘˚

= -

= +

= = -

&&

&&

q

q

.

( ) ( )

( . (

rad/s

m) 0.

2

a a a

a l

t n

t 88319 rad/s m/s

m)(0.19223 rad/s)

2 2

2

) .

( .

= -

= = =

0 6655

0 82a ln&q 00 02956

0 6655 0 02956 0 66622 2

.

( . ) ( . ) .

m/s

m/s

2

2a = + = a = 0 666. m/s2 �

18


PROBLEM 19.15

A 5-kg collar rests on but is not attached to the spring shown. It is observed that when the collar is pushed down 180 mm or more and released, it loses contact with the spring. Determine (a) the spring constant, (b) the position, velocity, and acceleration of the collar 0.16 s after it has been pushed down 180 mm and released.

SOLUTION

(a) x x t

x x

x x

x

m n

m

m

m

= += + == = +

=

=

sin( )

sin( ) .

cos( )

w ff

f

f p

0

0

0 0 180

0 0

2

m

&

00 180

0 1802

.

. sin

m

x tn= +ÊËÁ

ˆ¯

w p

When the collar just leaves the spring, its acceleration is gØ and v = 0.

&x t

v t

n n

n n

n

= +ÊËÁ

ˆ¯

= = +ÊËÁ

ˆ¯

( . ) cos

( . ) cos

0 1802

0 0 0 1802

w w p

w w p

w tt

a g t

g

n n

+ÊËÁ

ˆ¯

=

= - = - +ÊËÁ

ˆ¯

- = -

p p

w w p2 2

0 1802

0 180

2( . )( ) sin

( . )(( ).

..

(

w w

w

w

w

n n

n

n

n

k

m

k m

2

2

9 81

0 1807 382

5

=

=

=

=

=

m/s

m rad/s

kg)(

2

77.382 rad/s)2

= 272 5. N/m k = 273 N/m �

19



(b) wp

n

x t

=

= +ÈÎÍ

˘˚

7 382

0 180 7 3822

.

. sin ( . )

rad/s

At st = 0 16. :

Position.

x = +ÈÎÍ

˘˚

=0 1802

0 06838. . sin (7.382)(0.16) mp

x = 68 4. mm below equilibrium position �

Velocity.

&x = +ÈÎÍ

˘˚

= -

( . )( . )cos ( . )( . )

.

0 180 7 382 7 382 0 162

1 229

p

m/s v = ≠1 229. m/s �

Acceleration.

&&x = - +ÈÎÍ

˘˚

= -

( . )( . ) sin ( . )( . )

.

0 180 7 382 7 382 0 162

3 726

2 p

m/s2 a = ≠3 73. m/s2 �

20


PROBLEM 19.16

An 8-kg collar C can slide without friction on a horizontal rod between two identical springs A and B to which it is not attached. Each spring has a constant of 600 N/m. The collar is pushed to the left against spring A, compressing that spring 20 mm, and released in the position shown. It then slides along the rod to the right and hits spring B. After compressing that spring 20 mm, the collar slides to the left and hits spring A, which it compresses 20 mm. The cycle is then repeated. Determine (a) the period of the motion of the collar, (b) the position of the collar 1.5 s after it was pushed against spring A and released. (Note: This is a periodic motion, but not a simple harmonic motion.)

SOLUTION

(a) For either spring, t p

t p

t

n km

n

n

C

=

=

=

2

2

0 7255

6008

N/mkg

s.

Complete cycle is 1234, 4321.

Time from 1 to 2 is ( / ),t n 4 which is the same as time from 3 to 4, 4 to 3 and 2 to 1.

Thus, the time during which the springs are compressed is 4 4 0 7255( / ) .t tn n= = s.

Velocity at 2 or 3.

Use conservation of energy. v T V kx

V

T mv

1 1 12

1

2 22

0 01

2

1

2600

0 120

1

2

1

= = = =

=

= =

(

.

N/m)(0.020 m)

J

2

228 4

0

0 0 120 4 0 1732

22

2 22

2

1 1 2 2 22

2

( )

. .

kg)(

m

v T v

V

T V T V v v

=

=

+ = + + = = //s

Time from 2 to 3 is t2 30 020

0 17320 11545- = =( .

( ..

m)

m/s) s

and is the same as the time from 3 to 2.

Thus, total time for a complete cycle is t tc n t= += +

-2

0 7255 2 0 115452 3

. ( . )

= 0 9564. t c = 0 956. s �

21



(b) From (a), in 0.9564, the spring A is again fully compressed. Spring B is compressed the second time in 1.5 cycles or ( . )( . ) .1 5 0 9564 1 4346= s. At 1.5 s, the collar is still in contact with spring B moving to the left and is at a distance Dx from the maximum deflection of B equal to

D

D

x

x

= - -ÈÎÍ

˘˚

= -=

20 202

0 72551 5 1 4346

20 16 877

3 123

cos.

( . . )

.

.

p

mm

Thus, collar C is 60 3 123 56 877- =. . mm from its initial position.

56 9. mm from initial position �

22


PROBLEM 19.17

A 35-kg block is supported by the spring arrangement shown. The block is moved vertically downward from its equilibrium position and released. Knowing that the amplitude of the resulting motion is 45 mm, determine (a) the period and frequency of the motion, (b) the maximum velocity and maximum acceleration of the block.

SOLUTION

(a) Determine the constant k of a single spring equivalent to the three springs

P k

k

== + +=

d d d d16 8 8

32 kN/m

Natural frequency.

w

w

n

n

k

m=

= ¥ = ◊

=

32 10

351 1

30 237

3 N/m

kg N kg 1 m/s

rad/s

2( )

.

t pw

pn

n

= = =2 2

30 230 208

.. s �

fnn

= =14 81

t. Hz �

(b) x x t x x

x

m n m

n

= + = ===

sin( ) .

.

.

w fw

0 0 045

30 24

0 045

m

rad/s

sin(30.24tt + f)

&x t= +( . )( . )cos( . )0 045 30 24 30 24 f vmax .= 1 361 m/s �

&&x t= - +( . )( . ) sin( . )0 045 30 237 30 242 f amax .= 41 1 m/s2 �

23


PROBLEM 19.18

A 35-kg block is supported by the spring arrangement shown. The block is moved vertically downward from its equilibrium position and released. Knowing that the amplitude of the resulting motion is 45 mm, determine (a) the period and frequency of the motion, (b) the maximum velocity and maximum acceleration of the block.

SOLUTION

(a) Determine the constant k of a single spring equivalent to the two springs shown.

d d d= + = + =

= + =

1 2 16 161 1

16

1

168

P P P

k

kk

kN/m kN/m

kN/m

Period of the motion.

t p pn k

m

= = =¥

2 20 416

8 1035

3. s �

fnn

= = =1 1

0 4162 41

t .. Hz �

(b) w p pn nf= = =2 2 2 41 15 12( . ) . rad/s

x t

x t

= += +

0 045

0 045 15 12 15 12

. )

( . )( . )cos( . )

sin(15.12 ff& vmax .= 0 680 m/s �

&&x t= - +( . )( . ) sin( . )0 045 15 12 15 122 f amax .= 10 29 m/s2 �

24


PROBLEM 19.19

A 15 kg block is supported by the spring arrangement shown. If the block is moved from its equilibrium position 40 mm vertically downward and released, determine (a) the period and frequency of the resulting motion, (b) the maximum velocity and acceleration of the block.

SOLUTION

Determine the constant k of a single spring equivalent to the three springs shown.

Springs 1 and 2: d d d= +¢

= +1 21 1

1

1

2

, andP

k

P

k

P

k

Hence, ¢ =+

kk k

k k1 2

1 2

where ¢k is the spring constant of a single spring equivalent of springs 1 and 2.

Springs ¢k and 3 Deflection in each spring is the same.

So P P P P k P k P k= + = = ¢ =1 2 1 2 3, , ,and d d d

Now k k k

k k kk k

k kk

k

d d d= ¢ +

= ¢ + =+

+

=+

+ =

3

31 2

1 23

4 2 4

4 2 43 2 4 7

( )( . )

.. . kN/m ==

=

= = = =

4700

15

4700

15313 33 17 70122

N/m

kg

rad/s

m

k

mn nw w. .

25



(a) Period. t pwn

n

= 2 t n = 0 355. s �

Frequency. fnn=

wp2

fn = 2 82. Hz �

(b) Simple harmonic motion.

x t= 0 04. sin( ) in mw j+ xm = 40 mm �

&x t= +( . )( . )cos( )0 04 17 7012 w f

v xm m= =& 0 70805. m/s vm = 0 708. m/s �

&&x t= - +( . )( . ) sin( )0 04 17 7012 2 w f

a xm m= =&& 12 533. m/s2 am = 12 53. m/s2 �

26


PROBLEM 19.20

A 5-kg block, attached to the lower end of a spring whose upper end is fixed, vibrates with a period of 6.8 s. Knowing that the constant k of a spring is inversely proportional to its length, determine the period of a 3-kg block which is attached to the center of the same spring if the upper and lower ends of the spring are fixed.

SOLUTION

Equivalent spring constant.

¢ = + =k k k k2 2 4 (Deflection of each spring is the same.)

For case 1 , t

w pt

p

w

w

n

nn

n

n

k

m

k m

1

11

2

1

1 12

6 8

2 2

6 80 924

5 0

=

= = =

=

= =

.

..

( )( .

s

rad/s

9924 4 26892) .= N/m

For case 2 , wnk

m22

2

4 4 4 2689

35 6918= = =( )( . ). (rad/s)2

w

t pw

pn

nn

2

2

2 3857

2 2

2 3857

=

= =

.

.

rad/s

t n2 2 63= . s �

27


PROBLEM 19.21

A 15 kg block is supported by the spring arrangement shown. The block is moved from its equilibrium position 20 mm vertically downward and released. Knowing that the period of the resulting motion is 1.5 s, determine (a) the constant k, (b) the maximum velocity and maximum acceleration of the block.

SOLUTION

Since the force in each spring is the same, the constant ¢k of a single equivalent spring is

1 1

2

1 1

2 5¢= + + ¢ =

k k k kk

k (See Problem 19.19.)

.

(a) t p pn k

m

k12

1 52 2

1 52 5= = = Ê

ËÁˆ¯¢

. ;.

( )( . ) s m

k =

=

( )

( . )( )( . )

.

2

1 515 2 5

657 974

2

2

p s

kg

N/m k = 658 N/m �

(b) x x t

x x t

v x

m n

m n n

m n

= += +=

sin( )

cos( )

max

w fw w fw

&

w pt

pn

n

mx

v

= = =

= ==

2 2

1 54 189

20 0 02

0 02 4 1

..

.

( . )( .max

srad/s

mm m

m 889 rad/s) vmax .= 0 0838 m/s �

&&x x tm n n= - +w w f2 cos( )

| | ( . )( . )maxa xm n= =w2 20 02 4 189 m | | .maxa = 0 351 m/s2 �

28


PROBLEM 19.22

Two springs of constants k1 and k2 are connected in series to a block A that vibrates in simple harmonic motion with a period of 5 s. When the same two springs are connected in parallel to the same block, the block vibrates with a period of 2 s. Determine the ratio k k1 2/ of the two spring constants.

SOLUTION

Equivalent springs.

Series: kk k

k ks =+1 2

1 2

Parallel: k k kp = +1 2

t pw

p t pw

p

tt

ss

km

pp

k

m

s

p

p

s

s p

k

k

k k

= = = =

Ê

ËÁ

ˆ

¯˜ = Ê

ËÁˆ¯

= =+

2 2 2 2

5

2

2 21

;

22 1 22

1 2

1 2 12

1 2 22

1

1 2

1 2

6 25 2

4

( )

( )

( . )( )

( .

k kk k

k k

k k

k k k k k k

k

+

=+

= + +

=225 4 25 4

22

222

22) ( . )k k km -

k

k1

2

2 125 3 516 0 250 4 00= =. . . .m or k

k1

2

0 250 4 00= . .or �

29


PROBLEM 19.23

The period of vibration of the system shown is observed to be 0.6 s. After cylinder B has been removed, the period is observed to be 0.5 s. Determine (a) the weight of cylinder A, (b) the constant of the spring.

SOLUTION

m m

k

mk

A1 1

11

11

12

1

1 5 0 6

2 2 2

0 63 333

= + =

= = = =

= =

. .

..

t

t pw

w pt

p p

w

s

rad/s

mm mA1 12 21 5 3 333w p= + . ( . ) (1)

m m

k

mk m m

A2 2

22

2

22

22 2

2

0 5

2 2 2

0 54

= =

= = = =

= = =

t

t pw

w pt

p p

w w

.

.

s

rad/s2

AA( )4 2p (2)

(a) Equating the expressions found for k in Eqs. (1) and (2):

m mA A+ =1 5 3 333 42 2. ( . ) ( )p p

( . )( . )

. .

.

(

11 111 1 5 16

4 889 16 6665

3 40898

3

m m

m

m

W m g

A A

A

A

A A

+ ==== =

kg

.. )( . )40898 9 81 WA = 33 4. N �

(b) Eq. (1): k = +3 40898 1 5 3 333 2. . ( . )p rad/s

k = 538 22. N/m k = 538 N/m �

30


PROBLEM 19.24

The period of vibration of the system shown is observed to be 0.8 s. If block A is removed, the period is observed to be 0.7 s. Determine (a) the mass of block C, (b) the period of vibration when both blocks A and B have been removed.

SOLUTION

m m

W

k

mk m

C1 1

11

12

11

6 0 8

2 2

0 8

2

0 8

= + =

= = =

= =

kg s

s rad/s

tp

tp p

w w

.

. .

; 112

2

62

0 8= + Ê

ËÁˆ¯

( ).

mCp

(1)

m m

k

mk m

C2

2

22

22 2

2

3 0 7

2 2

0 7

2

0 7

= + =

= = =

= =

kg s

srad/s

2

2

t

w pt

p p

w w

.

. .

; == + ÊËÁ

ˆ¯

( ).

mC 32

0 7

2p (2)

Equating the expressions found for k in Eqs. (1) and (2):

( ).

( ).

.

.

m m

m

m

C C

C

C

+ ÊËÁ

ˆ¯

= + ÊËÁ

ˆ¯

++

= ÊËÁ

ˆ¯

62

0 83

2

0 7

6

3

0 8

0 7

2 2p p

22

; :solvefor mC mC = 6 80. kg �

w pt

w w pt

33

32

32

3

2

2

2

=

= = =ÊËÁ

ˆ¯

k

mk m m

CC C; (3)

Equating expressions for k from Eqs. (2) and (3),

( ).

m mC C+ ÊËÁ

ˆ¯

=ÊËÁ

ˆ¯

32

0 7

22

3

2p p

t

Recall mC = 6 8. : kg ( . ).

.6 8 32

0 76 8

22

3

2

+ ÊËÁ

ˆ¯

=ÊËÁ

ˆ¯

p pt

t t3

23

0 7

6 8

9 8 0 70 833

.

.

.;

..

ÊËÁ

ˆ¯

= = t3 0 583= . s �

31


PROBLEM 19.25

The period of vibration of the system shown is observed to be 0.2 s. After the spring of constant k2 4= kN/m is removed and block A is connected to the spring of constant k1, the period is observed to be 0.12 s. Determine (a) the constant k1 of the remaining spring, (b) the weight of block A.

SOLUTION

Equivalent spring constant for springs in series.

kk k

k ke =+1 2

1 2( )

For and k k1 2 , t p p= =+

2 2

1 2

1 2

km

k km k k

e

A A

( )( )( )

For alone,k1 ¢ =t p2

1kmA

(a) tt

tt

tt

¢=

+=

+

¢ÊËÁ

ˆ¯

= +

¢=

( )( )

( )

.

.

k k k

k k

k k

k

k k k

1 2 1

1 2

1 2

2

2

2

1 2

0 2

0 12==

=

1 6667

42

.

k kN/m

( )( . )4 1 6667 421 kN/m s kN/= +k m k1 7 11= . kN/m �

(b) kkmA

1 7 11112

1

= ¢ =. kN / m t p

mk

W m g

k

W

A A A

A

=¢

=

= =

=

( )

( );

. .

( .

tp

21

2

1

2

7 1111 7111 1

9 81

kN/ N/m

m/

m

ss s N/m2 2

2

25 4453 0 12 7111 1

2

) . ( . ) ( . )

( )

= N

p WA = 25 4. N �

32


PROBLEM 19.26

The 50-kg platform A is attached to springs B and D, each of which has a constant k = 2 kN/m. Knowing that the frequency of vibration of the platform is to remain unchanged when a 40-kg block is placed on it and a third spring C is added between springs B and D, determine the required constant of spring C.

SOLUTION

Frequency of the original system.

Springs B and D are in parallel. k k k

k

m

e B D

ne

A

n

= + = = =

= =

2 2 4 4000

4000

2

( ) kN/m kN/m

N/m

50 kg2

N / m

w

w == 80 2( )rad/s

Frequency of new system.

Springs A, B, and C are in parallel. ¢ = + + = +k k k k ke B D C c( )( )2 200 in N/m

( )( )

( )

( )( )

¢ =¢

+=

++

¢ =+

w

w

w

ne

A B

c

nc

n

k

m m

k

k

2

2

2000

50 40

2000

90

kg kg

22 2

802000

905200 5 2

= ¢

=+

= =

( )

( )

.

wn

c

C

k

k N/m kN / m kC = 5 20. kN/m �

33


PROBLEM 19.27

From mechanics of materials it is known that when a static load P is applied at the end B of a uniform metal rod fixed at end A, the length of the rod will increase by an amount d = PL AE/ , where L is the length of the undeformed rod, A is its cross-sectional area, and E is the modulus of elasticity of the metal. Knowing that L = 450 mm and E = 200 GPa and that the diameter of the rod is 8 mm, and neglecting the mass of the rod, determine (a) the equivalent spring constant of the rod, (b) the frequency of the vertical vibrations of a block of mass m = 8 kg attached to end B of the same rod.

SOLUTION

(a) P k

PL

AE

PAE

L

kAE

L

Ad

A

e

e

=

=

= ÊËÁ

ˆ¯

=

= =¥

= ¥

-

d

d

d

p p2 3 2

4

8 10

4

5 027 10

( )

.

m

--

-

=

= ¥

=¥ ¥

5 2

9 2

5 2 9 2

0 450

200 10

5 027 10 200 10

0

m

m

N/m

m N m

L

E

ke

.

( . )( / )

( .. )450 m

ke = ¥22 34 106. N/m ke = 22 3. MN/m �

(b) fn

km

e

=

=¥

2p

p

22 3 108

6

2

.

= 265 96. Hz fn = 266 Hz �

34


PROBLEM 19.28

From mechanics of materials it is known that for a cantilever beam of constant cross section, a static load P applied at end B will cause a deflection d B PL EI= 3 3/ , where L is the length of the beam, E is the modulus of elasticity, and I is the moment of inertia of the cross-sectional area of the beam. Knowing that L = 3 m, E = 230GPa, and I = 5 ¥ 106 mm4, determine (a) the equivalent spring constant of the beam, (b) the frequency of vibration of a 250 kg block attached to end B of the same beam.

SOLUTION

(a) Equivalent spring constant. kP

P k

PL

EI

PEI

L

eB

e B

B

B

=

=

=

= ÊËÁ

ˆ¯

dd

d

d

3

3

33

kEI

L

k

e

e

=

=¥ ¥ ¥

=

3

3 230 10 5 10 10

3

127 778 10

3

9 6 12

3

3

( )( )( )

( )

.

N/m m2 4-

¥ NN/m ke = 127 8. kN/m �

(b) Natural frequency.

f

k

n

km

e

e

=

=2

N/

p127 778 103. ¥ m

f

f

n

n

=

=

¥( . )

.

127 778 103

3 5981

N/m250 kg

2 Hz

p fn = 3 60. Hz �

35


PROBLEM 19.29

A 40-mm deflection of the second floor of a building is measured directly under a newly installed 4000-kg piece of rotating machinery, which has a slightly unbalanced rotor. Assuming that the deflection of the floor is proportional to the load it supports, determine (a) the equivalent spring constant of the floor system, (b) the speed in rpm of the rotating machinery that should be avoided if it is not to coincide with the natural frequency of the floor-machinery system.

SOLUTION

(a) Equivalent spring constant.

W k

kW

e s

e

=

=

= =

d

d( )( . )

( .

4000 9 81

0 04981 103

m)N / m¥ ke = 981 kN/m �

(b) Natural frequency. f

f

n

km

n

e

=

=

===

¥( )

2

Hz

Hz cycle/s

r

N/m)

p

p

(

.

981 104000

3

22 4924

1 1

60

kg

ppm

Speed Hzrpm

Hzrpm

=

=

( . )( )

.

2 492460

149 5 �

36


PROBLEM 19.30

The force-deflection equation for a nonlinear spring fixed at one end is F x= 5 1 2/ where F is the force, expressed in newtons, applied at the other end and x is the deflection expressed in meters. (a) Determine the deflection x0 if a 120-g block is suspended from the spring and is at rest. (b) Assuming that the slope of the force-deflection curve at the point corresponding to this loading can be used as an equivalent spring constant, determine the frequency of vibration of the block if it is given a very small downward displacement from its equilibrium position and released.

SOLUTION

(a) Deflection x0 . mg

mg

F mg

x

===

=

( . )( . )

.

/

0 120 9 81

1 177

5

2

01 2

kg m/s

N

x0

21 177

5

0 0554

= ÊËÁ

ˆ¯

=

.

. m x0 55 4= . mm �

Equivalent spring constant.

At x0 , dF

dxx

dF

dx

x

x

ÊËÁ

ˆ¯

= =

ÊËÁ

ˆ¯

=

- -

0

0

5

2

5

20 0554

10 618

01 2 1 2( ) ( . )

.

/ /

NN/m

N/mke = 10 618.

(b) Natural frequency. fn

km

e

=

=

2

2

10 6180 120

p

p

( . )( .

N/m kg)

fn = 1 4971. Hz fn = 1 497. Hz �

37


PROBLEM 19.31

If h = 700 mm and d = 500 mm and each spring has a constant k = 600 N/m, determine the mass m for which the period of small oscillations is (a) 0.50 s,(b) infinite. Neglect the mass of the rod and assume that each spring can act in either tension or compression.

SOLUTION

x xd

h

F kx kd

hx

s

s

=

= =2 2 2

+ S SM M Fd mgx mx hA A= - = -( ) : ( )eff 2 &&

2

20

2

2

2

22

2

kd

hx d mgx mxh

xkd

mh

g

hx

kd

mhn

ÊËÁ

ˆ¯

- = -

+ -È

ÎÍ

˘

˚˙ =

= -

&&

&&

w gg

h

k

m

d

h

g

hn

È

ÎÍ

˘

˚˙

= ÊËÁ

ˆ¯

-w222

(1)

Data: d

h

k

===

0 5

0 7

600

.

.

m

m

N/m

38



(a) For st = 0 5. : tn n

n= = =20 5

24

pw

pw

w p; .

Eq. (1): ( )( ) .

.

.

.4

2 600 0 5

0 7

9 81

0 72

2

p = ÊËÁ

ˆ¯

-m

m = 3 561. kg m = 3 56. kg �

(b) For infinite:t = tn

n= =20

pw

w

Eq. (1): 02 600 0 5

0 7

9 81

0 7

2

= ÊËÁ

ˆ¯

-( ) .

.

.

.m

m = 43 69. kg m = 43 7. kg �

39


PROBLEM 19.32

Denoting by dST the static deflection of a beam under a given load, show that the frequency of vibration of the load is

fg= 1

2p dST

Neglect the mass of the beam, and assume that the load remains in contact with the beam.

SOLUTION

kW

mW

g

k

m

gn

W

Wg

=

=

= = =

d

wd

d

ST

ST

ST2 Fg

nn= =

wp p d2

1

2 ST

�

40


PROBLEM 19.33*

Expanding the integrand in Equation (19.19) of Section 19.4 into a series of even powers of sinf and integrating, show that the period of a simple pendulum of length l may be approximated by the formula

t pq

= +ÊËÁ

ˆ¯

2 12

1

42l

gmsin

where qm is the amplitude of the oscillations.

SOLUTION

Using the Binomial Theorem, we write

1

11

2

11

2

22

2

21 2

2

- ( )= - Ê

ËÁˆ¯

ÈÎÍ

˘˚˙

= +

-

sin sinsin sin

sin

q f

qf

q

m

m

m

/

222sin f + ◊ ◊ ◊ ◊

Neglecting terms of order higher than 2 and setting sin ( cos ),2 12 1 2f f= - we have

tq

f fp

nml

gd

l

g

= + -( )ÈÎÍ

˘˚

ÏÌÓ

¸˝˛

= +

Ú4 11

2 2

1

21 2

4 11

4

2

0

2

2

sin cos

sinqq q

f fp m m d

2

1

4 222

0

2-Ï

ÌÓ

¸˝˛Ú sin cos

= + ÊËÁ

ˆ¯

-ÈÎÍ

˘˚˙

= +

41

4 2

1

8 22

42

1

4

2 2

0

2l

g

l

g

m mfq

fq

f

p

p

sin sin sin

si

/

nn2

2 20

q pmÊËÁ

ˆ¯

+ÈÎÍ

˘˚˙ t p

qn

ml

g= +Ê

ËÁˆ¯

2 11

4 22sin �

41


PROBLEM 19.34*

Using the formula given in Problem 19.33, determine the amplitude qm for which the period of a simple pendulum is 1

2 percent longer than the period of the same pendulum for small oscillations.

SOLUTION

For small oscillations, ( )t pnl

g0 2=

We want t t

p

n n

l

g

=

=

1 005

1 005 2

0. ( )

.

Using the formula of Problem 19.33, we write

t tq

tq

n nm

n

m

= +ÊËÁ

ˆ¯

=

= - =

( ) sin

. ( )

sin [ . ] .

02

0

2

11

4 2

1 005

24 1 005 1 0 022

20 02

28 130

sin .

.

q

q

m

m

=

= ∞ qm = ∞16 26. �

42


PROBLEM 19.35*

Using the data of Table 19.1, determine the period of a simple pendulum of length l = 750 mm (a) for small oscillations, (b) for oscillations of amplitude qm = ∞60 , (c) for oscillations of amplitude qm = ∞90 .

SOLUTION

(a) t pnl

g= 2 (Equation 19.18 for small oscillations):

t pn =

=

20 750

9 81

1 737

.

.

.

m

m/s

s

2

t n = 1 737. s �

(b) For large oscillations (Eq. 19.20),

tp

p

p

nk l

g

k

= ÊËÁ

ˆ¯

Ê

ËÁˆ

¯

=

22

21 737( . )s

For qm = ∞60 , k = 1 686. (Table 19.1)

tpn ( )

( . )( . )

.

602 1 686 1 737

1 864

∞ =

=

s

s t n ( ) .60 1 864= s �

(c) For qm = ∞90 , k = 1 854. tpn = =

2 1 854 1 7372 05

( . )( . ).

ss �

43


PROBLEM 19.36*

Using the data of Table 19.1, determine the length in mm of a simple pendulum which oscillates with a period of 2 s and an amplitude of 90°.

SOLUTION

For large oscillations (Eq. 19.20),

tp

pnk l

g= Ê

ËÁˆ¯

Ê

ËÁˆ

¯2

2

for qm = ∞90

k = 1 854. (Table 19.1)

( ) ( )( . )( ).

( ) ( . )

[( )( . )]

2 2 1 854 29 81

2 9 81

4 1 854

2

2

sm/s

s m/s

2

2

=

=

=

l

l

00 71349. m l = 713 mm �

44


PROBLEM 19.37

The 5-kg uniform rod AC is attached to springs of constantk = 500 N/m at B and k = 620 N/m at C, which can act in tension or compression. If the end C of the rod is depressed slightly and released, determine (a) the frequency of vibration, (b) the amplitude of the motion of Point C, knowing that the maximum velocity of that point is 0.9 m/s.

SOLUTION

F k x

k

F k x

k

B B B B

B B

C C C C

C

= += += += +

( ( ) )

( . ( ) )

( ( ) )

( . (

dq d

dq

ST

ST

ST

0 7

1 4 ddST ) )C

Equation of motion. + S SM MA A= ( ) :eff

( . )[ [( . ( ) ) ] . [ ( . ( ) )] ( . )(0 7 0 7 1 4 1 4 0 7k mg k IB B C Cq d q d a+ - + + = - -ST ST mmat ) (1)

But in equilibrium ( ),q = 0 + SM k mg kA B B C C= = - +0 0 7 1 4. [ ( ) ] . ( )d dST ST (2)

Substituting Equation (2) into Equation (1),

I ma k kt B Ca q q+ + + =0 7 0 7 1 4 02 2. ( . ) ( . )

Kinematics ( ):a q= && at = =0 7 0 7. .a q&&

[ ( . ) ] [( . ) ( . ) ]

( )( .

I m k k

I ml

B C+ + + =

=

=

0 7 0 7 1 4 0

1

121

125 1 4

2 2 2

2

&&q q

kg mm

kg m

)

.

2

20 8167= ◊

45



(. ) ( . )( )

.

( . ) ( . ) ( . )(

7 0 49 5

2 45

0 7 1 4 0 49 5

2 2

2

2 2

m

k kB C

=

= ◊

+ =

m kg

kg m

m2 000 1 96 620

245 1215 2

1460 2

0 8167 2 45

N/m m N/m

N m

2) ( . )( )

.

.

[ . . ]

+= += ◊

+ &&&

&&

&&

q q

q q

q q

+ =

+◊

◊=

+ =

1460 2 0

1460 2

3 2670

447 0

.

( . )

( . )

(

N m

kg m

N/k

2

gg m s 2◊ = - )

(a) Natural frequency. wn ==

-447

21 14

s

rad/s

2

.

fnn= = =

wp p2

21 14

23 36

.. Hz �

(b) Maximum velocity. q q w f

q q w w f

= +

= +m n

m n n

t

t

sin( )

( )( )cos( )&

Maximum angular velocity. &q q wm m n=

Maximum velocity at C. ( ) .

( . ) ( )( )

( . )

( . )( . )

& &xC m m

m n

m

==

=

=

1 4

1 4

0 9

1 4 21 14

qq w

q

m

m/s

m rad/s

00 03041. rad

Maximum amplitude at C. ( ) ( . )( )

( . )( . )

xC m m==

1 4

1 4 0 03041

m

m

q

( ) .xC m = 0 0426 m ( ) .xC m = 42 6 mm �

46


PROBLEM 19.38

The uniform rod shown weighs 7.5 kg and is attached to a spring of constant k = 800 N/m If end B of the rod is depressed 10 mm and released, determine (a) the period of vibration, (b) the maximum velocity of end B.

SOLUTION

k

m

==

800

7 5

N/m

kg.

where F k x

k

ma mr

I

= += +

= = =

=

( )

( . )

( . )( . )

dq d

a q q

a

ST

ST

m

0 5

7 5 0 12515

161

1

&& &&

227 5 0 75

45

128

2( . )( . ) &&

&&

q

q=

(a) Equation of motion.+ S SM M mg F I maC C= - = +( ) : ( . ) ( . ) ( . )eff m m m0 125 0 5 0 125a

mg k( . ) ( . )( . ) ( . )0 125 0 5 0 545

128

15

160 25- + = +q d q qST m && &&

But in equilibrium, we have + mg k( . ) ( . )0 125 0 5 0- =dST

47



Thus, - = +ÈÎÍ

˘˚

- =

+

k( . )

( )( . )

(

0 545

128

15

128

800 0 515

32

42

2

2

q q

q q

q

&&

&&

&& 66 667 0. )q =

Natural frequency and period.

ww

t pw

p

n

n

n

2 426 667

20 656

2 2

20 656

==

= =

.

.

.

rad/s

rad/s t = 0 304. s �

(b) At end B. xm = 10 mm = 0.01 m

v xm m n===

w( . )( . )

.

0 01 20 656

0 20656

m rad/s

m/s vm = 0 207. m/s �

48


SOLUTION

Spring deflection. x x x

x r

x

rx x

F k x k x

A A

A

A

S A

= +

=

=

== + = +

0 0

0

0

0

0

2

2

/

/

( ) ( )

q

q

d dST ST

S SM M rk x rw rmx IC = - + + ∞ = +( ) : ( ) sineff ST2 2 150 0d q&& && (1)

But in equilibrium, x0 0=

SM rk rwC = = - + ∞0 2 15dST sin (2)

Substituting Eq. (2) into Eq. (1) and noting that q q= =x

r

x

r0 0, && &&

rmx Ix

rrkx

I mr

mrx rkx

xk

m

&& &&

&&

&&

00

0

2

0 0

0

4 0

1

23

24 0

8

3

+ + =

=

+ =

+ ÊËÁ

ˆ¯

=x0 0

+

PROBLEM 19.39

A 15 kg uniform cylinder can roll without sliding on a 15° incline. A belt is attached to the rim of the cylinder, and a spring holds the cylinder at rest in the position shown. If the center of the cylinder is moved 50 mm down the incline and released, determine (a) the period of vibration, (b) the maximum acceleration of the center of the cylinder.

49



Natural frequency. wnk

m= = =8

3

8 6000

3 1532 6599 1( )( )

( )( ).

N / m

kgs-

(a) Period. t pw

pn

n

= =2 2

32 6599. t n = 0 1924. s �

(b) x x tm n0 0= +( ) sin( )w f

At t = 0, x x

x x t

t

x

m n n

m n

0 0

0 0

0

0 05 0

0

0

= == +==

.

( ) cos( )

( ) cos

m && w w f

w f

Thus, f p

f

=

== = =

20

0 0 05 10 0 0

t

x x xm m( ) . ( ) sin ( ) ( ) m

( ) .

( ) sin( )

( ) ( )

( )

max max

x

x x t

a x

x

m

m n n

0

0 02

0 0

0

0 05=

= - +=

= -

m

&&&&

w w f

mm nw2

1 20 05 32 6599

53 333

= -

=

-( . )( . )

.

m s

m/s2 ( ) .maxa0 53 3= m/s2 �

50


PROBLEM 19.40

A 7.5 kg slender rod AB is riveted to a 6 kg uniform disk as shown. A belt is attached to the rim of the disk and to a spring which holds the rod at rest in the position shown. If end A of the rod is moved 20 mm down and released, determine (a) the period of vibration, (b) the maximum velocity of end A.

SOLUTION

Equation of motion. + S SM M mgL

kxr I mL L

IB B AB= - = + ÊËÁ

ˆ¯

ÊËÁ

ˆ¯

+( ) : coseff Disk2 2 2q a a a (1)

where x r= +q dST and from statics, mgL

k r2

= dST

Assuming small angles (cos ),q ª 1 Equation (1) becomes

mgL

kr kr I mL

I

ImL

I

AB

AB

2 2

4

22

2

q q d a- - = + ÊËÁ

ˆ¯

+Ê

ËÁ

ˆ

¯˜

+ +

ST Disk

Disk

ÊÊ

ËÁˆ

¯+ =&&q qkr2 0

Data: m = 7 5. kg

m

L

r

k

disk

mm m

mm m

kN/m N/m

== == == =

6

900 0 9

250 0 25

6 6000

kg

.

.

51



I mL

I m r

AB =

=

= ◊

=

=

1

121

127 5 0 9

0 50625

1

21

26

2

2

2

( . )( . )

.

(

kg m2

Disk Disk

))( . )

.

0 25

0 1875

2

= ◊kg m2

0 506251

47 5 0 9 0 1875 6000 0 25 02 2. ( . )( . ) . ( )( . )+ +È

ÎÍ˘˚

+ =&&q q

2 2125 375 0. &&q q+ = or &&q q+ =169 492 0.

(a) Natural frequency and period. wn2 2169 492= . ( )rad/s

w

t pw

pn

n

=

= =

13 0189

2 2

13 0189

.

.

rad/s

t = 0 483. s �

(b) Maximum velocity. v xm n m= =w ( . )( . )13 0189 0 02rad/s m vm = 260 m/s �

52


PROBLEM 19.41

An 8-kg uniform rod AB is hinged to a fixed support at A and is attached by means of pins B and C to a 12-kg disk of radius 400 mm. A spring attached at D holds the rod at rest in the position shown. If Point B is moved down 25 mm and released, determine (a) the period of vibration, (b) the maximum velocity of Point B.

SOLUTION

(a)

Equation of motion. S SM M F kA A S= = +( ) : ( . )eff ST0 6q d

+ 0 6 1 2 0 6 1 2. ( ) . ( ) . ( )( ) . ( )( )m g F m g I I m a m aR S D R D R t D D t B- + = + + +a (1)

At equilibrium ( ),q = 0 F kS = dST

SM m g k m gA R D= = - +0 0 6 1 2. ( ( )) .dST (2)

Substituting Eq. (2) into Eq. (1), ( ) . ( ) . ( ) ( . )I I m a m a kR D R t D D t B+ + + + =a q0 6 1 2 0 6 02

a q

q

q

=

=

=

= =

=

&&&&&&

( ) .

( ) .

( )( . )

.

a

a

I m l

t B

t D

R R

0 6

1 2

1

12

1

128 1 2

0 960

2 2

kg m◊

53



I m RD D= = = ◊

+ + +

1

2

1

212 0 4 0 960

0 960 0 960 0 6 8 1

2 2

2

( )( . ) .

[ . . ( . ) ( ) (

kg m

.. ) ( )] ( . ) ( )

( . )

2 12 0 6 800 0

288

22 080

2 2&&&&

q q

q q

+ =

+ ◊◊

= N m

kg m2

(a) Natural frequency and period.

w

t pw

p

n

nn

=

=

= =

288

22 083 6116

2 2

3 6116

..

.

rad/s

t n = 1 740. s �

(b) Maximum velocity at B.

( ) ( . )( )

..

..

sin

max maxv

yB

mB

m

m

=

=

= =

=

1 2

1 20 025

1 20 02083

&q

q

q

q q

rad

(( )

cos( )

( . )( . ) .max

w f

q q w w f

q q w

n

m n n

m n

t

t

+

= +

= = =

&& 0 02083 3 612 0 075224 rad/s

( ) ( . )( ) ( . )( . )max maxvB = =1 2 1 2 0 07524&q m rad/s ( ) .maxvB = 0 09029 m/s ( ) .maxvB = 90 3 mm/s �

54


PROBLEM 19.42

Solve Problem 19.41, assuming that pin C is removed and that the disk can rotate freely about pin B.

SOLUTION

(a)

Note: This problem is the same as Problem 19.41, except that the disk does not rotate, so that the effective moment IDa = 0.

Equation of motion. S SM M F kA A S= = +( ) : ( . )eff ST0 60 d

+ ( . )( ) . ( . )( )( ) . ( )( )0 6 1 2 0 6 1 2m g F m g I m a m aR S D R R t D D t B- + = + +a (1)

At equilibrium ( ),q = 0 F kS = dST

+ SM m g m gA R D= = - +0 0 6 1 2. ( ) .dST (2)

Substituting Eq. (2) into Eq. (1), I m a m a kR R t D D t Ba q+ + + =0 6 1 2 0 6 02. ( ) . ( ) ( . )

a q

q

q

=

=

=

= =

=

&&&&&&

( ) .

( ) .

( )( . )

.

a

a

I m l

t B

t D

R R

0 6

1 2

1

12

1

128 1 2

0 960

2 2

kg m◊

55



[ . ( . ) ( ) ( . ) ( )] ( . ) ( )

(

0 960 0 6 8 1 2 12 0 6 800 0

288

2 2 2+ + + =

+ ◊

&&&&

q q

q N m))

.21 120

2 kg m◊=q


w

t pw

p

n

nn

=

=

= =

288

21 123 693

2 2

3 693

..

.

rad/s

t n = 1 701. s �

(b) Maximum velocity at B.

( ) ( . )( )

.

.

..

sin(

max maxvB

mB

m

=

= = =

=

1 2

1 2

0 025

1 200 02083

&q

qm

q q w

rad

nn

m n n

m n

t

t

+

= +

===

f

q q w w f

q q w

)

cos( )

( . )( . )

.

max

&&&&

0 02083 3 693

0 076994

1 2

1 2 0 07694

0 09233

rad/s

m/s

( ) ( . )( )

( . )( . )

.

max maxvB ===

&q

( ) .maxvB = 92 3 mm/s �

56


PROBLEM 19.43

A belt is placed around the rim of a 240-kg flywheel and attached as shown to two springs, each of constant k = 15 kN/m. If end C of the belt is pulled 40 mm down and released, the period of vibration of the flywheel is observed to be 0.5 s. Knowing that the initial tension in the belt is sufficient to prevent slipping, determine (a) the maximum angular velocity of the flywheel, (b) the centroidal radius of gyration of the flywheel.

SOLUTION

Denote the initial tension by T0.

Equation of motion. + S SM M T r T r I

T k r r T kr I

A B0 0

0 0

= - + =

- + + - =

( ) :

( ) ( )

eff&&

&&q

q q q q

&&q q

w

+ =

=

20

2

2

22

kr

I

kr

In (1)

Data: m k

r

nn

= == =

= = = =

240 15

0 5 0 45

2 2 2

0 54

kg kN/m

s m

rad/s

t

t pw

w pt

p p

. .

;.

(a) Maximum angular velocity. If Point C is pulled down 40 mm and released,

q q

q q w

m

m m n

=

= ÊËÁ

ˆ¯

= ¥

=

= ¥

-

-

max

.

( .

40

450

88 89 10

88 92 10

3

3

mm

mm

rad

r

&aad rad/s)( )4p &qm = 1 117. rad/s �

57



(b) Centroidal radius of gyration.

w

p

nkr

I

I

I

22

22

2

2

42 15 0 45

38 47

=

=

= ◊

( )( )( . )

.

rad/skN/m m

kg m

or since I mk

k

=

= ◊

2

2240 28 47( ) .kg kg m2

k = 0 4004. m k = 400 mm �

58


PROBLEM 19.44

A 75-mm-radius hole is cut in a 200-mm-radius uniform disk, which is attached to a frictionless pin at its geometric center O. Determine (a) the period of small oscillations of the disk, (b) the length of a simple pendulum which has the same period.

SOLUTION

Equation of motion.

S SM M0 0= ( ) :eff + - = - -m g I I mH D H H( . )sin ( . )0 1 0 1 2q q q q&& &&

m t R

t

t

m t r

t

D

H

=

==

=

==

r p

r ppr

r p

r p

2

2

2

2

0 2

0 04

0 075

0 00

( )( . )

( . )

( )( . )

( . 55625

1

2

1

20 04 2

800 10

1

21

2 2

6

2

)

( . )(. )

pr

pr

pr

t

I m R t

t

I m r

D D

H H

= =

= ¥

=

=

-

220 005625 0 75

15 82 10

2

6

( . )( . )

.

pr

pr

t

t= ¥ -

59



Small angles. sinq qª

[( . ( . ) ( . )]

( . )

800 10 15 82 10 0 1 0 005625

0 005625

6 6 2¥ - ¥ -+

- -p p p r qt &&pp r qt ( . )( . )9 81 0 1 0=

727 9 10 5 518 10 06 3. .¥ + ¥ =- -&&q q


w

w

t pw

n

n

nn

23

6

5 518 10

727 9 107 581

2 753

2

= ¥¥

==

=

-

-.

..

. rad/s

t pn = =2

2 7532 28

.. s �

(b) Length and period of a simple pendulum.

t p

tp

p

n

n

l

g

l g

l

=

= ÊËÁ

ˆ¯

= ÈÎÍ

˘˚

2

2

2 753

29 81

2

2( . )( . )m/s2 l = 1 294. m �

60


PROBLEM 19.45

Two small weights w are attached at A and B to the rim of a uniform disk of radius r and weight W. Denoting by t0 the period of small oscillations when b = 0, determine the angle b for which the period of small oscillations is 2 0t .

SOLUTION

a q

a q

=

= =

=

&&&&a r r

IW

gr

t

D1

22

Equation of motion.

S SM M wr wrw

gra I

wr

C C t= - - + = +

- -

( ) : sin( ) sin( )

[sin( ) si

eff b q b q a

b q

2

nn( )] sin cos

sin

( cos )

b q q bq q

q b q

+ = -ª

+ÊËÁ

ˆ¯

+

2

2

222 2

wr

w

gr

W

gr wr&& == 0

Natural frequency. w b b

b t pw

p

t

n W Ww

gr

n

wg

w r

g

r

Ww

=+( ) =

+

= = =

=

+

2

2

4

4

02 2

2

2

00

44

cos cos

( )

( )

pp t pbcos

( ) ( )4

0 44

24

+ +

= =Ww

Wwrg

r

(1)

cos b = ÊËÁ

ˆ¯

=1

2

1

4

2

b = ∞75 5. �

61


PROBLEM 19.46

Two 50 g weights are attached at A and B to the rim of a 1.5 kg uniform disk of radius r = 100 mm . Determine the frequency of small oscillations when b = ∞60 .

SOLUTION

a q

a q

=

= =

=

&&&&a r r

IW

gr

t

D1

22

Equation of motion.

S SM M wr wrw

gra I

wr

C C t= - - + = +

- -

( ) : sin( ) sin( )

[sin( ) si

eff b q b q a

b q

2

nn( )] sin cos

sin

( cos )

b q q bq q

q b q

+ = -ª

+ÊËÁ

ˆ¯

+

2

2

222 2

wr

w

gr

W

gr wr&& == 0

Natural frequency. w b bn W W

w

wg

w r

g

r=

+( ) =+

2

2

4

42

cos cos

( ) (1)

Data: W

w

m

mr

n

= =¥

=

= = ∞

= ∞+

-1 5

50 1030

100 60

4 9 81 60

4

3

.

( )( . )cos

(

mm = 0.1 m b

w330 0 1

2 4022

2

2 4022

2

)( . ).

.

=

= =

rad/s

fnwn

p p fn = 0 382. Hz �

62


PROBLEM 19.47

For the uniform square plate of side b = 300 mm,determine (a) the period of small oscillations if the plate is suspended as shown, (b) the distance c from O to a Point A from which the plate should be suspended for the period to be a minimum.

SOLUTION

(a) Equation of motion.

S SM M

I mb

a OG

OG b

a b

t

t

0 0

21

6

2

2

2

2

= =

=

=

=

=Ê

ËÁˆ

¯

( ) :

( )( )

eff a q

a

q

&&

&&

+ ( )(sin )( ) ( ) sinOG mg OG ma Itq a q q= - - ª

b m b mb b mg

b

2

2

2

2

1

6

2

20

1

2

1

6

2Ê

ËÁˆ

¯

Ê

ËÁˆ

¯+ +

Ê

ËÁˆ

¯=

+ÊËÁ

ˆ

&& &&q q q

( )¯

+Ê

ËÁˆ

¯=m mg&&q q2

20

63



&& &&q q q+( )( ) = + =

22

23

03 2

40

g

b

g

bor

Natural frequency and period.

w

w

t pw

n

n

nn

g

b02

0

00

3 2

4

3 2 9 81

4 0 334 68

5 889

2

= = =

=

=

( . )

( )( . ).

. rad/s

t n0 1 067= . s �

(b) Suspended about A. Let e OG c= -( )

a et = a

Equation of motion.

+S SM MA A= ( ) :eff mge ema I me I

m e b mge

tsin ( )q a a

q q

= - - = - +

+ÊËÁ

ˆ¯

+ =

2

2 21

60&&

Frequency and period. w

t pw

p

t p

n

nn

n

eg

e b

e b

eg

ge

b

e

22 1

62

22

2

2 2 16

2

22 2

4 4

4

6

=+

= =+

= +Ê

ËÁˆ

¯

( )

For t n to be minimum, d

dee

b

e+

Ê

ËÁˆ

¯=

2

60

16

0 66

2

2 60 29886

0 29886 300

2

2

2

2- = = =

= - = - =

=

b

e

b

ee

b

c OG e bb

b

c

.

( . )( mm)) c = 89 7. mm �

64


PROBLEM 19.48

A connecting rod is supported by a knife-edge at Point A; the period of its small oscillations is observed to be 0.87 s. The rod is then inverted and supported by a knife-edge at Point B and the period of small oscillations is observed to be 0.78 s. Knowing that r ra b+ = 250 mmdetermine (a) the location of the mass center G, (b) the centroidal radius of gyration k .

SOLUTION

Consider general pendulum of centroidal radius of gyration k .

Equation of motion. + S SM M mgr mr r mk0 02= - = +( ) : sin ( )eff q q q&& &&

&&q q++

ÈÎÍ

˘˚

=gr

r k2 20sin

For small oscillations, sin ,q qª we have

&&q q

w

t pw

p

++

ÈÎÍ

˘˚

=

=+

= = +

gr

r kgr

r k

r k

gr

n

n

2 2

22 2

2 2

0

22

For rod suspended at A, t pAa

a

r k

gr=

+2

2 2

g r r kA a at p2 2 2 24= +( ) (1)

For rod suspended at B, t pBb

b

r k

gr=

+2

2 2

g r r kB b bt p2 2 2 24= +( ) (2)

65



(a) Value of ra .

Subtracting Eq. (2) from Eq. (1), g r g r r r

g r g r r r r r

A a B b a b

A a B b a b a b

t t p

t t p

2 2 2 2 2

2 2 2

4

4

- = -( )- = + -( )( )

Applying the numerical data with r ra b+ = =250 0 25mm m.

( . )( . ) ( . )( . ) ( . )( )

.

9 81 0 87 9 81 0 78 4 0 25

7 4252 5

2 2 2r r r r

ra b a b

a

- = --

p.. . ( )9684 9 8696r r rb a b= -

3 9012 2 4444 0 6266

0 25 0 6266 0 15369

. . .

. . .

r r r r

r r rb a b a

a a a

= == + = m ra = 153 7. mm �

rb = -0 25 0 15369. . rb = 0 09631. m rb = 96 31. mm �

(b) Centroidal radius of gyration.

From Eq. (1), 4 4

9 81 0 87 0 15369 4 0 15369 0

2 2 2 2 2

2 2 2

p p

p

k g r rA a a= -

= - =

t

( . )( . ) ( . ) ( . ) ..

. .

20867

0 072703 72 7

2m

m mmk k= =

66


PROBLEM 19.49

For the uniform equilateral triangular plate of side l = 300 mm, determine the period of small oscillations if the plate is suspended from (a) one of its vertices, (b) the midpoint of one of its sides.

SOLUTION

For an equilateral triangle, h b A bh b= = =3

2

1

2

3

42,

I bh l l l

I hb

l

x

y

= =Ê

ËÁˆ

¯=

= - ÊËÁ

ˆ¯

=Ê

ËÁˆ

1

36

1

36

3

2

3

96

21

12 2

1

6

3

2

33

4

3

¯ÊËÁ

ˆ¯

=

= + =

= = = =

ll

I I I l

kI

Al

z x

z

2

3

96

3

481

12

1

120 300 0

34

44

2 2 2( . ) ..0075 2m

Let r be the distance from the suspension point to the center of mass.

Equation of motion. + SM0 = + S( ) : sin ( )M mge r mr I0 eff - = +q q q&& &&

where I mk= 2

For small angle q, - = +

++

=

mrg m r k

gr

r k

q q

q q

( )2 2

2 20

&&&&

Natural frequency: wngr

r k2

2 2=

+

67



(a) Plate is suspended from a vertex.

r h l

r

n

= = =

=

=

=

2

3

2

3

3

2

2

3

3

20 3

0 173205

0 03

9 81 0 1732

2

2

( . )

.

.

( . )( .

m

m

m2

w 005

0 03 0 007545 31

6 7313

2

)

. ..

.+

=

=

=

w

t pw

n

n

rad/s

t = 0 933. s �

(b) Plate is suspended at the midpoint of one side.

r h l

r

n

= =Ê

ËÁˆ

¯=

=

=

=

1

3

1

3

3

2

1

3

3

20 3

0 0866025

0 0075

9 81

2 2

2

( . )

.

.

( .

m

m

w ))( . )

. ..

.

0 0866025

0 0075 0 007556 64

7 5258+

=

=

=

w

t pw

n

n

rad/s

2 t = 0 835. s �

68


PROBLEM 19.50

A uniform disk of radius r = 250 mm is attached at A to a 650-mm rod AB of negligible mass, which can rotate freely in a vertical plane about B. Determine the period of small oscillations (a) if the disk is free to rotate in a bearing at A, (b) if the rod is riveted to the disk at A.

SOLUTION

I mr

mm

a lt

=

= =

= =

=

1

21

20 250

320 650

2

2( . )

.a a

a q&&(a) The disk is free to rotate and is in curvilinear translation.

Thus, I a = 0

S SM MB B= ( ) :eff +- = ªmgl lmatsin sinq q q

ml mgl

g

In

n

n

2

2

0

9 81

0 650

15 092

3 885

2

&&q q

w

w

t

- =

= =

==

=

.

.

.

.

m/s

m

rad/s

2

ppw

p

n

= 2

3 885. t n = 1 617. s �

69



(b) When the disk is riveted at A, it rotates at an angular acceleration a.

S SM MB B= ( ) :eff

+- = + =mgl I lma I mrtsinq a 1

22

1

20

9 81 0 650

2 2

2

222

mr ml mgl

gl

ln r

+ÊËÁ

ˆ¯

+ =

=+( )

=

&&q q

w

( . )( . )m/s m2

00 2502

220 650

14 053

3 749

2

2

3 749

. ( . )

.

.

.

( ) +ÈÎ

˘˚

==

=

=

w

t pw

p

n

nn

rad/s

t n = 1 676. s �

70


PROBLEM 19.51

A small collar weighing 1 kg is rigidly attached to a 3 kg uniform rod of length L = 1 m Determine (a) the distance d to maximize the frequency of oscillation when the rod is given a small initial displacement, (b) the corresponding period of oscillation.

SOLUTION

Equation of motion.

S SM MA A= ( )eff :

+

- - = + +WL

W d I mL

a m d aR C R R t R C t C2 2sin sin ( ) ( )q q a

sinq qª a q a q a q= = = = =&& && &&, ( ) , ( )aL L

a d dt R t C2 2

I mL

m d m gL

m gdR R C R C+ ÊËÁ

ˆ¯

+Ê

ËÁ

ˆ

¯˜ + +Ê

ËÁˆ¯

=2 2

02

2 &&q q

I m L

I mL m L

m Lm d m g

Lm

R R

R RR

RC R C

=

+ ÊËÁ

ˆ¯

=

+Ê

ËÁˆ

¯+ +

1

12

2 3

3 2

2

2 2

22 &&q ggdÊ

ËÁˆ¯

=q 0

71



&&

&&

q q

q

++( )+( ) =

= =

++( )

+

L mm

L mm

C

R

C

R

C

R

d g

d

m

mL

d

2

32

12

13

20

1

31

13

13

m

dd20

ÊËÁ

ˆ¯

=q

Natural frequency. wn

d g

d

d g

d2

12

13

13

13

2 2

1 5

1=

+( )+

=+( )

+.

( )

(a) To maximize the frequency, we need to take the derivative and set A equal to zero.

1 1 1 1 5 2

10

3 1 0

2 2

2 2

2

g

d

dt

d d d

d

d d

nw( )= + - +

+=

+ - =

( )( ) ( . )( )

( )

Solve for d. d = -0 3028 3 3028. .or

d = 0 303. m d = 303mm �

w

ww

n

n

n

22

2

1 5 0 3028 9 81

1 0 3028

16 200

4 0249

= ++

==

( . . )( . )

( . )

.

. rad/s wn = 4 02. rad/s �

(b) Period of oscillation. t pw

pn

n

= =2 2

4 0249. t n = 1 561. s �

72


PROBLEM 19.52

A compound pendulum is defined as a rigid slab which oscillates about a fixed Point O called the center of suspension. Show that the period of oscillation of a compound pendulum is equal to the period of a simple pendulum of length OA, where the distance from A to the mass center G is GA k r= 2/ . Point A is defined as the center of oscillation and coincides with the center of percussion defined in Problem 17.66.

SOLUTION

+ S SM M W r I ma rt0 0= - = +( ) : sineff q a

- = +mgr mk mrsinq q q2 2&& &&

&&q q++

=gr

r k2 20sin (1)

For a simple pendulum of length OA l= ,

&&q q+ =g

l0 (2)

Comparing Equations (1) and (2), lr k

r= +2 2

GA l rk

r= - =

2

Q.E.D. �

73


PROBLEM 19.53

A rigid slab oscillates about a fixed Point O. Show that the smallest period of oscillation occurs when the distance r from Point O to the mass center G is equal to k .

SOLUTION

See Solution to Problem 19.52 for derivation of

&&q q++

=gr

r k2 20sin

For small oscillations, sinq qª and

t pw

p pn

n

r k

gF gr

k

r= = + = +2

222 2 2

For smallest t n , we must have r kr+

2

as a minimum:

d r

dr

k

r

kr+( )

= - =

2

1 02

2

r k2 2= r k= Q.E.D. �

74


PROBLEM 19.54

Show that if the compound pendulum of Problem 19.52 is suspended from A instead of O, the period of oscillation is the same as before and the new center of oscillation is located at O.

SOLUTION

Same derivation as in Problem 19.52 with r replaced by R.

Thus, &&q q++

=gR

R k20

Length of the equivalent simple pendulum is

LR k

RR

k

R

L l rk

lkr

= + = +

= - + =

2 2 2

2

2( )

Thus, the length of the equivalent simple pendulum is the same as in Problem 19.52. It follows that the period is the same and that the new center of oscillation is at O. Q.E.D.

75


PROBLEM 19.55

The 8-kg uniform bar AB is hinged at C and is attached at A to a spring of constant k = 500 N/m. If end A is given a small displacement and released, determine (a) the frequency of small oscillations, (b) the smallest value of the spring constant k for which oscillations will occur.

SOLUTION

I ml

I

at

= = ÊËÁ

ˆ¯

= ◊

=

= =

1

12

1

128 0 250

0 04167

0 04

2 2( )( . )

.

.

kg m2

a q

a

&&00 04.

sin

&&qq qª

Equation of motion.

S SM MC C= ( )eff : - + = +( . ) . ( . )0 165 0 04 0 042 2k mg I mAq q q&& && ( . . ) ( . . )0 04167 0 01280 0 02722 0 32 0+ + - =&&q qk g (1)

(a) Frequency if N mk = ◊500 .

0 05447 10 47 0. ( . )&&q q+ =

fnn= =

( )wp p2 2

10 470 05447

..

fn = 2 21. Hz �

(b) For t n Æ or wn Æ 0, oscillations will not occur.

From Equation (1), wnk g2 0 02722 0 32

0 054470= - =. .

( . )

kg= =0 32

0 02722

0 32 9 81

0 02722

.

.

( . )( . )

( . ) k = 115 3. N/m �

76


PROBLEM 19.56

A 20-kg uniform square plate is suspended from a pin located at the midpoint A of one of its 0.4 m edges and is attached to springs, each of constant k = 1 6. kN/m If corner B is given a small displacement and released, determine the frequency of the resulting vibration. Assume that each spring can act in either tension or compression.

SOLUTION

a q

a q

q q

=

= -

ª

&&&&a

b bt 2 2

sin

Equation of motion.

+

S SM M mgb

kb Ib

m0 02

2

22

2= ( ) - + = + Ê

ËÁˆ¯eff

: q q a a

I mb

mb mb

mb+ ÊËÁ

ˆ¯

= + =2

1

6 4

5

12

22

22

5

12 22 0

12

10

24

50

2 2mb mgb

kb

g

b

k

m

&&

&&

q q

q q

+ +ÊËÁ

ˆ¯

=

+ +ÊËÁ

ˆ¯

=

Data: b m

k

= == =

0 4 20

1 6 1600

.

.

m; kg

kN/m N/m

&&q q+ +È

ÎÍ

˘

˚˙ =( )( . )

( )( . )

( )( )

( )( )

12 9 81

10 0 4

24 1600

5 200

&&qw w

+ =

= =

413 43 0

413 43 20 3332

.

. .n n rad/s

Frequency. fnn= =

wp p2

20 333

2

. fn = 3 24. Hz �

77


PROBLEM 19.57

Two uniform rods, each of mass m = 12 kg and length L = 800 mm, are welded together to form the assembly shown. Knowing that the constant of each spring is k = 500 N/m and that end A is given a small displacement and released, determine the frequency of the resulting motion.

SOLUTION

Equation of motion. S SM M0 0= ( )eff : + m gL

kL

I I mL

AC AC BD AC22

2 2

2 2

- ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

= + + ÊËÁ

ˆ¯

q a a( )

m m m

I I I mL

BD AC

BD AC

= =

= = = 1

22

1

6

1

42

2 202

2

+ÊËÁ

ˆ¯

+ ÊËÁ

ˆ¯

-È

ÎÍÍ

˘

˚˙˙

=mL kL

mgL&&q q

10

24 2 20

6

52

22mL

kL mgL kL mg

mLn&&q q w+ -

È

ÎÍ

˘

˚˙ = fi = -( )

Data: L m k= = = =800 0 8 12 500 mm m, kg, N/m.

Frequency. wn2 6 500 0 8 12 9 81

5 12 0 835 285= - =[( )( . ) ( )( . )]

( )( )( . ).

ww

pn nnf= =5 9401

2. rad/s fn = 0 945. Hz �

78


PROBLEM 19.58

The rod ABC of total mass m is bent as shown and is supported in a vertical plane by a pin at B and by a spring of constant k at C. If end C is given a small displacement and released, determine the frequency of the resulting motion in terms of m, L, and k.

SOLUTION

a q q

q q q

= =

= =

ª ª

&& &&aL

Im

LmL

t 2

1

12 2 241

22

sin cos

Equation of motion.

+ S SM Hmg L mg L

kL L Im

aB B= + - + = +( ) sin cos (eff ST: )cos2 2 2 2

2 22

q q q d q q&& ttL

2

mgL mgL

kL kLmL mL

4 4 12 42 2

2 2

q q d q q+ - - = +ST&& && (1)

But for equilibrium ( ),q = 0 SMm

gL

kLB = = -02 2

2dST (2)

79



Substituting Eq. (2) into Eq. (1), mgL

kLmL

kL mgL

mL

4 3

0

22

24

3

2

-ÊËÁ

ˆ¯

=

+-( )

=

q q

q q

&&

&&

Frequency. w

w

n

n

k

m

g

L

k

m

g

L

2 3 3

4

34

= -

= -

fnn=

wp2

fk

m

g

Ln = -3

2 4p �

80


PROBLEM 19.59

A uniform disk of radius r = 250 mm is attached at A to a 650-mm rod AB of negligible mass which can rotate freely in a vertical plane about B. If the rod is displaced 2° from the position shown and released, determine the magnitude of the maximum velocity of Point A, assuming that the disk (a) is free to rotate in a bearing at A, (b) is riveted to the rod at A.

SOLUTION

I mr= 1

22

Kinematics: a q

a q

=

= =

&&&&a l lt

(a) The disk is free to rotate and is in curvilinear translation.

Thus, I a = 0

Equation of motion. S SM M mgl lmaB B t= - = ª( ) sin , sineff : q q q

ml mgl2 0&&q q+ =

Frequency. w

w

n

n

g

l2

9 81

0 65015 092

3 8849

=

=

==

.

..

. rad/s

t pwn

n

= =21 617. s t n = 1 617. s �

81



(b) When the disk is riveted at A, it rotates at an angular acceleration a.

Equation of motion. S SM M mgl I lma I mrB B t= - = + = ª( ) sin , , sineff : q a q q1

22

1

202 2mr ml mgl+Ê

ËÁˆ¯

+ =&&q q

Frequency. w

w

n r

n

gl

l

2

22

12

2 2

2

9 81 0 650

0 250 0 650

14 053

3

=+( )

=+

==

( . )( . )

( . ) ( . )

.

..7487 rad/s

t pwn

n

= =21 676. s t n = 1 676. s �

82


PROBLEM 19.60

A 3-kg slender rod is suspended from a steel wire which is known to have a torsional spring constant K = ◊2 25. N m/rad. If the rod is rotated through 180° about the vertical and released, determine (a) the period of oscillation, (b) the maximum velocity of end A of the rod.

SOLUTION

Equation of motion. S SM M K IK

IK

I

G G

n n

= - = + =

+ = =

( )eff : q q q q

q w q w

&& &&

&&

0

02 2

Data: m

l

I ml

K

== =

= = ( )

= ◊= ◊

3

200 0 2

1

12

1

123 0 2

0 01

2 25

2 2

kg

mm m

kg m

N m/

2

.

( ) .

.

. rrad

rad/s

w

w

n

n

2 2 25

0 01225

15

= =

=

.

.

(a) Period of oscillation. t pw

pn

n

= =2 2

15 t n = 0 419. s �

Simple harmonic motion: q q w j

q w q w j

q w q

q w q

= +

= +

=

= =

m n

n m n

m n m

A m m A m

t

t

vl

l

sin ( )

cos( )

( )

&&

&2

1

2qq pm = ∞ =180 radians

(b) Maximum velocity at end A. ( ) ( . )( )( )vA m = ÊËÁ

ˆ¯

1

20 2 15 p ( ) .vA m = 4 71m/s �

83


PROBLEM 19.61

A homogeneous wire bent to form the figure shown is attached to a pin support at A. Knowing that r = 220 mm and that Point B is pushed down 20 mm and released, determine the magnitude of the velocity of B, 8 s later.

SOLUTION

Determine location of the centroid G.

Let r = mass per unit length

Then total mass m r r r= + = +r p r p( ) ( )2 2

About C: mgc rr

g r g= + ÊËÁ

ˆ¯

=02

2 2p rp

r

yr

r c r cr

=

+ = =+

2

2 22

22

p

r p rp

for a semicircle

( ) ,( )

Equation of motion. S SM M a c c

mgc I mcat

n

0 0= = = =- = + ª

( ) :

sineff

sin

a q a qq a q q

&& &&

(( )I mc mgc I mgc+ + = + =200 0&& &&q q q q

Moment of inertia. I mc I

I I I m r mr

+ =

= + = +

20

0 0 02

22

1( ) ( )

( )semicirc. line semicirc. line 22

22

02

m r m rm

r

I r

semicirc. line = = =+

= ◊

rp r rp

r p

( )

rr rr mr+ ◊

È

ÎÍ

˘

˚˙ =

+( ) +ÈÎÍ

˘˚

23 2

2

3

2 2

pp

mrmg

r2

2

2

3

2

20

( ) ( )++È

ÎÍ˘˚

++

=p

p qp

q&&

+

84



Frequency. wp p

w w

n

n n

g

r2

23

23

2 2

2 2 9 81

0 220

23 418 4 8392

=+( ) =

+( )= =-

( )( . )

( . )

. .s rad/s

q q w f q= + =m n Bt y rsin( )

At t = 0, mm,

mm

y y

y y

y y

B B

B B m n

B B m

= =

= = + =

= =

20 0

0 02

20

&& ( ) cos( ),

( ) si

w f f p

nn ,

( sin .

02

20

202

4 8

+ÊËÁ

ˆ¯

=

= +ÊËÁ

ˆ¯

=

p

w p w

( ) mm

mm)

y

y t

B m

B n n& 3392

202

20

rad/s

mm&y t tB n n n= +ÊËÁ

ˆ¯

= -w w p w wcos ( ) sin

At t = 8 s, &yB = - = -( )( . )sin[( . )( )] ( . )( . )20 4 8392 4 8392 8 96 88 0 8492

= -82 2. mm/s vB = 82 2. mm/s � �

85


PROBLEM 19.62

A homogeneous wire bent to form the figure shown is attached to a pin support at A. Knowing that r = 400 mm and that Point B is pushed down 40 mm and released, determine the magnitude of the acceleration of B, 10 s later.

SOLUTION

Determine location of the centroid G.

Let r = mass per unit length

Then total mass m r r r= + = +r p r p( ) ( )2 2

About C: mgc rr

g r g= + ÊËÁ

ˆ¯

=02

2 2p rp

r

yr

r c r cr

=

+ = =+

2

2 22

22

p

r p rp

for a semicircle

( )( )

Equation of motion. S SM M a c c

mgc I mcat

n

0 0= = = =- = + ª

( ) :

sineff

sin

a q a qq a q q

&& &&

(( )I mc mgc I mgc+ + = + =200 0&& &&q q q q

Moment of inertia. I mc I+ =20

I I I m r mr

m

0 0 02

22

12= + = +( ) ( )

( )semicirc. line semicirc. line

semicirrc. line= = =+

= ◊ + ◊È

ÎÍ

˘

˚˙ =

+

rp r rp

r pp

p

r m rm

r

I r rr r mr

22

2

3 202

2 2

( )

( )++È

ÎÍ˘˚

++È

ÎÍ˘˚

++

=

2

3

2

2

3

2

20

2mrmg

r

( ) ( )pp q

pq&&

+

86



Frequency. wp p

w w

n

n n

g

r2

23

23

2 2

2 2 9 81

0 4

12 8799 3

=+( ) =

+( )= =

( )( . )

( . )

. ) (rad/s ..

sin( )

sin( ) ( ) sin(

5889 rad/s

q q w f qq w f w

= + == + =

m n B

B m n B m

t y r

y r t y nnt + f)

At t = 0, mm

cos(0 )

mm

y y

t y y

y y

B B

B B m

B B m

= =

= = = + =

= =

40 0

0 02

40

&&( ): ( )

( ) sin

f f p

002

40

402

3 5889

+ÊËÁ

ˆ¯

=

= +ÊËÁ

ˆ¯

=

p

w p w

( )

( )sin .

y

y t

B m

B n n

mm

mm radd/s

&

&&

y t t

y

B n n n n

B n

= +ÊËÁ

ˆ¯

= -

= -

( )( )cos ( ) sin

( )

402

40

40 2

w w p w w

w(( ) +ÊËÁ

ˆ¯

=sin cosw p w wn n nt t2

40 2

At t = 10 s, ( ) ( )( . ) cos( . )( ) .

(

a y

vB t B

B

= = - = -&& 40 3 5889 3 5889 10 122 1242 mm/s2

)) ( )( . )sin ( . )( ) .= = - [ ] = -&yB 40 3 5889 3 5889 10 139 465 mm/s

a av

rB B tB= +

È

ÎÍ

˘

˚˙

È

ÎÍÍ

˘

˚˙˙

= + -È( ) ( . )

( . )22 2

12

22

122 124139 465

400ÎÎÍ

˘

˚˙

È

ÎÍÍ

˘

˚˙˙

=2

12

122 3. mm/s2 �

87


PROBLEM 19.63

A uniform disk of radius r = 120 mm is welded at its center to two elastic rods of equal length with fixed ends at A and B. Knowing that the disk rotates through an 8∞ angle when a 500-mN m◊ couple is applied to the disk and that it oscillates with a period of 1.3 s when the couple is removed, determine (a) the mass of the disk, (b) the period of vibration if one of the rods is removed.

SOLUTION

Torsional spring constant.

kT

k

= = ◊

( )= ◊

q p0 5

3 581

.

.

N m

(8)

N m/rad180

Equation of motion.

S SM M K JK

J0 0 0= - = + =( ) :eff q q q q&& &&

Natural frequency and period. wnK

J2 =

Period. t pw

pnn

J

K= =2

2

Mass moment of inertia. JK

J mr

n= = ◊

= ◊ ◊ = =

tp p

2

2

2

2

2 2

2

1 35 3581

2

0 15331

2

( )

( . ) ( )

( )

.

N m/r

N m s11

20 120 2m( . ) m

(a) Mass of the disk. m =◊ ◊( . )( )

( . )

0 1533 2

0 120

2

2

N m s

m m = 21 3. kg �

(b) With one rod removed: ¢ = = = ◊KK

2

3 581

21 791

.. N m/rad

Period. t p p=¢

=◊ ◊

◊2 2

0 1533 2J

K

( . )N m s

1.791 N m/rad t = 1 838. s �

88


PROBLEM 19.64

A 5 kg uniform rod CD of length l = 0 75. m is welded at C to two elastic rods, which have fixed ends at A and B and are known to have a combined torsional spring constant K = ◊30 N m/rad. Determine the period of small oscillation if the equilibrium position of CD is (a) vertical as shown, (b) horizontal.

SOLUTION

(a) Equation of motion.

+

a q a q

q q a

q

= = =

= - - = +

-

&& &&al l

M M K mgl

Il

ma

K

t

C C t

2 2

2 2S S( ) : ( ) sin ( )eff

-- = +1

2

1

42mgl I mlsinq q q&& &&

I ml mgl K

ml ml K

+ÊËÁ

ˆ¯

+ + =

+ÊËÁ

ˆ¯

+ +

1

4

1

20

1

12

1

4

1

2

2

2 2

&&

&&

q q q

q q

sin

mmgl

ml K mgl

K

ml

g

l

q

q q

q q

=

+ +ÊËÁ

ˆ¯

=

+ +ÊËÁ

ˆ¯

=

0

1

3

1

20

3 3

20

2

2

&&

&&

89



Data: K m l= ◊ = =

+ +

30 5 0 75

3 30

5 0 75

3 9 812

N m/rad kg m, , .

( )( )

( )( . )

( )( . )&&q(( )( . )

.

2 0 750

51 62 0

È

ÎÍ

˘

˚˙ =

+ =

q

q q&& Frequency. w wn n

2 51 62 7 1847= =. . rad/s

Period. t pw

p= =2 2

7 1847n . t n = 0 875. s �

(b) If the rod is horizontal, the gravity term is not present and the equation of motion is

&&q q

w

+ =

= = =

30

3 3 30

5 0 7532

2

22 2

K

mlK

mln

( )( )

( )( . )

w t pw

pn

n

= = =5 99382

5 65685.

. rad/s

2 t n = 1 111. s �

90


PROBLEM 19.65

A 1.8-kg uniform plate in the shape of an equilateral triangle is suspended at its center of gravity from a steel wire which is known to have a torsional constant K = ◊35 mN m/rad. If the plate is rotated 360° about the vertical and then released, determine (a) the period of oscillation, (b) the maximum velocity of one of the vertices of the triangle.

SOLUTION

Mass moment of inertia of plate about a vertical axis:

h b A bh b= = =3

2

1

2

3

42

For area, I I bhb

I I I b

x y

z x y

= = =

= + =

1

36

3

96

3

48

34

4

For mass, Im

AI

bb mb

I

z=

= ÊËÁ

ˆ¯

Ê

ËÁˆ

¯=

=

( )

( . )( . )

area

m4

3

3

48

1

12

1

121 8 0 150

4 2

22 3 23 375 10= ¥ ◊-. kg m

Equation of motion. + = - =

+ =

S SM M K I

K

I

G G( ) :eff q q

q q

&&

&& 0

Frequency. w

w

n

n

K

I2

3

3

35 10

3 375 1010 37

3 2203

= = ¥¥

=

=

-

-..

. rad/s

91



(a) Period. t pw

p= =2 2

3 2203n . t = 1 951. s �

Maximum rotation. q pm = ∞ =360 2 rad

Maximum angular velocity. &q w q pm n m= = =( . )( ) .3 2203 2 20 234 rad/s

(b) Maximum velocity at a vertex.

v r h bm m m= = = = ÊËÁ

ˆ¯

Ê

ËÁˆ

¯& &q q2

3

2

3

3

2

2

3

3

20 150 20 234( . )( . )

vm = 1 752. m/s �

92


PROBLEM 19.66

A horizontal platform P is held by several rigid bars which are connected to a vertical wire. The period of oscillation of the platform is found to be 2.2 s when the platform is empty and 3.8 s when an object A of unknown moment of inertia is placed on the platform with its mass center directly above the center of the plate. Knowing that the wire has a torsional constant K = ◊30 N m/rad, determine the centroidal moment of inertia of object A.

SOLUTION

Equation of motion. S SM M K I

K

I

K

I

G G

n

= - =

+ = =

( ) :eff q q

q q w

&&

&& & 0 2

Case 1. The platform is empty. w pt

p

w

n

n

IK

11

11

2 2

2 2

2 22 856

30

2 8563 6779

= = =

= = = ◊

..

( . ).

rad/s

kg m2

Case 2. Object A is on the platform. w pt

p

w

n

n

IK

22

22

2 2

2 2

3 81 653

30

1 65310 9793

= = =

= = = ◊

..

( . ).

rad/s

kg m22

Moment of inertia of object A. I I IA = -2 1 I A = ◊7 30. kg m2 �

93


PROBLEM 19.67

A thin rectangular plate of sides a and b is suspended from four vertical wires of the same length l. Determine the period of small oscillations of the plate when (a) it is rotated through a small angle about a vertical axis through its mass center G, (b) it is given a small horizontal displacement in a direction perpendicular to AB, (c) it is given a small horizontal displacement in a direction perpendicular to BC.

SOLUTION

(a) Plate is rotated about vertical axis.

Similarly, for B, D, and C,

S SM M Tr

lr J T

mg

mgr

JlJ m a b

r

G G= - ◊ = =

= = = +

( ) :

( )

eff 44

01

12

22 2

q q

q q

&&

&&

22 2 21

4= +( )a b

t pw

p

t p p

nn

n

Jl

mgr

m a b l

mg a b

l

g

= =

=++

=

22

2 23

2

112

2 2

14

2 2

( )

( ) �

(b) Plate is moved horizontally. The plate is in curvilinear translation.

sin

cos

cos

q qq

q q q

ªª

ª =

1

l l a&& &&

94



+SF T mg

Tmg

y = = - =

=

0 4 0

4

( cos )q

+®S SF F T ma

l g

H H= - =

+ =

( ) : sineff 4

0

q

q q&&

t pnl

g= 2 �

(c) Since the oscillation about axes parallel to AB (and CD) is independent of the length of the sides of the plate, the period of vibration about axes parallel to BC (and AD) is the same.

t pnl

g= 2 �

95


PROBLEM 19.68

A circular disk of radius r = 0 8. m is suspended at its center C from wires AB and BC soldered together at B. The torsional spring constants of the wires are K1 100= ◊ N m/rad for AB and K2 50= ◊ N m/rad for BC. If the period of oscillation is 0.5 s about the axis AC, determine the mass of the disk.

SOLUTION

Spring constant. Let T be the torque carried by the wires.

q q

q q q

B A C B

C A B A C B

T

K

T

K

K KT

T

K

K K

/ /

/ / /eq

eq

= =

= + = +ÊËÁ

ˆ¯

=

=

1 2

1 2

1

1 1

1 1 ++ =+

=+

= ◊1 100 50

100 5033 333

2

1 2

1 2KK

K K

K Keq N m/rad( )( )

.

Equation of motion. S SM M K I

K

IK

II

K

C C

nn

= - =

+ =

= =

( ) :eff eq

eq

eq eqor

q q

q q

ww

&&

&& 0

22

But t n = 0 5. s: w pt

pn

n

= = =2 2

0 512 566

.. rad/s

Then I = = ◊33 333

12 5660 21109

2

.

( . ). kg m2

For a circular disk, I mr= 1

22

mI

r= =2 2 0 21109

0 82 2

( )( . )

( . ) m = 0 660. kg �

96


PROBLEM 19.69

Determine the period of small oscillations of a small particle which moves without friction inside a cylindrical surface of radius R.

SOLUTION

Datum at 2 :

Positionj

T

V WR m

1

1

0

1

== -( cos )q

Small oscillations:

( cos ) sin1 22 2

2

22

1

2

- = ª

=

qq q

q

mm m

mVWR

Positionk

v R T mv mR

V

m m m m= = =

=

& &q q22 2 2

2

1

2

1

20

Conservation of energy. T Y T V

WR mR

W mg

mgR mR

mm m n m

mn

1 1 2 2

22 2

22

02

1

20

2

1

2

+ = +

+ = + =

=

=

qq q w q

qw

& &

22 2qm

nWg

R=

Period of oscillations. t pw

pnn

R

g= =2

2 �

97


PROBLEM 19.70

A 400 g sphere A and a 300 sphere C are attached to the ends of a rod AC of negligible weight which can rotate in a vertical plane about an axis at B. Determine the period of small oscillations of the rod.

SOLUTION

Datum at j:

Positionj

T

V W h W h

h BC

h BA

C C A A

C m

A m

1

1

0

1

1

== -= -= -

( cos )

( cos )

qq

Small angles.

12

2

0 3 0 2

2

1

2

1

- =

= -

= -

cos

[( )( ) ( )( )]

( . )( . ) (

qq

q

mm

C AmV m g BC m g BA

V kg m 00 4 0 1252

9 81

0 0981

2

1

. )( . ) ( . )

( . )

kg m m/s2[ ]ÊËÁ

ˆ

¯=

qm

V

Positionk

V T m v m v v

T m

C m A A m C m m

C m

2 2 02 2

22 2

01

2

1

20 2

1

20 2

= = + =

=

( ) ( ) , ( ) .

( . )

&

&

q

q ++ =

= +

1

20 125 0 125

1

20 3 0 2 0 4 0 1

2 2

2

m v

T

A m A m m( . ) ( ) .

( . )( . ) ( . )( .

& &q q

2 225

1

20 01825

2 2 2

22 2

)

( . )

2ÈÎ ˘ =

=

W

T

n m m n m

n m

q q w q

w q

&

98



Conservation of energy. T V T V mn m

n

1 1 2 2

22 2

2

0 0 09812

0 01825

20 0981

0 0182

+ = + + =

=

: ..

( . )

( .

qw q

w55

5 3753)

.=


pn

n

= =2 2

5 3753. t n = 2 71. s �

99


PROBLEM 19.71

A 1.8-kg collar A is attached to a spring of constant 800 N/m and can slide without friction on a horizontal rod. If the collar is moved 70 mm to the left from its equilibrium position and released, determine the maximum velocity and maximum acceleration of the collar during the resulting motion.

SOLUTION

Datum at j:

Positionj

T V kxm1 120

1

2= =

Positionk

T mv V v x

x x

T V T V kx mx

m

m n m

m m

2 22

2 2

1 1 2 22 2

1

20

01

2

1

20

1

2

= = =

=

+ = + + = +

&&

&w

kkx m xk

mm n m n

n n

2 2 2

2 2

1

2

800

1 8

444 4 21 08

= = =

= =-

w w

w w

2 N/m

kg

s r

.

. . aad/s

&x xm n m= = =-w ( . )( . ) .21 08 0 070 1 4761 s m m/s �

&&x xm n m= = =-w2 221 08 0 070 31 1( . )( . ) . s m m/s2 �

100


PROBLEM 19.72

A 1.5-kg collar A is attached to a spring of constant 1 kN/m and can slide without friction on a horizontal rod. The collar is at rest when it is struck with a mallet and given an initial velocity of 1 m/s. Determine the amplitude of the resulting motion and the maximum acceleration of the collar.

SOLUTION

m k= = =1 5 1 1000. kg kN/m N/m

Positionj Immediately after collar is struck:

x v

V T mv

= =

= = = = ◊

0 1

01

2

1

21 5 1 0 751 1

2 2

,

, ( . )( ) .

m/s

N m

Positionk v = 0, x is maximum, i.e., x x

V kx x x T

m

m m m

=

= = = =22 2 2

21

2

1

21000 500 0( )

Conservation of energy. T V T V

xm

1 1 2 2

20 75 0 0 500

+ = +

+ = +.

Amplitude of motion. xm = 0 03873. m xm = 38 7. mm �

Maximum force: F kxm m= = =( )( . ) .1000 0 03873 38 73 N

Maximum acceleration. aF

mmm= = 38 73

1 5

.

.

am = 25 82. m/s2 am = 25 8. m/s2 �

101


PROBLEM 19.73

A uniform rod AB can rotate in a vertical plane about a horizontal axis at C located at a distance c above the mass center G of the rod. For small oscillations, determine the value of c for which the frequency of the motion will be maximum.

SOLUTION

Find wn as a function of c.

Datum at k:

Positionj T V mgh

V mgc

V mgc

m

mm m

m

1 1

1

22

1

2

0

1

1 22 2

2

= == -

- = ª

=

( cos )

cos sin

q

qq q

q

Positionk T I

I I mc ml mc

T ml

c

C m

C

m

22

2 2 2

2

22 2

1

21

12

1

2 12

=

= + = +

= +Ê

ËÁˆ

¯

&

&

q

q V

T V T V mgc ml

cm m

m n m

2

1 1 2 2

2 22

2

0

02 12 2

0

=

+ = + + = +Ê

ËÁˆ

¯+

=

q q

q w q

&

&

ggc ml

c

gc

c

n

n l

= +Ê

ËÁˆ

¯

=+( )

22 2

2

122

12

2

w

w

Maximum c, when d

dc

g c c g

c

nl

l

w212

2 2

122

02

0

2

2= =

+( ) -

+( ) =

l

c2

2

120- = c

l=12

�

102


PROBLEM 19.74

A homogeneous wire of length 2 l is bent as shown and allowed to oscillate about a frictionless pin at B. Denoting by t0 the period of small oscillations when b = 0, determine the angle b for which the period of small oscillations is 2 0t .

SOLUTION

We denote by m the mass of half the wire.

Positionj Maximum deflections:

T V mgl

mgl

mgl

m m

m m

1 102 2

2

= = - - - +

= - +

, cos( ) cos( )

(cos cos sin

q b q b

q b q ssin cos cos sin sin )

cos cos

b q b q b

b q

+ -

= -

m m

mV mgl1

For small oscillations, cos

cos cos

q q

b b q

m m

mV mgl mgl

ª -

= - +

11

21

2

2

12

Positionk Maximum velocity:

T I I mlB m B22 21

22

1

3= = Ê

ËÁˆ¯

&q but

Thus, T ml

V mgl

mgl

m22 2

2

1

2

2

3

22

= ÊËÁ

ˆ¯

= - ÊËÁ

ˆ¯

= -

&q

b bcos cos

103



Conservation of energy. T V T V1 1 2 2+ = +

- + = -mgl mgl ml mglm mcos cos cosb b q q b1

2

1

32 2 2&

Setting &q q wm m n= , 1

2

1

32 2 2 2mgl mlm m ncos b q q w=

w b t pw

pbn

n

g

l

l

g2 3

2

22

2

3= = =cos

cos (1)

But for b = 0, t p0 22

3= l

g

For t t= 2 0 , 22

32 2

2

3p

bpl

g

l

gcos=

Ê

ËÁˆ

¯

Squaring and reducing,

1

41

4coscos

bb= = b = ∞75 5. �

104


PROBLEM 19.75

The inner rim of an 40 kg flywheel is placed on a knife edge, and the period of its small oscillations is found to be 1.26 s. Determine the centroidal moment of inertia of the flywheel.

SOLUTION

Datum at j:

Positionj

T J Vm1 02

11

20= =&q

Positionk

T V mgh

h r r

r

V mgr

mm

m

m

2 2

2

2

2

2

0

1 22

2

2

= =

= - =

ª

=

( cos ) sinqq

q

q

Conservation of energy.

T V T V I mgrmm

1 1 2 2 02

21

20 0

2+ = + + = +: &q q

For simple harmonic motion, &q w q

w q q w t pw

pm n m

n m m n nn

I mgrmgr

J

I

mgr

=

= = = =02 2 2 2

0

22

2

204 4( )

Moment of inertia. I I mr I mrmgr

Imgr

mr

n

n

02 2

2

2

2

22

2

4

4

1 26 40

= + + =( )

=( )

- =

t

pt

p

( )

( ) ( . ) ( )(( . ) .( )

.9 81

4

0 35

240

0 35

22

2

pÊËÁ

ˆ¯

- ÊËÁ

ˆ¯

I = -2 7615 1 225. . I = ◊1 537. kg m2 �

105


PROBLEM 19.76

A connecting rod is supported by a knife edge at Point A; the period of its small oscillations is observed to be 1.03 s. Knowing that the distance ra is 150 mm, determine the centroidal radius of gyration of the connecting rod.

SOLUTION

Positionj Displacement is maximum.

T V mgr mgra m a m1 120 1

1

2= = - ª, ( cos )q q

Positionk Velocity is maximum.

( )v r

T mv I mr mk

m r k

G m m

G m a m m

a

=

= + = +

= +

qq

q q q

&

& & &2

2 2 2 2 2 2

2

1

2

1

2

1

2

1

21

222 2

2 0

( )=

&qm

V

For simple harmonic motion, &q w qm n m=


01

2

1

202 2 2 2 2

22 2

22

+ = +( ) +

=+

= -

mgr m r k

gr

r kk

grr

a m a n m

na

a

a

na

q w q

ww

or 22

Data: t w pt

pn n

n

ar g

= = = =

= = =

1 032 2

1 036 1002

150 0 15 9 81

..

.

. .

s rad/s

mm m m/s2

kk 22

29 81 0 15

6 10020 15 0 039543 0 0225 0 017043= - = - =( . )( . )

( . )( . ) . . . mm2

k = 0 13055. m k = 130 6. mm �

106


PROBLEM 19.77

The rod ABC of total mass m is bent as shown and is supported in a vertical plane by a pin at B and a spring of constant k at C. If end C is given a small displacement and released, determine the frequency of the resulting motion in terms of m, L, and k.

SOLUTION

Positionj T I V kB m12

121

2

1

2= =&q d( )ST

Positionk T Vm

ghm

gL

k L

hL L

m m

m

2 22

2

02 2 2

1

2

21 2

2

= = - - + +

= - =

sin ( )

( cos ) sin

q q d

q

ST

qq

q q q q d

q q

m

m m m

m m

hmL

Vm

gL m

gL

k L

2

4 2 4 2 2

1

2

2

22 2ª = - - + +

ª

( )

sin

ST


T V T V

I km

gL mg L

k LB mm

m m

1 1 2 2

2 22

2 21

2

1

20

2 4 2 2

1

2

+ = +

+ = - - + +&q dq

q q dST STT ST2 2+È

ÎÍ˘˚L md q (1)

When the rod is in equilibrium,

+

SMm

gL

kLB = = -02 2

dST (2)

107



Substituting Eq. (2) into Eq. (1)

I kLmgL

I kLmgL

B m m m n m

B n m m

& &q q q w q

w q q

2 2 2

2 2 2 2

4

4

= -ÊËÁ

ˆ¯

=

= -ÊËÁ

ˆ¯

ww

w

n

mgL

B

B

n

mgL

mL

kL

I

Im

LmL

kL k

m

22

4

22

22

4

3

21

3 2 3

3 32

=-

= ÊËÁ

ˆ¯

=

=-

= - gg

L4

Frequency of oscillation.

&f k

m

g

Lnn= = -

wp p2

3

2 4 �

108


PROBLEM 19.78

A 7.5 kg uniform cylinder can roll without sliding on an incline and is attached to a spring AB as shown. If the center of the cylinder is moved 10 mm down the incline and released, determine (a) the period of vibration, (b) the maximum velocity of the center of the cylinder.

SOLUTION

(a) Positionj

T V k r m1 120

1

2= = +( )d qST

Positionk

T J mv

V mgh k

m m22 2

22

1

2

1

21

2

= +

= +

&q

d( )ST


T V T V k r J mv mgh k

k

m m m1 1 2 22 2 2 20

1

2

1

2

1

2

1

2+ = + + + = + + +: ( ) ( )d q q d

d

ST ST&

SST ST ST2 2 2 2 2 22 2+ + = + + +k r kr J mv mgh km m m md q q q d& (1)

When the disk is in equilibrium,

+SM mg r k rc = = -0 sin b dST

Also, h r m= sin b q

Thus,

mgh k r- =dST 0 (2)

109



Substituting Eq. (2) into Eq. (1)

kr J mv

v r r

kr J mr

m m m

m n m m m n m

m m n

2 2 2 2

2 2 2 2 2

q q

q w q q w q

q q w

= +

= = =

= +

&& &

( )

wwnkr

J mr2

2

2=

+ J mr= 1

22

wnkr

mr mr

k

m2

2

2 212

2

3=

+=

t pw

p pn

n

= = = =2 2

23

9007 5

2

800 702

( )( . )

.N/mkg

s �

(b) Maximum velocity.

v r

v r r

m m

m m n

m m n m

=

=

= = =

&&

q

q q w

q w q 10

10000 01. m

vm = =( . ) .0 01 80 0 0894m/s m/s �

110


PROBLEM 19.79

Two uniform rods, each of weight m = 600 g and length l = 200 mm, are welded together to form the assembly shown. Knowing that the constant of each spring is k = 120 N/m and that end A is given a small displacement and released, determine the frequency of the resulting motion.

SOLUTION

Mass and moment of inertia of one rod. m = 600 g = 0.6 kg

I ml= = = ¥ ◊-1

12

1

120 6 0 2 2 102 2 3( . )( . ) kg m2

Approximation.

sin tan

cos sin

q q q

qq

q

m m m

mm

m

ª ª

- = ª1 22

1

22 2

Spring constant: k = 120 N/m

Positionj T I ml

m m

m

12

2

3 2

21

2

1

2 2

21

22 10

= ÊËÁ

ˆ¯

+ ÊËÁ

ˆ¯

= ÊËÁ

ˆ¯

¥ +-

& &

&

q q

q( ) ( )11

20 6 0 1

5 10

0

2

3 2

1

ÊËÁ

ˆ¯ ( )

= ¥=

-

( . ) . &

&q

q

m

m

V

111



Positionk T

VWl

kl

m m

2

2

2

0

21 2

1

2 2

0 6 9 81 0

=

= - - + ÊËÁ

ˆ¯

ÊËÁ

ˆ¯

ª -

( cos )

( . )( . )( .

q q

22

2

1

22

1

2120

0 2

2

1 4943

22

2

)( ) ( )

.

.

q q

q

m m

m

ÊËÁ

ˆ¯

+ ÊËÁ

ˆ¯

ÊËÁ

ˆ¯

ª


5 10 0 0 1 4943

17 2876

3 2 2¥ + = +

=

- &&

q q

q qm m

m m

.

.

Simple harmonic motion. &q w qm n m=

wn = 17 2876. rad/s

Frequency. fnn=

wp2

fn = 2 75. Hz �

112


PROBLEM 19.80

A slender 8-kg rod AB of length l = 600 mm is connected to two collars of negligible mass. Collar A is attached to a spring of constant k = 1.2 kN/m and can slide on a vertical rod, while collar B can slide freely on a horizontal rod. Knowing that the system is in equilibrium and that q = 40°, determine the period of vibration if collar B is given a small displacement and released.

SOLUTION

Vertical rod.

y l

y l

y l

x l

x l

x l

==

== -= -

= -

sin

cos

cos

cos

sin

sin

qd qdq

d qdqq

d qdq

d qd

&

& &&qy

yx

x= =2 2

Positionj (Maximum velocity, dq&m)

T J m x y

T ml

m m m

m

12 2 2

12

1

2

1

2

1

2

1

12

= + +ÈÎ ˘

= ÊËÁ

ˆ¯

( ) ( ) ( )

( )

dq d d

dq

& &

& 222 2

2

12

12 2

1

2

1

12

+ ÊËÁ

ˆ¯

+ ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

=

ml l

T ml

msin cos ( )q q dq&

++ÈÎÍ

˘˚

=

= +

1

4

1

2

1

3

1

2

2 2 2

12

( ) ( )

( )

dq dq

d

& &m mml

V k mg yST

Positionk (Zero velocity, maximum dqm )

T

V k y mg y ym

2

22

0

1

2

=

= + + -( ) ( )d d dST

113



Conservation of energy. T V T V1 1 2 2+ = + :

1

2

1

3

1

20

1

21

3

2 2 2 2

2

ml k mgy y mg y y

ml

m m+ + + = + + + -( ) ( ) ( )

(

dq d d d d

d

&ST ST

&&q d d d d d dm xk mgy k y y mg y y) ( ) ( )2 2 2 22+ + = + + + -ST ST ST

(1)

But when the rod is in equilibrium,

+

SM mgx

k x mg kB = - = =2

0 2d dST ST (2)

Substituting Eq. (2) into Eq. (1),

ml k y y l

ml kl

m m m m

m m

2 2 2

2 2 2 2 2

1

31

3

( ) cos

( ) cos ( )

dq d d q dq

dq q dq

&

&

= =

=

For simple harmonic motion,

dq dq w f

dq dq w

dq w q dq

w

= +

=

=

=

m n

m m n

m n m

n

t

m k

k

m

sin( )

( ) cos ( )

&1

3

3

2 2 2 2

2 ccos( )

cos

.

2 2

2

3 1200

840

264 07

q

w

= ∞

=

N/m

kg

n

Period of vibration. t pw

pn

n

= =2 2

264 07. t n = 0 387. s �

114


PROBLEM 19.81

A slender rod AB of length l = 600 mm and negligible mass is connected to two collars, each of mass 8 kg. Collar A is attached to a spring of constant k = 1 2. kN/m and can slide on a vertical rod, while collar B can slide freely on a horizontal rod. Knowing that the system is in equilibrium and that q = ∞40 , determine the period of vibration if collar A is given a small displacement and released.

SOLUTION

Horizontal rod.

Y l

l

l

x l

l

l

Y

g

x

x

==

=

== -

= -

sin

cos

cos

cos

sin

sin

qd qdq

d qdq

qd qdq

d qdq

&

&

Positionj (Maximum velocity, dq&m )

T m y m x

T m l l

T

m m

m

12 2

12 2 2

1

1

2

1

21

2

= +

= +

( ) ( )

[( cos ) ( sin ) ]( )

d d

q q dq

& &

&

==

=

1

20

2 2

1

ml

V

m( )dq&

Positionk (Zero velocity, maximum dq )

T

V k ym

2

22

0

1

2

=

= d


ml k y

y l

ml kl

m m

m m

m

1 1 2 2

2 2 2

2 2

1

20

1

2

+ = +

+ =

=

=

( )

cos

( )

dq d

d qdq

dq

&

& 22 2 2cos ( )q dqm

115



Simple harmonic motion. dq dq w f

dq dq w

dq w q dq

w

= +

=

=

=

m n

m m n

m n m

n

t

ml kl

k

sin ( )

( ) cos ( )

&2 2 2 2 2 2

2

mm

n

cos

cos .

2

2 2 21200

840 88 02

q

w = = -N/m

kgs


pn

n

= = =2 2

88 020 6697

.. s t n = 0 670. s �

116


PROBLEM 19.82

A 3-kg slender rod AB is bolted to a 5-kg uniform disk. A spring of constant 280 N/m is attached to the disk and is unstretched in the position shown. If end B of the rod is given a small displacement and released, determine the period of vibration of the system.

SOLUTION

r

l

==

0 08

0 3

.

.

m

m

Positionj T J J

V

J m r

J m l

m A m

A AB

12 2

1

02

1

2

1

20

1

21

3

= +

=

=

=

disk rod

disk

rod

& &q q( )

( ) 22

Positionk T2 0=

V k r m gl

V kr

m AB m

mm m

22

22

22

1

2 21

1 22 2

1

2

= + -

- = ª

=

( ) ( cos )

cos sin

q q

qq q

qmmAB

l

m

m g2 2 2

2+

( )q

117




T V T V m r m l kr m gAB m m AB1 1 2 2 02 2 2 2 21

2

1

2

1

30 0

1

2

1

2+ = + +Ê

ËÁˆ¯

+ = + +: &q q llm22q

For simple harmonic motion,

&q w q

w q q

w

m n m

AB n m AB m

n

m t m l kr m gl

=

+ÊËÁ

ˆ¯

= +ÊËÁ

ˆ¯

=

1

2

1

3 202 2 2 2 2 2

2 kkr m gl

m r m lAB

AB

n

2

12 0

2 13

2

22280 0 08 3 9 81

++

=+

w( )( . ) ( )( .N/m m kg m/s22 m

kg m kg m

)

( )( . ) ( )( . )

.

.

.0 32

12

2 13

2

2

5 0 08 3 0 300

6 207

0 1065

( )+

= =wn 88 55.


pn

n

= =2 2

58 55. t n = 0 821. s �

118


PROBLEM 19.83

A 400-g sphere A and a 300-g sphere C are attached to the ends of a 600-g rod AC, which can rotate in a vertical plane about an axis at B. Determine the period of small oscillations of the rod.

SOLUTION

Positionj

T m m m I

I

A C m AC m AC m12 2 2 21

20 125

1

20 2

1

20 0375

1

2= + + +( . ) ( . ) ( . )& & &q q q

AAC ACm

T

=

= + + +

1

120 325

1

20 4 0 125 0 3 0 2 0 6 0 0375

2

12 2 2

( . )

. ( . ) . ( . ) . ( . )11

120 6 0 325

1

20 024375

0

2 2

12

1

( . )( . )

( . ) ( )

ÈÎÍ

˘˚

= ◊

=

&

&

q

q

m

mT

V

N m

119



Positionk

T

V W W WA m C m AC m

2

2

0

0 125 1 0 2 1 0 0375 1

== - - + - + -. ( cos ) . ( cos ) . ( cos )q q q

11 2 22

0 4 9 81 0 125 0 3 9 81

22

2

- = ª

= - +

cos sin

( . )( . )( . ) ( . )( . )(

q qq

mmm

V

/

00 2 0 6 9 81 0 03752

0 318825

2

2

2

2

. ) ( . )( . )( . ) ( )

.

+[ ] ◊

=

q

q

m

mV

N m

Conservation of energy. T V T V m m1 1 2 22 21

20 024375 0 0

0 318825

2+ = + + = +: ( . )

.&q q

Simple harmonic motion.

&q w q

w

t pw

p

m n m

n

nn

=

= =

= =

2 0 318825

0 02437513 08

2 2

13 08

.

..

. t n = 1 737. s �

120


PROBLEM 19.84

Three identical rods are connected as shown. If b l= 34 , determine the

frequency of small oscillations of the system.

SOLUTION

l = length of each rod

m = mass of each rod

Kinematics: vl

v b

m m

BE m m

=

=2&

&q

q( )

Positionj T

V mgl

mgb

V mg l b

m m

m

1

1

1

0

22

=

= - -

= - +

cos cos

( )cos

q q

q

Positionk V mgl

mgb

mg l b

T I mv m vm m BE m

2

22 2 2

22

21

2

1

2

1

2

= - -

= - +

= +ÈÎÍ

˘˚

+

=

( )

( )&q

11

12 2

1

2

1

3

1

2

2 22

2

22 2

ml ml

m b

T l b m

m m m

m

& & &

&

q q q

q

+ ÊËÁ

ˆ¯

+

= +ÊËÁ

ˆ¯

( )

22

121



Conservation of energy. T V T V mg l b l b m mg l b

mg

m m1 1 2 22 2 20

1

3

1

2+ = + - + = +Ê

ËÁˆ¯

- +: ( )cos ( )

(

q q&

ll b l b mm m+ - = +ÊËÁ

ˆ¯

)( cos )11

3

1

22 2 2q q&

For small oscillations, ( cos )

( )

11

21

2

1

3

1

2

2

2 2 2 2

- =

+ = +ÊËÁ

ˆ¯

q q

q q

m m

m mmg l b l b m &

But for simple harmonic motion, &q w q q w q

w

m n m m n m

n

mg l b l b m

gl b

l

= + = +ÊËÁ

ˆ¯

= +

: ( ) ( )1

2

1

3

1

2

1

2

2 2 2 2

213

22 12

2+ b

or wn gl b

l b2

2 23

2 3= +

+ (1)

For b l= 3

4, we have w

w

n

n

gl l

l l

gl

l

g

l

g

l

234

2 34

2

74

5916

2

32 3

3

1 4237

1 1932

=+

+ ( )=

=

=

.

.

f

g

l

nn=

=

wp

p

2

1 1932

2

. f

g

ln = 0 1899. �

122


PROBLEM 19.85

An 800-g rod AB is bolted to a 1.2-kg disk. A spring of constant k = 12 N/m is attached to the center of the disk at A and to the wall at C. Knowing that the disk rolls without sliding, determine the period of small oscillations of the system.

SOLUTION

Positionj

T I ml

r I mG AB m AB m G m12

221

2

1

2 2

1

2

1

2= + -Ê

ËÁˆ¯

+ +( ) ( )& & &q q qdisk diskk

2 kg m

r

I ml

ml

r

m

G AB

AB

2 2

2 21

12

1

120 8 0 6 0 024

2

&q

( ) ( . )( . ) .= = = ◊

-ÊËÁ

ˆ¯

= - = ◊

= =

22

2

0 8 0 3 0 25 0 002

1

2

1

21

( . )( . . ) .

( ) ( .

kg m2

disk diskI m rG 22 0 25 0 0375

1 2 0 25 0 0750

1

2

2

2 2

1

)( . ) .

. ( . ) .

= ◊

= = ◊

=

kg m

kg m

2

disk2m r

T 00 024 0 002 0 0375 0 0750

1

20 1385

0

2

12

1

. . . .

[ . ]

+ + +[ ]

=

=

&

&

q

q

m

mT

V

Positionk T

V k r m gl

m AB m

mm m

2

22

22

0

1

2 21

1 22 2

=

= + -

- = ª

( ) ( cos )

cos sin

q q

qq q

(smalll angles)

N/m m kg m/s 2V m2

2 21

212 0 25 8 9 81

0 6= +( )( . ) ( )( . ).q mm

2ÊËÁ

ˆ¯

qm2

2

123



V

T V T V

m

m

m n

22

2

1 1 2 2

2

1

20 750 2 354

1

23 104

= +

= ◊

+ = +

=

[ . . ]

( . )

q

q

q w q

N m

2&mm

m n m

n

2

2 2

2

1

20 1385 0 0

1

23 104

0 1385

( . ) ( )

( . )

( .

q w q

w

+ = +

=◊

3.104

N m

2

kkg m

s

2◊

= -

)

.22 41 2 t pw

pn

n

= = =2 2

22 411 327

.. s �

124


PROBLEM 19.86

Two uniform rods AB and CD, each of length l and mass m, are attached to gears as shown. Knowing that the mass of gear C is m and that the mass of gear A is 4 m, determine the period of small oscillations of the system.

SOLUTION

Kinematics: 2

2

2

r rA C

A C

A C

q qq q

q q

==

=& &

Let q qq q

q q

A m

m C m

m C m

==

=

2

2

( )

( )& &

Positionj T I I I I

ml

A m C m AB m CD m

AB m

12 2 2 21

2

1

22

1

2

1

22

1

2 2

= + + +

+ Ê

& & & &

&

q q q q

q

( ) ( )

ËËÁˆ¯

+ ÊËÁ

ˆ¯

2 21

2 22m

lCD m

&q

I m r mr

I m r mr

I ml I

A

C

AB CD

= =

= =

=

1

24 2 8

1

2

1

21

12

2 2

2 2

2

( )( )

( )( )

==

= +Ê

ËÁˆ

¯+ + + +

È

ÎÍÍ

˘

˚˙˙

=

1

12

1

28

24

12 3 4

1

2

2

12

2 2 2 22

1

ml

T m rr l l l

l

T m 1105

3

0

2 2 2

1

r l

V

m+ÈÎÍ

˘˚

=

&q

125



Positionk T

V mgl mgl

m m

1

1

0

21

21 4

=

= - + -( cos ) ( cos )q q

For small angles, 1 22 2

1 4 2 2 2

1

2 22

22

2 2

1

2

- = ª

- = =

= +

cos sin

cos sin

qq q

q q q

qq

mm m

m m m

mV mgl mmm

m n m

mgl

T V T V

m r

22

1 1 2 22 2 2

2

1

2

5

2

1

210

Ê

ËÁˆ

¯=

+ = + =

+

q

q w q &55

30 0

1

2

5

22 2 2

2

l mgln mmÈ

ÎÍ˘˚

+ = +w qq

wn

gl

r l

gl

r l

2522 5

32

2 2

10

3

12 2

=+

=+

t pw

pnn

r l

gl= = +2

212 2

3

2 2

�

126


PROBLEM 19.87

Two uniform rods AB and CD, each of length l and mass m, are attached to gears as shown. Knowing that the mass of gear C is m and that the mass of gear A is 4m, determine the period of small oscillations of the system.

SOLUTION

Kinematics: 2 2

2

r rA C A C

A C

q q q q

q q

= =

=

& &

Let q q q q

q qA m m C m

m C m

= =

=

2

2

( )

( )& &

Positionj T I I I I ml

A m C m AB m CD m AB m12 2 2 21

2

1

22

1

2

1

22

1

2 2= + + + + Ê& & & & &q q q q q( ) ( )

ËËÁˆ¯

+ ÊËÁ

ˆ¯

2 21

2 22m

lCD m

&q

I m r mr

I m r mr

I ml I

A

C

AB CD

= =

= =

=

1

24 2 8

1

2

1

21

12

2 2

2 2

2

( )( )

( )( )

==

= +Ê

ËÁˆ

¯+ + + +

È

ÎÍÍ

˘

˚˙˙

1

12

1

28

24

12 3 4

2

12

2 2 2 22 2

1

ml

T m rr l l l

l

T

m&q

== +ÈÎÍ

˘˚

=1

210

5

302 2 2

1m r l Vm&q

127



Positionk T

V mgl mgl

m m

2

2

0

21

21 2

=

= - - + -( cos ) ( cos )q q

For small angles, 1 22 2

1 2 2 2

2 2

22

2 2

2

2

- = ª

- = ª

= - +

cos sin

cos sin

qq q

q q q

q

mm m

m m m

mV mgl mgl

222

1

2

3

2

2

2

1 1 2 2

q

q

q w q

m

m

m n m

mgl

T V T V

=

+ = + = &

1

210

5

30 0

1

2

3

2

10

9

2 2 2 2 2

2322 5

32

m r l mgl

gl

r l

m n m

n

+ÈÎÍ

˘˚

+ = +

=+

=

q w q

w

ggl

r l60 102 2+ t pw

pnn

r l

gl= = +2

260 10

9

2 2

�

128


PROBLEM 19.88

A 5-kg uniform rod CD is welded at C to a shaft of negligible mass which is welded to the centers of two 10-kg uniform disks A and B. Knowing that the disks roll without sliding, determine the period of small oscillations of the system.

SOLUTION

W W WA B= = disk

Positionj T I m r

I ml

r

A m m

CD m CD

12 2

2

1

22

1

22

1

2

1

2 2

= +

+ + -ÊË

( ) ( )( )disk disk& &

&

q q

q ÁÁˆ¯

= = = ◊

=

22

2 21

2

1

210 0 5 1 25

1

12

&qm

A

CD

I m r

I

( ) ( )( . ) .disk disk2kg m

mm l

T

CD2 2

1

1

125 1 5 0 9375

1

22 5 5 0 9375 0 3125

= = ◊

= + + +[ ]

( )( . ) .

. . .

kg m2

&qq

q

m

mT V

2

12

11

28 75 0= =( . ) &

Positionk T

V Wl

CD m

2

2

0

21

=

= -( cos )q

129



Small angles: 1 22 2

1

2 2

1

25 9 81 1 5

21

2

22

2

2

2

- = ª

=

=

=

cos sin

( )( . )( . )

q q q

q

q

mm

CDm

m

m

V W l

(( . )36 7875 2qm

Conservation of energy and simple harmonic motion.

T V T V

m n m

n m m

n

1 1 2 2

2 2 2

2

1

28 75 0

1

236 7875

36 7

+ = +

=

+ =

=

&q w q

w q q

w

( . ) ( . )

. 8875

8 754 2043

..=


pn

n

= =2 2

4 2043. t n = 3 06. s �

130


PROBLEM 19.89

Four bars of the same mass m and equal length l are connected by pins at A, B, C, and D and can move in a horizontal plane. The bars are attached to four springs of the same constant k and are in equilibrium in the position shown ( ).q = ∞45 Determine the period of vibration if corners A and C are given small and equal displacements toward each other and released.

SOLUTION

Kinematics: BC xl

xl

xl

yl

yl

G G

G

G G

= = -

= -

= =

2 2

2

2 2

cos sin

sin

sin cos

q d q dq

d q dq

q d q

& &

ddq

d q dqyl

G =2

cos &

x l x l

x l

x l x l

x

C C

C

B B

B

= = -= -

= ==

cos sin

sin

sin cos

q d q dqd q dq

q d q dqd

& &

& ll cosq dq&y

y l y l

y l

C

B B

B m

== =

=

0

sin cos

cos

q d q dqd q dq& &

The kinetic energy is the same for all four bars.

131



Positionj T I m x y

I ml

T ml

m G m G m12 2 2

2

12

41

2

1

2

1

12

2

= + +ÈÎÍ

˘˚

=

=

( ) [( ) ( ) ]dq d d& & &

11

12

1

4

2

30

2 2 2

12

1

+ +ÈÎÍ

˘˚

=

=

(sin cos )q q d qm m m

T ml

V

&

Positionk T

V k x k y

V k l l

m m

m

2

22 2

22 2 2 2 2

0

21

22

1

2

=

= +

= +

=

( ) ( ) ( ) ( )

( sin cos )

d d

q q d q

kk l m2 2d q


ml k lm m

1 1 2 2

2 2 2 22

30 0

+ = +

+ = +( ) ( )dq dq&

Simple harmonic motion. dq w dq

w

&m n m

nk

m

=

= ÊËÁ

ˆ¯

2 3

2

Period of vibration. t pwn

n

= 2 t pn

m

k= 2

2

3 �

132


PROBLEM 19.90

The 10-kg rod AB is attached to two 4-kg disks as shown. Knowing that the disks roll without sliding, determine the frequency of small oscillations of the system.

SOLUTION

Positionk Positionj

r = 0 15. m

Masses and moments of inertia. m m

I I m r

m

A B

A B A A

AB

= =

= = =

= ◊=

4

1

2

1

24 0 15

0 045

10

2 2

2

kg

kg m

kg

( )( . )

.

Kinematics: v r

v

m A m m

AB m

= =

=

& &&

q q

q

0 15

0 05

.

.

Positionj (Maximum displacement) T

V WAB m m

m

1

1

0

0 1 10 9 81 0 1

9 81

== - = -= -

( . cos ) ( )( . )( . cos )

. cos

q qq

Positionk (Maximum speed) T m v I m v I m vA m A m B m B m AB AB22 2 2 2 21

2

1

2

1

2

1

2

1

2

21

24 0 15

= + + + +

=

& &q q

( )( . && & &

&q q q

q

m m m

m

V

) ( . ) ( )( . )

.

2 2 2

2

2

1

20 045

1

210 0 05

0 1475

+ÈÎÍ

˘˚

+

== -WWAB ( . ) ( )( . )( . ) .0 1 10 9 81 0 1 9 81= - = -

133




m m

m m

1 1 2 2

2

2

9 81 0 1475 9 81

66 5085 1

+ = +

- = -

= -

. cos . .

. ( cos )

q q

q q

&&

ªª ÊËÁ

ˆ¯

==

66 50851

2

33 2542

5 7666

2

2

.

.

.

q

qq q

m

m

m m

Simple harmonic motion. &q w qm n m= wn = 5 7666. rad/s

Frequency. fnn= =

wp p2

5 7666

2

. fn = 0 918. Hz �

134


PROBLEM 19.91

An inverted pendulum consisting of a sphere of mass m and a rigid bar ABC of length l and negligible weight is supported by a pin and bracket at C. A spring of constant k is attached to the bar at B and is undeformed when the bar is in the vertical position shown. Determine (a) the frequency of small oscillations, (b) the smallest value of a for which these oscillations will occur.

SOLUTION

(a) Positionj T m l ml

V

m m12 2 2

1

1

2

1

20

= =

=

( )& &q q

Positionk T

V k a mlm m

2

22

0

1

21

=

= - -( sin ) ( cos )q q

For small angles, sin

cos sin

[

q q

qq

q

q q

m m

mm

m

m mV ka ml

ka ml

ª

- =

ª

= -

= -

1 22

2

1

2 2 21

2

2

2

22

2 2

2 ]]qm2

T V T V

ml ka mlm m

1 1 2 2

2 2 2 21

20 0

1

2

+ = +

+ = + -&q q[ ]

135



&q w q

w q

w

m n m

n m

n mg

mm

g

m

gl ka ml

ka ml

l

g

l

ka

ml

=

=

= -

= -

( )= -

È

ÎÍ

2 2 2 2

22

2

2

1˘

˚˙ f

g

l

ka

mln n= = -Ê

ËÁˆ

¯1

2

1

21

2

pw

p �

(b) fn = 0

ka

ml

2

1 0- � aml

k� �

136


PROBLEM 19.92

For the inverted pendulum of Problem 19.91 and for given values of k, a, and l, it is observed that f = 1.5 Hz when m = 1 kg and that f = 0.8 Hz when m = 2 kg. Determine the largest value of W for which small oscillations will occur.

SOLUTION

See Solution to Problem 19.91 for the frequency in terms of W, k, a, and l.

fg

l

ka

Wln = -Ê

ËÁˆ

¯1

21

2

p

f Wg

l

ka

gln = = = -Ê

ËÁˆ

¯1 5 1

1

21

2

. ( ) ( )Hz g N 1.5p

(1)

f Wg

l

ka

gln = = = -Ê

ËÁˆ

¯0 8 2

1

2 21

2

. ( ) ( )Hz g N 0.8p

(2)

Dividing Eq. (1) by Eq. (2) and squaring, 1 5

0 8

1

1

2

2

2

2

2

2

2

2

.

.

( )

( )ÊËÁ

ˆ¯

=-( )-( ) = -

-

kagl

kagl

ka gl

ka gl

225

642 2

3 3196

1

2

3 31961

2 2

2

( ) ( )

.

.

ka gl ka gl

ka gl

fg

l

gl

Wln

- = -

=

= -ÊËp ÁÁ

ˆ¯

=

- =

f

g

W

n 0

3 31961 0

. W = 32 6. N �

137


PROBLEM 19.93

A uniform rod of length L is supported by a ball-and-socket joint at A and by a vertical wire CD. Derive an expression for the period of oscillation of the rod if end B is given a small horizontal displacement and then released.

SOLUTION

Positionj (Maximum deflection)

Looking from above:

Horizontal displacement of C: x bC m= q

Looking from right: f qmC

mx

h

b

h= =

y h h

y hb

h

b

h

y yAG

ACy

C m m

C m m

m G

= - ª

= ÊËÁ

ˆ¯

=

= = -

( cos )11

2

1

2

1

2

2

2 22

f f

q q

CC m

m m

L

b

b

h

ybL

h

=Ê

ËÁˆ

¯

= ◊

12

22

2

1

2

1

4

q

q

We have T

V mgymgbL

hm m

1

12

0

1

4

=

= = q

Positionk (Maximum velocity)

Looking from above: T I mv

mL mL

T mL

m m

m

22 2

2 22

2

1

2

1

2

1

2

1

12

1

2 2

1

6

= +

= ÊËÁ

ˆ¯

+ ÊËÁ

ˆ¯

=

&

& &

q

q q

22 2

2 0

&qm

V =

Conservation of energy. T V T VmgbL

hmLm m1 1 2 2

2 2 21

4

1

6+ = + + =: 0 q q&

138



But for simple harmonic motion, &q w q

q w q

w

m n m

m n m

n

mgbL

hmL

bg

hL

=

=

=

1

4

1

63

2

2 2 2

2

( )

Period of vibration. t pwn

n

= 2 t pn

hL

bg= 2

2

3 �

139


PROBLEM 19.94

A 2-kg uniform rod ABC is supported by a pin at B and is attached to a spring at C. It is connected at A to a 2-kg block DE, which is attached to a spring and can roll without friction. Knowing that each spring can act in tension or compression, determine the frequency of small oscillations of the system when the rod is rotated through a small angle and released.

SOLUTION

Positionj T I m C m x

I m l

R m R m E m

R R

12 2

12

2

1

2

1

2

1

21

121

122

= + +

=

=

( ) ( ) ( )

(

& & &q q

kg)(00.9 m)

kg m

kg(0.15 m) kg m

2

2

2 2

= ◊

= = ◊

=

0 135

2 0 0225

0 6

2

1

.

.

.

m C

x

l

q mm x m

x m x m

T

E m m

E m m

( ) ( )

. ( ) .

& && & &

12 2

1 12 2 2

1

2

0 6 0 72

=

= =

=

kg)(0.6q

q q11

20 135 0 0225 0 72

1

20 9

0

2

1

1

[ . . . ]

( . )

+ +

=

=

&

&

q

q

m

T

V

m

140



Positionk T

V k x k x m gc

x x

m m R m

m m

2

2 12

22

1 2

0

1

2

1

21

0 6 0 3

=

= + - -

= =

( ) ( ) ( cos )

( ) . .

q

q qmm

m m

m

m

V

1 22

1

250 50

2 2

22

- = ª

= +

cos sin

[( (

q q q

q N/m)(0.6 m) N/m)(02 ..3 m)

kg)(9.81 m/s

2

2

q

q

q

m

m

V

2

2

22

2 0 15

1

218 4 5 2 943

-

= + -

( )( . )

[ . . ] mm

V m221

219 55= ( . )q

Conservation of energy. T V T V m m1 1 2 22 20 9 0 0

1

219 55+ = + + = +:

1

2( . ) ( . )&q q

Simple harmonic motion. &q w q w q q

w

m n m n m m

n

= =

= = -

2 2 2 2 2

2

0 9 19 55

19 55

0 921 73

( . )( ) ( . )

.

.. s 2

Frequency of oscillations. fnn= =

wp p2

21 73

2

. fn = 0 742. Hz �

141


PROBLEM 19.95

A 750-g uniform arm ABC is supported by a pin at B and is attached to a spring at A. It is connected at C to a 1.5-kg mass m which is attached to a spring. Knowing that each spring can act in tension or compression, determine the frequency of small oscillations of the system when the weight is given a small vertical displacement and released.

SOLUTION

Positionj k

kC

A

==

300

400

N/m

N/m

T I ml

I ml

BC m BCBC

m

BA m BAAB

12

22

2

1

2

1

2 2

1

2

1

2 2

= + ÊËÁ

ˆ¯

+ + ÊËÁ

( )& &

&

q q

q ˆ¯

+

+ ÊËÁ

ˆ¯

= + =

22 2

22 2

1

2

2

1

12

1

4

1

3

& &qm m

BC BBC

BC BC BC BC

my

I ml

m l m l mm l

I ml

m l m l m l

BC BC

BA BAAB

AB AB AB AB AB AB

2

22 2 2

2

1

12

1

4

1

3+ Ê

ËÁˆ¯

= + =

142



m m

m m

m l

BC ABC

BA ABC

BC BC

= =

= =

=

3

50 45

2

50 3

1

3

1

30 45 0 32

.

.

( . )( .

kg

kg

kg m)) kg m

) m kg m

2 2= ◊

= = ◊

=

0 0135

1

3

1

30 3 0 2 0 0042 2 2

.

( . ( . ) .m l

y l

BA BA

m B& CC m

m BC m m

m

my ml

T

&& & &

&

q

q q

q

2 2 2 2

2

1

1 5 0 3

0 135

1

20

= =

=

=

( . )( .

.

[ .

kg m)2

00135 0 004 0 135

0 15252

1

2

1

2

2

2

12

+ +

=

= +

. . ]

.

( ) ( )

&

&q

q

d d

m

C C A

m

V k kST ST AA2

Positionk T

V Wl W lW

k l

BC m BC BC mBA

l m

C BC m

AB

2

2

2

0

2 1

1

2

=

= + -Ê

ËÁ

ˆ

¯˜ -

+ +

q q q

q

( cos )

( (( ) ) ( ( ) )d q dST STC A AB m Ak l2 21

2+ +

when the system is in equilibrium ( ).q = 0

SM Wl Wl

k l k lB BC BCBC

C C BC A A AB= = + - -02

( ) ( )d dST ST (1)

1 22 2

2

22

2

- - =

= + ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

-

cos sinqq q

q

mm m

BC BCBC

m BAV Wl Wl

Wll

k l

k l

AB mC BC m

C C BC m

2 2

1

2

1

2

22 2Ê

ËÁˆ¯

Ê

ËÁˆ

¯

È

ÎÍÍ

˘

˚˙˙

+

- +

qq

d q( )ST kk

k l k l k

C C

A AB m A A AB m A A

( )

( ) ( )

d

q d q d

ST

ST ST

2

2 2 21

2

1

2+ - +

143



Taking Equation (1) into account,

V Wl

k l

k k l

BAAB m

C BC m

C C A AB

2

22 2

2 2

2 2

1

2

1

2

1

2

= - ÊËÁ

ˆ¯

+È

ÎÍÍ

+ +

qq

d( )ST qq dm A Ak

V

2 2

22

1

2

1

20 3 9 81

0 2

2300 0 3

+ ˘˚

= - ÊËÁ

ˆ¯

+

( )

( . )( . ).

( )( . )

ST

++ÈÎÍ

˘˚

+ +

= - +

400 0 2

1

2

1

21

20 2943

2 2

2 2

2

( . )

( ) ( )

[ .

q

d d

m

C C A Ak k

V

ST ST

227 161

2

1

21

242 7057

1

2

2 2 2

22

+ + +

= +

] ( ) ( )

[ . ] (

q d d

q d

m C C A A

m C

k k

V k

ST ST

SST ST) ( )C A Ak2 21

2+ d


T V T V k km C C A A1 1 2 22 2 21

20 1525

1

2

1

2

01

242

+ = + + +

= +

: ST ST( . ) ( ) ( )

(

&q d d

.. ) ( ) ( )70571

2

1

22 2 2q d dm C C A Ak k+ +ST ST

Simple harmonic motion. &q w q

w q q

w

m n m

n m m

n

=

=

= =

0 1525 42 7057

42 7057

0 1525280 037

2 2 2

2

. .

.

.. (raad/s)2

Frequency. fnn= =

wp p2

280 037

2

. fn = 2 66. Hz �

144


PROBLEM 19.96*

Two uniform rods AB and BC, each of mass m and length l, are pinned together at A and are pin-connected to small rollers at B and C. A spring of constant k is attached to the pins at B and C, and the system is observed to be in equilibrium when each rod forms an angle b with the vertical. Determine the period of small oscillations when Point A is given a small downward deflection and released.

SOLUTION

x l

x l

xl

xl

yl

yl

C

C

G

G

G

G

==

=

= -

=

=

sin

cos

cos

sin

sin

bd bdb

b

d bdb

b

d

2

2

2

2

& &

ccos

cos

bdb

d bdb& &yl

G =2

Positionj T I m x y

mlml

m m C m G12 2 2

22

2

21

2

1

2

1

12 4

= + +ÈÎÍ

˘˚

= +

( ) ( ) ( )

(sin

db d d& & &

bb b db

db

+È

ÎÍ

˘

˚˙

=

=

cos )2 2

2

1

1

30

m

mml

V

&

Positionk T

V k x

l

l

m

m n m

2

22

2 2

2 2

0

1

22

1

24

2

=

=

=

=

=

( )

( cos )

cos

d

b

b

db w db&

145


PROBLEM 19.96* (Continued)

T V T V1 1 2 2+ = +

Conservation of energy. 1

30 0 2

6

2 2 2 2 2 2

2 2

ml kl

k

m

n n n

n

w db bdb

w b

+ = +

=

cos

cos


p

bn

n km

= =( )

2 2

6 cos2 t p

bnm

k= 2

6cos �

146


PROBLEM 19.97*

As a submerged body moves through a fluid, the particles of the fluid flow around the body and thus acquire kinetic energy. In the case of a sphere moving in an ideal fluid, the total kinetic energy acquired by the fluid is 1

42rVv , where r is the mass

density of the fluid, V is the volume of the sphere, and v is the velocity of the sphere. Consider a 500-g hollow spherical shell of radius 80 mm, which is held submerged in a tank of water by a spring of constant 500 N/m. (a) Neglecting fluid friction, determine the period of vibration of the shell when it is displaced vertically and then released. (b) Solve Part a, assuming that the tank is accelerated upward at the constant rate of 8 m/s2.

SOLUTION

This is not a damped vibration. However, the kinetic energy of the fluid must be included.

(a) Positionk T

V kxm

2

22

0

1

2

=

=

Positionj T T T m v Vv

V

s m m12 2

1

1

2

1

40

= + = +

=

spere fluid r


T V T V m v Vv kxs m m m1 1 2 22 2 21

40 0

1

2+ = + + + = +:

1

2r

v l x

m V x kx

k

m V

m m m n

s m n m

n

s

n

= =

+ÊËÁ

ˆ¯

=

=+

=

w

r w

wr

w

1

2

1

2

1

2

12

500

2 2 2

2

2 NN/m

kg)

kgm

( .

( . )

0 512

1

2

1

2

1000 4

30 08

33

+ ÊËÁ

ˆ¯

= ÊËÁ

ˆ¯

ÊË

r

r p

V

Vm

ÁÁˆ¯

=

= -

1 0723

3182

. kg

s 2wn


pn

n

= =2 2

318

(b) Acceleration does not change mass. t n = 0 352. s �

147


PROBLEM 19.98*

A thin plate of length l rests on a half cylinder of radius r. Derive an expression for the period of small oscillations of the plate.

SOLUTION

( sin )sin

( cos )

r r

r r

m m m

mm

q q q

qq

ª

- ª

2

2

12

Positionj (Maximum deflection) T

V Wy

mgr

m

m

1

1

2

0

2

==

=q

Positionk ( ):q = 0 T I

ml

T ml

m

m

m n m

n

22

2 2

22 2

1

21

2

1

12

1

2

1

12

=

= ÊËÁ

ˆ¯

=

= ÊËÁ

ˆ¯

&

&

&

q

q

q w q

w qmm

T V T V

2

1 1 2 2+ = +

01

2

1

2

1

12

12

22

12

2 2 2 2

22

2

+ = ÊËÁ

ˆ¯

=

= =

mgr ml

gr

l

l

gr

m n m

n

nn

q w q

w

t pw

p t pn

l

gr=

3 �

148


PROBLEM 19.99

A 50-kg block is attached to a spring of constant k = 20 kN/m and can move without friction in a vertical slot as shown. It is acted upon by a periodic force of magnitude P P tm f= sin ,w where w f = 18 rad/s. Knowing that the amplitude of the motion is 3 mm, determine the value of Pm .

SOLUTION

Equation of motion. mx kx P tm f&&+ = sinw

The steady state response is xm

Pkm

f

n

=( )

- ( )1ww

2

or P kxm mf

n

= -ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

12w

w

Data: k

m

= ¥=

20 10

50

3 N/m

kg

w

ww

n

f

n

m

k

m

x

= = ¥ =

= =

= = ¥

20 10

5020

18

200 9

3

3

rad/s

rad/s

rad/s

mm 3

.

110 m3-

Pm = ¥ ¥ --( )( )[ ( . ) ]20 10 3 10 1 0 93 3 2 Pm = 11 40. N �

149


PROBLEM 19.100

A 5 kg collar can slide on a frictionless horizontal rod and is attached to a spring of constant 500 N/m. It is acted upon by a periodic force of magnitude P P tm f= sin ,w where Pm = 15 N. Determine the amplitude of the motion of the collar if (a) w f = 5 rad/s, (b) w f = 10 rad/s.

SOLUTION

Eq. (19.33): xm

Pkm

f

n

=( )

- ( )12w

w

P k

m

k

mP

k

m

n

m

= ==

= = =

= =

15 500

5

500

510

50 0

N, N/m

kg

rad/s

N

500 N/m

w

. 33 m

Amplitude of vibration.

(a) w f = 5 rad/s xm =-

=0 03

1 100 04

2

.

( ). m | | 0.04 m(in phase)xm = �

(b) w f = 10 rad/s w wf n mx= fi xm �

150


PROBLEM 19.101

A 5-kg collar can slide on a frictionless horizontal rod and is attached to a spring of constant k. It is acted upon by a periodic force of magnitude P P tm f= sin ,w where Pm = 10 N and w f = 5 rad/s. Determine the value of the spring constant k knowing that the motion of the collar has an amplitude of 150 mm and is (a) in phase with the applied force, (b) out of phase with the applied force.

SOLUTION

Eq. (19.33): xm

Pkm

f

n

=- ( )1

2ww

wnk

m2 =

xP

k m

kP

xm

mm

f

m

mf

=-

= +

w

w

2

2

Data: Pm = 10 N, m = 5 kg

w f

m

m

m

m

kP

x

P

x

=

= +

= +

5

5 5

125

2

rad/s

( )( )

(a) (In phase) xm = =150 0 15mm m.

k = +10

0 15125

. k = 191 7. N/m �

(b) (Out of phase) xm = - = -150 0 15mm m.

k =-

+10

0 15125

. k = 58 3. N/m �

151


PROBLEM 19.102

A collar of mass m which slides on a frictionless horizontal rod is attached to a spring of constant k and is acted upon by a periodic force of magnitude P P tm f= sin .w Determine the range of values of w f for which the amplitude of the vibration exceeds two times the static deflection caused by a constant force of magnitude Pm .

SOLUTION

Circular natural frequency. wnk

m=

For forced vibration, the equation of motion is

mx kx p tm f&&+ = +sin ( )w jThe amplitude of vibration is

xm

pkm

f

n

f

n

=- ( )

=-( )1 1

2 2ww

ww

dST

For w wf n� and xm = 2 dST , we have

21

11

22

2

dd w

www

STST or=

- ( )-

ÊËÁ

ˆ¯

=f

n

f

n

w wf nk

m2 21

2

1

2= = w f

k

m=

2 (1)

For k

m f n2� �w w , | |xm exceeds ST2d

For w wf n� and xm = 2dST , we have

21

11

22

2

2d

dw w

w

wSTST or=

- -- =

( )f n

f

n

w wf nk

m2 23

2

3

2= = w f

k

m= 3

2 (2)

For w wn fk

m� �

3

2 | | exceeds STxm 2d

From Eqs. (1) and (2), Range: k

m

k

mf2

3

2� �w �

152


PROBLEM 19.103

An 8-kg uniform disk of radius 200 mm is welded to a vertical shaft with a fixed end at B. The disk rotates through an angle of 3∞ when a static couple of magnitude 50 N m◊ is applied to it. If the disk is acted upon by a periodic torsional couple of magnitude T T tm f= sin ,w where Tm = ◊60 N m, determine the range of values of w f for which the amplitude of the vibration is less than the angle of rotation caused by a static couple of magnitude Tm .

SOLUTION

Mass moment of inertia: I mr= = = ◊1

2

1

28 0 200 0 162 2 2( )( . ) . kg m

Torsional spring constant: KT

T

K

=

= ◊= ∞ =

=

= ◊

q

q50

3 0 05236

0 50

0 05236954 93

N m

rad

N m/rad

.

.

..

Natural circular frequency: wnK

I= = =954 93

0 1677 254

.

.. rad/s

For forced vibration, qq

ww

ww

m

TKm

f

n

f

n

=- ( )

=- ( )1 1

2 2ST

For the amplitude | |qm to be less than qST, we must have w wf n� .

Then | | STSTq

qq

ww

mf

n

=

( ) -2

1�

ww

f

n

ÊËÁ

ˆ¯

-2

1 1�

ww

w wf

nf n

ÊËÁ

ˆ¯

=2

2 2 2 77 254� � ( )( . )

w f �109 3. rad/s �

153


PROBLEM 19.104

For the disk of Problem 19.103 determine the range of values of w f for which the amplitude of the vibration will be less than 3 5. .∞

SOLUTION

Mass moment of inertia: I mr= = = ◊1

2

1

28 0 200 0 162 2 2( )( . ) . kg m

Torsional spring constant: KT

T

K

=

= ◊ = ∞ =

= = ◊

qq50 3 0 05236

50

0 05236954 93

N m rad

N m/rad

, .

..

Natural circular frequency: wnK

I= = =954 93

0 1677 254

.

.. rad/s

For forced vibration, qq

q

ww

ww

m

TK

m

m

m

f

n

f

n

T

T

K

=- ( )

=- ( )

= ◊

= = =

1 1

60

60

954 930 062

2 2ST

ST

N m

.. 8832 rad

| | rad STq qm � �3 5 0 061087. .∞ =

For the amplitude | |qm to be less than qST , we must have w wf n� .

| | STqq

ww

ww

mf

n

f

n

=

( ) -=

( ) -2 2

1

0 062832

10 061087

..�

ww

f

n

ÊËÁ

ˆ¯

- =2

10 062832

0 0610871 02857�

.

..

ww

w wf

nf n

ÊËÁ

ˆ¯

2

2 02857 2 02857� �. . w f �110 0. rad/s �

154


PROBLEM 19.105

An 8-kg block A slides in a vertical frictionless slot and is connected to a moving support B by means of a spring AB of constant k = 1.6 kN/m. Knowing that the displacement of the support is d d w= m f tsin , where dm = 150 mm, determine the range of values of w f for which the amplitude of the fluctuating force exerted by the spring on the block is less than 120 N.

SOLUTION

Natural circular frequency: wnk

m= = ¥ =1 6 10

814 1421

3.. rad/s

Eq. (19.33' ): xmm

f

n

=- ( )

dww1

2

Spring force: F K x km m m mf

n

= - - = - -- ( )

È

Î

ÍÍÍÍ

˘

˚

˙˙˙˙

( )d dww

11

12

=( )- ( )

k m

f

n

f

n

d

ww

ww

2

21

= ¥( )- ( )

=( )- ( )

( . )( . )1 6 10 0 1501

2401

3

2

2

2

2

ww

ww

ww

ww

f

n

f

n

f

n

f

n

Limit on spring force: | | NFm �120

2401

1201

1

2

2

2

2

2

ww

ww

ww

ww

f

n

f

n

f

n

f

n

( )- ( )

( )- ( )

� �or

155



In phase motion.

ww

ww

f

n

f

n

( )- ( )

2

21

1

2�

ww

ww

ww

ww

w w

f

n

f

n

f

n

f

n

f n

ÊËÁ

ˆ¯

-ÊËÁ

ˆ¯

ÊËÁ

ˆ¯

2 2

2

1

2

1

2

3

2

1

2

1

3

1

3

�

� �

� w f � 8 16. rad/s �

Out of phase motion.

ww

ww

ww

ww

f

n

f

n

f

n

f

n

( )( ) -

ÊËÁ

ˆ¯

ÊËÁ

ˆ¯

-

2

2

2 2

1

1

2

1

2

1

2

�

�

ww

f

n

ÊËÁ

ˆ¯

-2

1

2� No solution for w f .

156


PROBLEM 19.106

Rod AB is rigidly attached to the frame of a motor running at a constant speed. When a collar of mass m is placed on the spring, it is observed to vibrate with an amplitude of 15 mm. When two collars, each of mass m, are placed on the spring, the amplitude is observed to be 18 mm. What amplitude of vibration should be expected when three collars, each of mass m, are placed on the spring? (Obtain two answers.)

SOLUTION

(a) One collar: ( ) ( )xk

mm n1 1215= =mm w

(b) Two collars: ( ) ( ) ( )xk

mm n n2 22

1218

2

1

2= = =mm w w

ww

wwn n

ÊËÁ

ˆ¯

=ÊËÁ

ˆ¯2 1

2

(c) Three collars:

( ) ( ) ( ) ,xk

mm n nn n

3 32

12

3 13

1

33= = =

ÊËÁ

ˆ¯

=ÊËÁ

ˆ¯

unknown, w w ww

ww

We also note that the amplitude dm of the displacement of the base remains constant.

Referring to Section 19.7, Figure 19.9, we note that, since ( ) ( ) ,( ) ( )x xm mn n

2 12 1

� � and ww

ww we must

have ww( )n 1

1� and ( ) .xm 1 0� However, ww( )n 2

may be either � �1 or 1, with ( )xm 2 being correspondingly

either � �0 0or .

1. Assuming ( ) :xm 2 0�

For one collar,

( )xmm

n

m

n

1

1

2

1

2

1

15

1

=

-ÊËÁ

ˆ¯

+ =

-ÊËÁ

ˆ¯

d

ww

d

ww

j mm (1)

For two collars,

( )xmm

n

m

n

2

2

2

1

2

1

18

1 2

=

-ÊËÁ

ˆ¯

+ =

-ÊËÁ

ˆ¯

d

ww

d

ww

j mm (2)

157



Dividing Eq. (2) by Eq. (1), member by member:

1 2

1

1 2

1

71

2

1

21

2

. ;=-

ÊËÁ

ˆ¯

-ÊËÁ

ˆ¯

ÊËÁ

ˆ¯

=

ww

ww

ww

n

n

n

we find

Substituting into Eq. (1), dm = -ÊËÁ

ˆ¯

=( )15 11

7

90

7mm mm

For three collars,

( )xmm

n

3

1

2

1 3

907

1 317

90

4=

-ÊËÁ

ˆ¯

=

ÊËÁ

ˆ¯

- ÊËÁ

ˆ¯

=d

ww

mmmm, ( ) .xm 3 22 5= mm �

2. Assuming ( ) :xm 2 0�

For two collars, we have - =

-ÊËÁ

ˆ¯

18

1 21

2mm

d

ww

m

n

(3)

Dividing Eq. (3) by Eq. (1), member by member:

- =-

ÊËÁ

ˆ¯

-ÊËÁ

ˆ¯

- +ÊËÁ

ˆ¯

= -Ê

1 2

1

1 2

1 2 2 4 1

1

2

1

2

1

2

.

. .

ww

ww

ww

ww

n

n

n nËËÁˆ¯

ÊËÁ

ˆ¯

= =

1

2

1

22 2

3 4

1 1

1 7

wwn

.

.

.

.

Substitute into Eq. (1), dm = -ÊËÁ

ˆ¯

=( ).

. .15 1

1 1

1 7

9

1 7mm mm

158



For three collars,

( ).

.

..

xmm

n

3

1

2

1 3

91 7

1 31 11 7

9

1 6=

-ÊËÁ

ˆ¯

=

ÊËÁ

ˆ¯

- ÊËÁ

ˆ¯

=-

d

ww

mm, ( ) .xm 3 5 63= - mm �

( )out of phase

Points corresponding to the two solutions are indicated below:

159


PROBLEM 19.107

A cantilever beam AB supports a block which causes a static deflection of 50 mm at B. Assuming that the support at A undergoes a vertical periodic displacement d d w= m f tsin , where dm = 12 mm, determine the range of values of w f for which the amplitude of the motion of the block will be less than 25 mm. Neglect the weight of the beam and assume that the block does not leave the beam.

SOLUTION

For the static condition. mg k= dST

Natural circular frequency. wd

d

w

n

n

k

m

g

g

= =

= = =

= =

ST

2STm/s mm m

rad/s

9 81 50 0 05

9 81

0 0514 007

. , .

.

..

From Eqs. (19.31 and 19.33' ): ( )xm Bm

f

n

=- ( )

dww1

2

Conditions: | | m mxm B m� 0 025 0 012. .d =

In phase motion. 0 012

10 025

10 012

0 025

2

2

..

.

.

- ( )-

ÊËÁ

ˆ¯

ww

ww

f

n

f

n

�

�

0 52 0 522

. .� �ww

w wf

nf n

ÊËÁ

ˆ¯

w f �10 10. rad/s �

Out of phase motion. 0 012

10 025

2

..

ww

f

n( ) -�

ww

ww

f

n

f

n( ) - Ê

ËÁˆ¯

ÊËÁ

ˆ¯

-2 2

10 012

0 0251 0 48� �

.

..

ww

w wf

nf n

ÊËÁ

ˆ¯

2

1 48 1 48� �. . w f �17 04. rad/s �

160


PROBLEM 19.108

A variable-speed motor is rigidly attached to a beam BC. When the speed of the motor is less than 600 rpm or more than 1200 rpm, a small object placed at A is observed to remain in contact with the beam. For speeds between 600 and 1200 rpm the object is observed to “dance” and actually to lose contact with the beam. Determine the speed at which resonance will occur.

SOLUTION

Let m be the unbalanced mass and r the eccentricity of the unbalanced mass. The vertical force exerted on the beam due to the rotating unbalanced mass is

P mr t P tf f m f= =w w w2 sin sin

Then from Eq. 19.33, xm

Pk

mr

km

f

n

f

f

n

=- ( )

=- ( )1 1

2 2

2

ww

w

ww

For simple harmonic motion, the acceleration is

a xm f m

mr

kf

f

n

= - =- ( )

ww

ww

22

4

1

When the object loses contact with the beam is | |.am Acceleration is greater than g. Let w1 600 62 832= =rpm rad/s..

| |aU

m

mrk

mr Uk

n

n

1 2 2

14

1

4 4

1 1=

- ( )=

-

w

ww

w

(1)

where Un

=ww

1 .

Let w w2 11200 125 664 2= = =rpm rad/s.

| |aU

m

mrk

mr Uk

n

n

2 2

2

2

24

2

4 4

1 4 1=

( ) -=

-

w

ww

w ( )

(2)

161



Dividing Eq. (1) by Eq. (2),

14 1

16 116 16 4 1

20 1717

20

17

20

20

1

2

22 2

2

1

= --

- = -

= =

= =

U

UU U

U U

nn

( )or

ww

w77

1 084651 1w w= .

wn = ( . )( )1 08465 600 rpm wn = 651 rpm �

162


PROBLEM 19.109

An 8-kg block A slides in a vertical frictionless slot and is connected to a moving support B by means of a spring AB of constant k = 120 N/m. Knowing that the acceleration of the support is a a tm f= sin ,w where am = 1 5. m/s2

and w f = 5 rad/s, determine (a) the maximum displacement of block A, (b) the amplitude of the fluctuating force exerted by the spring on the block.

SOLUTION

(a) Support motion.

a a t

at

a

m f

m

ff

mm

f

= =

= -Ê

ËÁ

ˆ

¯˜

=-

= -

&&d w

dw

w

dw

sin

sin

.

(

2

2

21 5

5

m/s

rad/ssm

).

20 06= -

From Equation (19.31 and 19 33. ):¢

xk

mmm

nf

n

=-

= = =d

www

1

120152

2

2 2N/m

8 kgrad/s( )

xm = --

=0 06

10 09

2515

.

( ). m xm = 90 0. mm �

(b) x is out of phase with d for w f = 5 rad/s.

Thus,

F k xm m m= + = +( ) ( . . )d 120 0 09 0 06N/m m m

= 18 N Fm = 18 00. N �

163


PROBLEM 19.110

A 20 g ball is connected to a paddle by means of an elastic cord AB of constant k = 7 5. N/m. Knowing that the paddle is moved vertically according to the relation d d w= m f tsin , where dm = 200 mm, determine the maximum allowable circular frequency w f if the cord is not to become slack.

SOLUTION

SF ma k x mx xk

mx= - = + Ê

ËÁˆ¯

= ( )d d&& &&

From Equation (19.31 and 19 33. ):¢ xmm

f

n

=-Ê

Ëˆ¯

dww

12

2

Data: m = =20 0 02kg kg.

k = 7 5. N/m dm = =200 0 2mm m.

wnk

m=

=

=

7 5

0 0219 3649

.

.. rad/s

The cord becomes slack if xm m- d exceeds dST, where

dST m= = ¥ =m

k

0 02 9 81

7 50 02616

. .

..

164



Then 0 2

10 2 0 02616

2

.. .

- ( )-

ww

f

n

�

0 2 0 2 0 2 0 02616 0 02616

0 22616

2 2

. . . . .

.

- +ÊËÁ

ˆ¯

-ÊËÁ

ˆ¯

ww

ww

w

f

n

f

n

f

�

ww

ww

n

f

n

ÊËÁ

ˆ¯

=

2

0 02616

0 02616

0 226160 3401

�

�

.

.

..

Maximum allowable circular frequency.

w f � ( . )( . )0 3401 19 3649 rad/s w f � 6 59. rad/s �

165


PROBLEM 19.111

A simple pendulum of length l is suspended from a collar C, which is forced to move horizontally according to the relation x tC m f= d wsin . Determine the range of values of w f for which the amplitude of the motion of the bob is less than dm . (Assume that dm is small compared with the length l of the pendulum.)

SOLUTION

Geometry. x x l

x x

l

C

C

= +

=-

sin

sin

q

q

+ SF ma T mg T mgy y= ª - = ª0 0: cosq

+ SF max x= : - =T mxsinq &&

mxmg x x

l

xg

lx

g

lx

C

C

&&

&&

+-

=

+ =

( )0

Using the given motion of xC,

&&x g

lx

g

ltm f+ = d wsin

Circular natural frequency. wng

l=

&&x x tn n m f+ =w w d w2 2 sin

The steady state response is xmm

f

n

=- ( )

dww1

2

xmm

mf

n

22

2 22

1

=

- ( )ÈÎÍ

˘˚

dd

ww

�

Consider xm m2 2= d .

166



Then 1 1 2 12 2 2 4

-ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

= -ÊËÁ

ˆ¯

+ÊËÁ

ˆ¯

=

ÊËÁ

ww

ww

ww

ww

f

n

f

n

f

n

f

n

ˆ¯

=ÊËÁ

ˆ¯

=

ÊËÁ

ˆ¯

= =

2 2

0 2

0 2

and

and

ww

ww

ww

f

n

f

n

f

n

For 0 2� � �ww

df

nm mx, | |

For ww

df

nm nx� �2, | |

Then w wf ng

l� 2

2= w fg

l�

2 �

167


PROBLEM 19.112

The 1.2-kg bob of a simple pendulum of length l = 600 mm is suspended from a 1.4-kg collar C. The collar is forced to move according to the relation xC = d wm f tsin , with an amplitude dm = 10 mm and a frequency ff = 0.5 Hz. Determine (a) the amplitude of the motion of the bob, (b) the force that must be applied to collar C to maintain the motion.

SOLUTION

(a)

S

S

F ma

T mx

F T mg

x x

y

=- =

= - =sin

cos

qq

&&0

For small angles cos .q ª 1 Acceleration in the y direction is second order and is neglected.

T mg

mx mg

x x

l

mxmg

lx

g

lx

mg

lm t

g

l

c

c f

n

== -

=-

+ = =

=

&&

&&

sin

sin

sin

q

q

d w

w2

&&&x x m tn n f+ =w w d w2 2 sin

From Equation ( . ):19 33¢

xm

mf

n

=-

dww

12

2

So w p p p

w

f n

n

f

g

l

2 2 2 2 2 2

2 2

2 4 0 5

9 81

0 60016 35

= = =

= = =

-

-

( ) ( . )

. /

..

s

m s

ms

2

xm =¥-

=-10 10

10 02523

3

16 35

2

mm

p.

. xm = 25 2. mm �

168



(b)

a x m tc c f f= = -&& d w w2 sin

+ SF m ax c c=

F T m ac c- =sinq

From Part (a): T mgx x

lc= =

-, sinq

Thus, F mgx x

lm x

m x m x m x

m x t

cc c

n n c c c

n m f

= --È

ÎÍ˘˚

+

= - + +

= -

&&

&&w w

w w

2 2

2 sin ++ -

= - +

m t m tn m f c f m fw d w w d w2 2

1 2 16 35 0 02523 1 2

sin sin

[ ( . )( . )( . ) ( . )(( . )( ) ( . ) ( )]sin

. sin

16 35 10 10 1 4 10 10

0 437

3 2 3¥ - ¥= -

- -p pp

t

t

F t= -0 437. sin ( )p N �

169


PROBLEM 19.113

A motor of mass M is supported by springs with an equivalent spring constant k. The unbalance of its rotor is equivalent to a mass m located at a distance r from the axis of rotation. Show that when the angular velocity of the motor is w f , the amplitude xm of the motion of the motor is

xr

m

mM

f

n

f

n

=( )( )- ( )

ww

ww

2

21

where wnk

M= .

SOLUTION

Rotor Motor

+SF ma P t kx Mxm f= - =sinw &&Mx kx P t

xk

mx

P

Mt

k

M

m f

mf

n

&&

&&

+ =

+ =

=

sin

sin

w

w

w2

From Equation (19.33):

xm

Pkm

f

n

=- ( )1

2ww

But P

k

mr

kk M

P

kr

m

M

m fn

m f

n

= =

= ÊËÁ

ˆ¯

ÊËÁ

ˆ¯

ww

ww

22

2

Thus, xr

m

mM

f

n

f

n

=( )( )- ( )

ww

ww

2

21

Q.E.D. �

170


PROBLEM 19.114

As the rotational speed of a spring-supported 100-kg motor is increased, the amplitude of the vibration due to the unbalance of its 15-kg rotor first increases and then decreases. It is observed that as very high speeds are reached, the amplitude of the vibration approaches 3.3 mm. Determine the distance between the mass center of the rotor and its axis of rotation. (Hint: Use the formula derived in Problem 19.113.)

SOLUTION

Use the equation derived in Problem 19.113 (above).

xr r

m

mM

mM

f

n

f

nf

n

=( )( )- ( )

=( )

-( )

ww

ww

ww

2

2 11

12

For very high speeds, 1

2ww

f

n( ) ® 0 and xm ®

rm

M,

thus, 3 315

100. mm = Ê

ËÁˆ¯

r r = 22 mm �

171


PROBLEM 19.115

A motor of mass 200 kg is supported by springs having a total constant of 240 kN/m. The unbalance of the rotor is equivalent to a 30-g mass located 200 mm from the axis of rotation. Determine the range of allowable values of the motor speed if the amplitude of the vibration is not to exceed 1.5 mm.

SOLUTION

Let M m r= = =mass of motor, unbalance mass, eccentricity

M

m

r K

== =

= = = = ¥

200

30 0 03

200 0 2 240 240 103

kg

g kg

mm m kN/m N/m

.

.


Mrm

M

= =¥

=

=

= ¥ -

240 10

20034 641

0 2 0 03

200

0 3 10

3

5

N/mrad/s.

( . )( . )

. mm mm= 0 03.

From the derivation given in Problem 19.113,

xm

rmM

f

n

f

n

f

n

f

n

=( )( )

- ( )=

( )- ( )

ww

ww

ww

ww

2

2

2

21

0 03

1

.mm

In phase motion with mm| | .xm �1 5

0 03

11 5

0 03 1 5 1 5

2

2

2 2

..

. . .

ww

ww

ww

ww

f

n

f

n

f

n

f

n

( )- ( )ÊËÁ

ˆ¯

-ÊËÁ

ˆ¯

�

�

11 53 1 5

1 5

1 530 99015

2

. .

.

..

ww

ww

f

n

f

n

ÊËÁ

ˆ¯

=

�

�

w f � ( . )( . ) .0 99015 34 641 34 30= rad/s w f � 328 rpm �

172



Out of phase motion with mm| | .xm = 1 5

0 03

11 5

0 03 1 5 1 5

2

2

2 2

..

. . .

ww

ww

ww

ww

f

n

f

n

f

n

f

n

( )( ) -

ÊËÁ

ˆ¯

ÊËÁ

ˆ¯

-

�

�

11 5 1 472

. .�ww

f

n

ÊËÁ

ˆ¯

ww

f

n

�1 5

1 471 01015

.

..=

w f � ( . )( . ) .1 01015 34 641 34 992= rad/s w f � 334 rpm �

173


PROBLEM 19.116

As the rotational speed of a spring-supported motor is slowly increased from 300 to 500 rpm, the amplitude of the vibration due to the unbalance of its rotor is observed to increase continuously from 1.5 to 6 mm. Determine the speed at which resonance will occur.

SOLUTION

Let ¢m be the mass of rotor of the motor and m the total mass of the motor. Let e be the distance from the rotor axis to the mass center of the rotor. The magnitude of centrifugal force due to unbalance of the rotor is

P m em f= ¢ w2

and in rotation, the unbalanced force is

P t m e tm f f fsin sinw w w= ¢ 2

The equation of motion of the spring mounted motor is

mx kx m e t P tf f m f&&+ = ¢ =w w w2 sin sin

The steady state response is x

xm e

k

C

m

Pk

mf f

m

f

n

f

n

f

n

=- ( )

=¢

◊- ( )

=- ( )

1

1

1 1

2

2

2

2

2

ww

ww

ww

w w

Case 1. w w p pf

mx x

= = = =

= =

1

1

3002 300

6010

1 5

rpm rad/s

| | mm

( )

.

Case 2. w w p pf

mx x

1 2

2

5002 500

60

50

36

= = = =

= =

rpm rad/s

| | mm

( )

x

x

n

z

n

2

1

22

2

12

2

12 2

6

1 54

1

1

4 1

1

2

= = =+ ( )È

ÎÍ˘˚

+ ( )ÈÎÍ

˘˚

-

.

w

w

www

ww

ww

ÊÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

= -ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

2

22 1

2

1 1wwwn

174



Multiplying by wn2 and transposing terms,

( )

( )( . )(

4 3 0

3

4

3 986 96 274

12

22 2

12

22

2 12

22

12

22

w w w w w

ww w

w w

- - =

=-

=

n

n11 56

3947 84 2741 566729 3

. )

. .. (

-= rad/s)2

wn = 82 032. rad/s wn = 783 rpm �

175


PROBLEM 19.117

A100 kg motor is bolted to a light horizontal beam. The unbalance of its rotor is equivalent to a mass of 60 g located 100 mm from the axis of rotation. Knowing that resonance occurs at a motor speed of 400 rpm, determine the amplitude of the steady-state vibration at (a) 800 rpm, (b) 200 rpm, (c) 425 rpm.

SOLUTION

From Problem 19.113:

xr

m

mM

f

n

n

f

=( )( )- ( )

ww

ww

2

21

Resonance at 400 rpm means that wn = 400 rpm.

rm

MÊËÁ

ˆ¯

= ¥Ê

ËÁˆ

¯=

-( ) .100

60 10

1000 06

3

mm mm

(a) ww

f

n

ÊËÁ

ˆ¯

= ÊËÁ

ˆ¯

=2 2800

4004

xm =-

0 06 4

1 4

. ( ) xm = - 0 080. mm �

(b) ww

f

n

ÊËÁ

ˆ¯

= ÊËÁ

ˆ¯

=2 2200

400

1

4

xm =( )

-0 06

1

14

14

. xm = 0 020. mm �

(c) ww

f

n

ÊËÁ

ˆ¯

= ÊËÁ

ˆ¯

=2 2425

4001 1289.

xm =-

0 06 1 1289

1 1 1289

. ( . )

. xm = - 0 525. mm �

176


PROBLEM 19.118

A 180-kg motor is bolted to a light horizontal beam. The unbalance of its rotor is equivalent to a 28-g mass located 150 mm from the axis of rotation, and the static deflection of the beam due to the weight of the motor is 12 mm. The amplitude of the vibration due to the unbalance can be decreased by adding a plate to the base of the motor. If the amplitude of vibration is to be less than 60 mm for motor speeds above 300 rpm, determine the required mass of the plate.

SOLUTION

Before the plate is added, M m

r1

3180 28 10

150 0 150

= = ¥= =

-kg, kg

mm m.

Equivalent spring constant: kW M g

k

= =

=¥

= ¥-

1 1

33180 9 81

12 10147 15 10

d dST ST

N/m( )( . )

.

Let M2 be the mass of motor plus the plate.

Natural circular frequency. wnk

M=

2

Forcing frequency: w

ww

w

f

f

n

f M

k

M

= =

ÊËÁ

ˆ¯

= =¥

300 31 416

31 416

147 15

22

22

rpm rad/s

2

.

( . )

. 1100 006707

3= . M2

From the derivation in Problem 19.113,

xm

rmM

f

n

f

n

=( )( )

- ( )2

2

21

ww

ww

For out of phase motion with xm = - ¥ -60 10 6 m,

- ¥ =ÈÎÍ

˘˚

--

¥ -

60 100 006707

1 0 0067076

0 150 28 102

2

3

2

( . )( ) ( . )

.

M M

M

- ¥ + ¥ = ¥

¥ =

- - -

-

60 10 60 10 0 006707 28 170 10

402 49 10 88

6 62

6

92

( )( . ) .

.

M

M ..

.

170 10

219 10

6

2

¥=

-

M kg

Added mass: DM M M= - = -2 1 219 10 180. DM = 39 1. kg �

177


PROBLEM 19.119

The unbalance of the rotor of a 200-kg motor is equivalent to a 100-g mass located 150 mm from the axis of rotation. In order to limit to 1-N the amplitude of the fluctuating force exerted on the foundation when the motor is run at speeds of 100 rpm and above, a pad is to be placed between the motor and the foundation. Determine (a) the maximum allowable spring constant k of the pad, (b) the corresponding amplitude of the fluctuating force exerted on the foundation when the motor is run at 200 rpm.

SOLUTION

Mass of motor. M = 200 kg

Unbalance mass. m = =100 0 1g kg.

Eccentricity. r = =150 mm 0.15 m

Equation of motion: Mx kx P t mr tm f f f&&+ = =sin sinw w w2

( )- + =

=-

M k x mr

xmr

k M

f m f

mf

f

w w

w

w

2 2

2

2

Transmitted force. F kxkmr

k Mm m

f

f

= =-

w

w

2

2

For out of phase motion, | |Fkmr

M km

f

f

=-

w

w

2

2 (1)

(a) Required value of k.

Solve Eq. (1) for k. |F M k kmrm f f| ( )w w2 2- =

k mr F F Mf m m f( | |) | |w w2 2+ =

kF M

mr F

m f

f m

=+

| |

| |

w

w

2

2

Data: | |Fm = 1 N w f = =100 10 472rpm rad/s.

k =+

=( )( )( . )

( . )( . )( . ).

1 200 10 472

0 1 0 15 10 472 18292 3

2

2N/m k = 8 292. kN/m �

178



(b) Force amplitude at 200 rpm. w f = 20 944. rad/s

From Eq. (1), | |( . )( . )( . )( . )

( )( . ) .Fm =

-8292 3 0 1 0 15 20 944

200 20 944 8292 3

2

2

| | .Fm = 0 687 N �

179


PROBLEM 19.120

A 180-kg motor is supported by springs of total constant 150 kN/m. The unbalance of the rotor is equivalent to a 28-g mass located 150 mm from the axis of rotation. Determine the range of speeds of the motor for which the amplitude of the fluctuating force exerted on the foundation is less than 20 N.

SOLUTION

Equation derived in Problem 19.113: xm

rmM

f

n

f

n

=( )( )

- ( )ww

ww

2

21

Force transmitted: F kxm m

kmrM

f

n

f

n

= =( )( )

- ( )ww

ww

2

21

Data: k

r

m g

M

kr

= = ¥= =

= = ¥=

-

150 150 10

150 0 150

28 28 10

180

3

3

kN/m N/m

mm m

kg

kg

.

mm

M

k

Mn

= ¥ ¥ =

= = ¥ =

-( )( . )( ).

.

150 10 0 150 28 10

1803 5

150 10

18028 8

3 3

3

N

w 668 rad/s

In phase motion with N| | .Fm � 20

3 5

120

3 5 20 20

23 5

2

2

2 2

.

.

.

ww

ww

ww

ww

w

f

n

f

n

f

n

f

n

( )- ( )ÊËÁ

ˆ¯

-ÊËÁ

ˆ¯

�

�

ff

nwÊËÁ

ˆ¯

2

20�

ww

w

f

n

f

�

�

20

23 5

20

23 528 868 26 63

.

.( . ) .= rad/s w f � 254 rpm �

180



Out of phase motion with N.| |Fm � 20

3 5

120

3 5 20 20

20 16

2

2

2 2

.

.

ww

ww

ww

ww

f

n

f

n

f

n

f

n

( )( ) -

ÊËÁ

ˆ¯

ÊËÁ

ˆ¯

-

�

�

� ..52w

wf

n

ÊËÁ

ˆ¯

ww

wf

nf� �

20

16 5

20

16 528 868 31 78

. .( . ) .= rad/s w f � 304 rpm �

181


PROBLEM 19.121

A vibrometer used to measure the amplitude of vibrations consists essentially of a box containing a mass-spring system with a known natural frequency of 120 Hz. The box is rigidly attached to a surface, which is moving according to the equation y tm f= d wsin . If the amplitude zm of the motion of the mass relative to the box is used as a measure of the amplitude dm of the vibration of the surface, determine (a) the percent error when the frequency of the vibration is 600 Hz, (b) the frequency at which the error is zero.

SOLUTION

x t

y t

z

z x y

m

f

m f

f

n

=Ê

ËÁÁ

ˆ

¯˜˜

=

=

= =

dw

d w

ww

12

2-

-

sin

sin

relative motion

ddd w

dd

ww

ww

w

mm

f

m m

m

f

n

f

n

f

t

z

1

1

11

2

2

2

2

2

--

--

È

ÎÍÍ

˘

˚˙˙

=È

ÎÍÍ

˘

˚˙˙

=

sin

wwww

n

f

n

2

2

21-

(a) zm

m

f

n

f

n

d

ww

ww

= =( )

( )= =

2

2

2

21 1

25

241 0417

600120

2

600120

2- -

. �

Error = 4 17. %

(b) zm

m

f

n

f

n

d

ww

ww

= =11

2

2

2

2-

1 2

2

2

2

2120 84 853

2

2=

= = =

w

wf

n

f ff n ( ) . Hz fn = 84 9. Hz �

182


PROBLEM 19.122

A certain accelerometer consists essentially of a box containing a mass-spring system with a known natural frequency of 2200 Hz. The box is rigidly attached to a surface, which is moving according to the equation y tm f= d wsin . If the amplitude zm of the motion of the mass relative to the box times a scale factorwn

2 is used as a measure of the maximum acceleration am m f= d w2 of the vibrating surface, determine the percent error when the frequency of the vibration is 600 Hz.

SOLUTION

x t

y t

z

z x y

m

f

m f

f

n

=Ê

ËÁÁ

ˆ

¯˜˜

=

=

= =

dw

d w

ww

12

2-

-

sin

sin

relative motion

ddd w

dd

ww

ww

w

mm

f

m m

m

f

n

f

n

f

t

z

1

1

11

2

2

2

2

2

--

--

È

ÎÍÍ

˘

˚˙˙

=È

ÎÍÍ

˘

˚˙˙

=

sin

wwww

n

f

n

2

2

21-

The actual acceleration is am f m= -w d2

The measurement is proportional to zm nw2.

Then z

a

zm n

m

m

m

n

f

f

n

wd

ww

ww

2 2

2

6002200

2

1

1

1

1

=Ê

ËÁ

ˆ

¯˜

=

( )=

( )

-

-

= 1 0804. Error = 8 04. % �

183


PROBLEM 19.123

Figures (1) and (2) show how springs can be used to support a block in two different situations. In Figure (1), they help decrease the amplitude of the fluctuating force transmitted by the block to the foundation. In Figure (2), they help decrease the amplitude of the fluctuating displacement transmitted by the foundation to the block. The ratio of the transmitted force to the impressed force or the ratio of the transmitted displacement to the impressed displacement is called the transmissibility. Derive an equation for the transmissibility for each situation. Give your answer in terms of the ratio w wf n/ of the frequency wf of the impressed force or impressed displacement to the natural frequency wn of the spring-mass system. Show that in order to cause any reduction in transmissibility, the ratio w wf n/ must be greater than 2.

SOLUTION

(1) From Equation (19.33): xm

Pkm

f

n

=

( )12

- ww

Force transmitted: ( )P kx kT m m

Pkm

f

n

= = ( )È

Î

ÍÍÍ

˘

˚

˙˙˙1

2- w

w

Thus, Transmissibility = =

( )( )P

PT m

m f

n

1

12

- ww

�

(2) From Equation ( . ):19 33¢

Displacement transmitted: xmm

f

n

=

( )d

ww1

2-

Transmissibility = =

( )xm

m f

n

d ww

1

12

- �

For ( )PP

xT m

m

m

mor to be less than 1,d

1

1

12

- ww

f

n( )�

1 12

� - ww

f

n( )

ww

f

n

ÊËÁ

ˆ¯

2

2� ww

f

n

� 2 Q.E.D. �

184


PROBLEM 19.124

A 30-kg disk is attached with an eccentricity e = 0 15. mm to the midpoint of a vertical shaft AB, which revolves at a constant angular velocity w f . Knowing that the spring constant k for horizontal movement of the disk is 650 kN/m, determine (a) the angular velocity w f at which resonance will occur, (b) the deflection r of the shaft when w f = 1200 rpm.

SOLUTION

G describes a circle about the axis AB of radius r e+ .

Thus, a r en f= +( )w2

Deflection of the shaft is F kr=

Thus, F ma

kr m r e

k

mm

k

krk

r e

re

n

f

nn

nf

f

n

f

=

= +

= =

= +

=

( )

( )

w

ww

ww

www

2

22

22

2

2

2

1-wwn

2

185



(a) Resonance occurs when w wf n r= , i.e., Æ

wnk

m=

=

==

( )

.

.

650 1000

30147 196

1405 6

¥

rad/s

rpm w wn f= = 1406 rpm �

(b) r =( )

( )( . ) .

.

0 15

1

12001405 6

2

12001405 6

2

mm

-

= 0 02964. mm r = 0 403. mm �

186


PROBLEM 19.125

A small trailer and its load have a total mass of 250 kg. The trailer is supported by two springs, each of constant 10 kN/m, and is pulled over a road, the surface of which can be approximated by a sine curve with an amplitude of 40 mm and a wavelength of 5 m (i.e., the distance between successive crests is 5 m and the vertical distance from crest to trough is 80 mm). Determine (a) the speed at which resonance will occur, (b) the amplitude of the vibration of the trailer at a speed of 50 km/h.

SOLUTION

Total spring constant k = ¥

= ¥

2 10 10

20 10

3

3

( )N/m

N/m

(a) w

l

d

dl

n

m

m

k

m

yx

x

23

2

3

20 1080

5

40 40 10

= =¥

=

=

= = ¥

=

N/m

250 kg s

m

mm m

-

-

sin

==

= ÊËÁ

ˆ¯

= =

vt

yv

tvm fd

lt l w p

tsin , ,and

2

so y tv v

f f= ¥ = =40 102 2

53- sinw w p

lp

where

From Equation ( . ):19 33¢ xmm

f

n

=ÊË

ˆ¯

dww

12

2-

Resonance: w p wf nv= = =2

580 1s- ,

v = 7 1176. m/s v = 25 6. km/h �

(b) Amplitude at v

f

f

= =

= =

=

50 13 8889

2 13 8889

517 4533

3042

km/h m/s

rad/s

.

( . ).

.

w p

w 660 2 s-

xm = ¥ = ¥40 10

114 246 10

3

304 6280

3-

-

--

.. m xm = -14 25. mm �

187


PROBLEM 19.126

Block A can move without friction in the slot as shown and is acted upon by a vertical periodic force of magnitude P P tm f= sin ,w where w f = 2 rad/s and Pm = 20 N. A spring of constant k is attached to the bottom of block A and to a 22-kg block B. Determine (a) the value of the constant k which will prevent a steady-state vibration of block A, (b) the corresponding amplitude of the vibration of block B.

SOLUTION

In steady state vibration, block A does not move and therefore, remains in its original equilibrium position.

Block A: +SF = 0

kx P tm f= - sinw (1)

Block B: SF m xB= &&

m x kx

x x t

k m

B

m n

n B

&&+ ==

=

0

2

sinw

w /

From Eq. (1): kx t P t

kx P

k

m

k m

m n m f

n f

m m

nB

B n

sin sinw w

w w

w

w

=

= =

=

=

=

-

-

2

2

rad/s

(a) Required spring constant. k = ( )( )22 2 2 k = 88 0. N/m �

(b) Corresponding amplitude of vibration of B.

k x P

xP

k

x

m m

mm

m

=

=

=

-

-

- 20

88

N

N/m xm = -0 227. N/m �

+

188


PROBLEM 19.127

Show that in the case of heavy damping ( ),c cc� a body never passes through its position of equilibrium O (a) if it is released with no initial velocity from an arbitrary position or (b) if it is started from O with an arbitrary initial velocity.

SOLUTION

Since c cc� , we use Equation (19.42), where

l l1 20 0� �,

x c e c et t= +1 21 2l l (1)

vdx

dtc e c et t= = +1 1 2 2

1 2l ll l (2)

(a) t x v= = =0 00, , :x

From Eqs. (1) and (2): x c c

c c0 1 2

1 1 2 20

= += +l l

Solving for c1 and c2 , c x

cx

x

12

2 10

21

2 10

=

=

ll l

ll

--

-

Substituting for c1 and c2 in Eq. (1), xx

e et t= ÈÎ ˘2

2 12 1

1 2

l ll ll l

--

For x = 0: when t , we must have

l lll

l l l l1 2

2

1

2 1 2 10e e et t t- -= = ( ) (3)

Recall that

l l1 20 0� �, . Choosing l l1 and 2 so that l l1 2 0� � , we have

0 1 02

11� � �

ll

l land 2 -

Thus a positive solution for t � 0 for Equation (3) cannot exist, since it would require that e raised to a positive power be less than 1, which is impossible. Thus, x is never 0.

�

The x t- curve for this case is as shown.

189



(b) t x v v= = =0 0 0, , : Equations (1) and (2) yield

0 1 2

0 1 1 2 2

= += +

c c

v c cl l

Solving for c c1 2 and , cv

cv

10

2 1

22

2 1

=

=

--

-

l l

l l

Substituting into Eq. (1), xv

e et t= 0

2 1

2 1

l ll l

--[ ]

For x = 0, t π a

e et tl l2 1=

For c cc� , ;l l1 2- thus, no solution can exist for t, and x is never 0.

�

The x t- curve for thiscase is as shown.

190


PROBLEM 19.128

Show that in the case of heavy damping ( ),c cc� a body released from an arbitrary position with an arbitrary initial velocity cannot pass more than once through its equilibrium position.

SOLUTION

Substitute the initial conditions, t x x v v= = =0 0 0, , in Equations (1) and (2) of Problem 19.127.

x c c v c c0 1 2 0 1 1 2 2= + = +l l

Solving for c c1 2 and , cv x

cv x

10 2 0

2 1

20 1 0

2 1

=

=

--

--

-

( )

( )

ll l

ll l

And substituting in Eq. (1) x v x e v x et t= ÈÎ ˘1

2 10 1 0 0 2 0

2 1

l ll ll l

-- - -( ) ( )

For x t= 0, : ( ) ( )

( )

( )( )

v x e v x e

ev x

v x

t

t t

t

0 1 0 0 2 0

0 2 0

0 1 0

2 1

2 1

1

- ---

-

l lll

l l

l l

=

=

=(( )

lnl l

ll2 1

0 2 0

0 1 0---

v x

v x

This defines one value of t only for x = 0, which will exist if the natural log is positive,

i.e., if v xv x

0 2 0

0 1 01-

-ll � . Assuming l l1 2 0� � ,

this occurs if v x0 1 0� l .

�

191


PROBLEM 19.129

In the case of light damping, the displacements x1, x2, x3, shown in Figure 19.11 may be assumed equal to the maximum displacements. Show that the ratio of any two successive maximum displacements xn and xn+1 is constant and that the natural logarithm of this ratio, called the logarithmic decrement, is

ln( )

( )

x

x

c c

c c

n

n

c

c+=

12

2

1

p /

/-

SOLUTION

For light damping,

Equation (19.46): x x e tcm t= +( )

0 02-

sin( )w f

At given maximum displacement, t t x x

t

x x e

n n

n

ntc

m n

= =+ =

= ( )

,

sin( )w0

0

1f- 2

At next maximum displacement, t t x x

t

x x e

n n

n

ntc

m n

= =+ =

=

+ +

+

+( ) +

1 1

0 1

1 0

1

1

,

sin( )w f- 2

But w w pp

w

D n D n

n nD

t t

t t

+

+

=

=

1

1

2

2

-

-

Ratio of successive displacements: x

x

x e

x e

e e

n

n

t

t

t t

cm n

cm n

cm n n

cm D

+

( ) +

=

= =

+

+

1

0

0

2

2 1

2 1 22

-

-

- - pw

Thus, lnx

x

c

mn

n D+=

1

pw

(1)

From Equations (19.45) and (19.41): w w

w

D nc

Dc

c

c

c

m

c

c

=ÊËÁ

ˆ¯

= ÊËÁ

ˆ¯

1

21

2

2

-

-

Thus, lnx

x

c

m

m

c cc

n

n c

c

+=

ÊËÁ

ˆ¯

12

2 1

1

p

-

lnx

xn

n

cc

cc

c

c

+=

( )( )1

2

2

1

p

-Q.E.D. �

192


PROBLEM 19.130

In practice, it is often difficult to determine the logarithmic decrement of a system with light damping defined in Problem 19.129 by measuring two successive maximum displacements. Show that the logarithmic decrement can also be expressed as ( ) ln ( ),1/ /k x xn n k+ where k is the number of cycles between readings of the maximum displacement.

SOLUTION

As in Problem 19.129, for maximum displacements xn and xn k+ at t tn n k n and t + + =, sin( )w0 1f

and sin( ) .wn n kt + + =f 1 x x e

x x e

nt

n kt

cm n

cm n k

=

=

( )

+( ) +

0

0

2

2

-

- ( )

Ratio of maximum displacements: x

x

x e

x een

n k

t

t

t tcm n

cm n k

cm n n k

+

( )( )

( )= =+

+0

0

2

2

2

-

-

- -

But w w pp

w

D n k D n

n n kD

t t k

t t k

+

+

=

=

-

-

( )2

2

Thus, x

x

c

m

kn

n k D+= +

ÊËÁ

ˆ¯2

2 pw

lnx

xk

c

mn

n k D+= p

w (2)

But from Problem 19.129, Equation (1):

log ln decrement = =+

x

x

c

mn

n D1

pw

Comparing with Equation (2), log ln decrement Q.E.D.=+

1

k

x

xn

n k

�

193


PROBLEM 19.131

In a system with light damping ( ),c cc� the period of vibration is commonly defined as the time interval t p wd d= 2 / corresponding to two successive points where the displacement-time curve touches one of the limiting curves shown in Figure 19.11. Show that the interval of time (a) between a maximum positive displacement and the following maximum negative displacement is 1

2 t d , (b) between two successive zero displacements is 1

2 t d , (c) between a maximum positive displacement and the following zero displacement is greater than 1

4 t d .

SOLUTION

Equation (19.46): x x e tcm t

D= +( )0

2-sin( )w f

(a) Maxima (positive or negative) when &x = 0:

&x xc

me t x e t

cm

cmt

D Dt

D= ÊËÁ

ˆ¯

+ + +( ) ( )0 02

2 2- - -

sin( ) cos( )w w wf f

Thus, zero velocities occur at times when

&x tm

cDD= + =0

2, tan( )or w

wf (1)

The time to the first zero velocity, t1, is

t

mc

D

D

1

1 2

=( )È

Î˘˚tan- -w

w

f (2)

The time to the next zero velocity where the displacement is negative is

¢ =( ) +È

Î˘˚t

mc

D

D

1

1 2tan- -w p

w

f (3)

194



Subtracting Eq. (2) from Eq. (3),

¢ = =◊

=t tD

D D1 1 2 2

- pw

p tp

tQ.E.D.

(b) Zero displacements occur when

sin( )wq t + =f 0 or at intervals of

w p p pDt n+ =f , 2

Thus, ( ) ( )

( )( )t

tD D

1 01 0

2=¢ =

pw

pw

- -f fand

Time between ¢ = ¢ = = =02

2 21 0 1 0sD

D Dt t( ) ( )- -p pw

ptp

tQ.E.D.

Plot of Equation (1)

(c) The first maxima occurs at 1: ( )wDt1 + f

The first zero occurs at ( ( ) )w pD t1 0 + =f

From the above plot, ( ( ) ) ( )w w pD Dt t1 0 1 2

+ +f f- �

or ( ) ( )t t t tD

D1 0 1 1 0 12 4

- -� �pw

tQ.E.D.

Similar proofs can be made for subsequent maximum and minimum.

195


PROBLEM 19.132

The block shown is depressed 30 mm from its equilibrium position and released. Knowing that after 10 cycles the maximum displacement of the block is 1.25 mm, determine (a) the damping factor c/c, (b) the value of the coefficient of viscous damping. (Hint: See Problems 19.129 and 19.130.)

SOLUTION

From Problems 19.130 and 19.129:

1 2

12k

x

x kn

n

cc

cc

c

c

ÊËÁ

ˆ¯ +

ÊËÁ

ˆ¯

=

( )ln

p

-

where number of cyclesk = = 10

(a) First maximum is x1 30= mm

Thus, n = 1 x

x

cc

cc

c

c

1

1

2

10

30

1 252 4

1

102 4 0 08755

2

1

+= =

=

=ÊËÁ

ˆ¯

..

ln . .

p

-

Damping factor. 12

0 08755

2 2 2

-c

c

c

cc c

ÊËÁ

ˆ¯

= ÊËÁ

ˆ¯

ÊËÁ

ˆ¯

p.

c

c

c

c

c

c

ÊËÁ

ˆ¯

ÊËÁ

ˆ¯

+È

ÎÍÍ

˘

˚˙˙

=

ÊËÁ

ˆ¯

=+( )

2 2

2

2

0 087551 1

1

5150 1

p.

== 0 0001941. c

cc

= 0 01393. �

196



(b) Critical damping coefficient. ck

mc = 2 m Eq 19.41)( .

or c km

c

c

c

c

c

=

== ◊

2

2 175

52 915

(

.

N/m)(4 kg)

N s/m

From Part (a), c

c

cc

=

=

0 01393

0 01393 52 915

.

( . )( . )

Coefficient of viscous damping. c = ◊0 737. N s/m �

197


PROBLEM 19.133

A loaded railroad car of mass 15,000 kg is rolling at a constant velocity v0 when it couples with a spring and dashpot bumper system (Figure 1). The recorded displacement-time curve of the loaded railroad car after coupling is as shown (Figure 2). Determine (a) the damping constant, (b) the spring constant. (Hint: Use the definition of logarithmic decrement given in Problem 19.129.)

SOLUTION

Mass of railroad car: m = 15000 kg

The differential equation of motion for the system is

mx cx kx&& &+ + = 0

For light damping, the solution is given by Eq. (19.44):

x e c t c tcm t

d d= +( )- 21 2( sin cos )w w

From the displacement versus time curve,

t

w pt

pd

dd

=

= = =

0 41

2 2

0 4115 325

.

..

s

rad/s

At the first peak, x t t1 112 5= =. . mm and

At the second peak, x t t d2 13= = + mm and t .

Forming the ratio xx

2

1,

x

x

e

ee

x

xe

cm d

cm

cm

c dm

t

t

d2

1

1

2

2 1

2 1

2

2

= =

=

( ) +

( )( )

-

--

( )tt

t

(1)

198



(a) Damping constant.

From Eq. (1): c d

m

x

x

t2

1

2

=ÊËÁ

ˆ¯

ln

cm x

xd

=

=

= ◊

2

2 15000

0 41

12 5

3

104 423

1

2

3

tln

( )( )

.ln

.

. ¥10 N s/m c = ◊104 4. kN s/m �

(b) Spring constant.

Equation for wd : wdk

m

c

m2

2

2= Ê

ËÁˆ¯

-

k mc

md= +

= +

=

w22

23 2

4

15000 15 325104 423 10

4 15000

3 70

( )( . )( . )

( )( )

.

¥

446 106¥ N/m k = 3 70 106. ¥ N/m �

199


PROBLEM 19.134

A 4-kg block A is dropped from a height of 800 mm onto a 9-kg block B which is at rest. Block B is supported by a spring of constant k = 1500 N/m and is attached to a dashpot of damping coefficient c = ◊230 N s/m. Knowing that there is no rebound, determine the maximum distance the blocks will move after the impact.

SOLUTION

Velocity of Block A just before impact.

v ghA =

==

2

2 9 81 0 8

3 962

( . )( . )

. m/s

Velocity of Blocks A and B immediately after impact.

Conservation of momentum.

m v m v m m v

v

v

A A B B A B+ = + ¢+ = + ¢

¢ =

( )

( )( . ) ( )

.

4 3 962 0 4 9

1 219 m/s &x0 1 219= + . m/s Ø = &x0

Static deflection (Block A): xm g

kA

0

4 9 82

15000 02619

=

=

=

-

-

-

( )( . )

. mx = 0, Equivalent position for both blocks:

c kmc =

== ◊

2

2 1500 13

279 3

( )( )

. N s/m

Since c cc< , Equation (19.44): x e c t c t

c

m

cm t

D D= +

=

=

( )-

-

21 2

1

2

230

2 13

8 846

[ sin cos ]

( )( )

.

w w

s

200



Expression forwD : w

w

D

D

k

m

c

m

x

22

2

2

1500

13

230

2 13

6 094

= ÊËÁ

ˆ¯

=ÊËÁ

ˆ¯

=

=

-

-( )( )

. rad/s

ee c t c tt-8 8461 26 094 6 094. ( sin . cos . )+

Initial conditions: x

t x

x e c c

0

0

00

1 2

0 02619

0 1 219

0 02619 0 1

== = +

= = +

-

-

.

( ) .

. [ ( ) (

m

m/s&))]

.

( ) . [ ( ) ( . )( )]( . )

c

x e c

2

8 846 01

0 02619

0 8 846 0 0 02619 1

=

= +

+

-

- --&ee c c( . )( )[ . ( ) ( )] .

. ( . )( .

-

- -

8 846 01 26 094 1 0 1 219

1 219 8 846 0 026

+ == 119 6 094

0 16202

0 16202 6 094 0 02619

1

1

8 846

) .

.

( . sin . . c.

+=

=

c

c

x e tt- - oos . )6 094t

’

Maximum deflection occurs when &x = 0

&x e t ttm m

m= =

+

0 8 846 0 16202 6 094 0 02619 6 0948 846- --. ( . sin . . cos . ).

ee t ttm m

m-

-

8 846 6 094 0 1620 6 094 0 02619 6 094

0

. [ . ][ . cos . . sin . ]

[(

+= 88 846 0 16202 6 094 0 02619 6 094

8 846 0

. )( . ) ( . )( . )]sin .

[( . )( .

++

tm

- - 002619 6 094 0 1620 6 094

0 1 274 6 094 1 219

) ( . )( . )]cos .

. sin . . c

+= +

t

tm

- oos .

tan ..

..

6 094

6 0941 219

1 2740 957

t

t = =

Time at maximum deflection s= = =

=

t

x e

m

m

tan .

..

-1 0 957

6 0940 1253

--

-

( . )( . )[ . sin( . )( . )

. cos( .

8 846 0 1253 0 1620 6 094 0 1253

0 02619 6 0944 0 1253

0 3301 0 1120 0 0189 0 307

)( . )]

( . )( . . ) .xm = =- m

Blocks move, static deflection + xm Total distance m mm= + = =0 02619 0 307 0 0569 56 9. . . . �

201


PROBLEM 19.135

Solve Problem 19.134, assuming that the damping coefficient of the dashpot is c = ◊300 N s/m.

SOLUTION

Velocity of Block A just before impact.

v ghA =

==

2

2 9 81 0 8

3 962

( . )( . )

. m/s

Velocity of Blocks A and B immediately after impact.

Conservation of momentum.

m v m v m m v

v

v

A A B B A B+ = + ¢+ = + ¢

¢ =

( )

( )( . ) ( )

.

4 3 962 0 4 9

1 219 m/s

&x0 1 219= + . m/s Ø = &x0

Static deflection (Block A). xm g

kA

0

4 9 82

15000 02619

=

=

=

-

-

-

( )( . )

. m

x = 0, Equivalent position for both blocks:

c kmc =

== ◊

2

2 1500 13

279 3

( )( )

. N s/m

Since c cc� , the system is heavily damped.

Eq. 19.42: x c e c et t= +1 21 2l l

202



Eq. 19.40: l = ± ÊËÁ

ˆ¯

= ± ÊËÁ

ˆ¯

= ±

- -

- -

-

c

m

c

m

k

m2 2

300

26

300

26

1500

13

11 538 4

2

2

. .2213

15 751

7 325

1

11

21

115 751

27 325

l

l

=

=

= +

=

-

-

-

-

-

- -

.

.. .

s

s

x c e c e

x

t t

& 55 751 7 325111 751

27 325. .. .c e c et t- --

Initial conditions: x

x0

0

0 02619

1 219

==

- .

.

m

m/s&Then -

- -0 02619

1 219 15 751 7 3251 2

1 2

.

. . .

= +=

c c

c c

Solving the simultaneous equations, c

c1 0 1219

0 09571

==

- .

.

m

m2

Then x e e

x

t t= +

=

-

- -

- -0 1219 0 09571

15 751 0 1219

15 751 7 325. . ( )

( . )( . )

. . m

& ee e

e e

t t

t

- -

-

-

-

15 751 7 325

15 751

7 325 0 09571

1 92 0 70108

. .

.

( . )( . )

. .= --7 325. t

The maximum value of x occurs when &x = 0

0 1 92 0 70108

1 92 0 70108

15 751 7 325

15 751 7

=

=

. .

. .

. .

.

e e

e e

t t

t

m m

m

- -

- -

-..

( . . )

.

.

.

.

.

325

15 751 0 7325

8 426

1 92

0 70108

2 7386

8 426

t

t

t

m

m

m

e

e

t

=

=

-

mm

m

m

t

x e

==

= +

ln( . )

.

. .( . )( . )

2 7386

0 11956

0 1219 0 015 751 0 11956

s

- - 99571

0 018542 0 039867

0 02132

7 325 0 11956e-

-

( . )( . )

. .

.

= += m

Total deflection static deflection + == +=

xm

0 02619 0 02132

0

. .

.004751 m Total deflection = 47.5 mm �

203


PROBLEM 19.136

The barrel of a field gun weighs 750 kg and is returned into firing position after recoil by a recuperator of constant c = ◊18 kN s/m Determine (a) the constant k which should be used for the recuperator to return the barrel into firing position in the shortest possible time without any oscillation, (b) the time needed for the barrel to move back two-thirds of the way from its maximum-recoil position to its firing position.

SOLUTION

(a) A critically damped system regains its equilibrium position in the shortest time.

c c

mk

m

c==

=

18000

2

Then km

cc

=( )

=( )

=2

218000

2

2

750108 kN/m �

(b) For a critically damped system, Equation (19.43):

x c c t e nt= +( )1 2-w

We take t = 0 at maximum deflection x0 .

Thus, &xx x

( )

( )

0 0

0 0

==

Using the initial conditions, x x c e c x

x x c t e nt

( ) ( ) ,

( )

0 00 10

1 0

0 2

= = + =

= +

so -w

and &&

x x c t e c e

x x c c xn

t t

n n

n n= + += = + =

--

- -ww w

w w( )

( ) ,0 2 2

0 2 2 00 0 so

Thus, x x t entn= +0 1( )w w-

For xx

t ek

mnt

nn= = + =0

3

1

31, ( ) ,w ww- with (1)

We obtain wn = =108

75012

31¥ -10

s( )

rad /

From solution of (1) wnt = 2 289. ,

t = =2 289

120 19075

.. s t = 0 1908. s �

204


PROBLEM 19.137

A uniform rod of mass m is supported by a pin at A and a spring of constant k at B and is connected at D to a dashpot of damping coefficient c. Determine in terms of m, k, and c, for small oscillations, (a) the differential equation of motion, (b) the critical damping coefficient cc.

SOLUTION

In equilibrium, the force in the spring is mg.

For small angles, sin cosq q q

d q

d q

ª ª 1

2y

l

y l

B

C

=

=

(a) Newton’s Law: S SM MA A= ( )eff

+mgl

kl

mgl

cl l I mal

t2 2 2 2- -q q a+Ê

ËÁˆ¯

= +&

Kinematics: a q

a q

=

= =

&&&&a

l lt 2 2

I ml

cl kl

I ml

+ ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

+ + ÊËÁ

ˆ¯

=

+ ÊËÁ

ˆ¯

=

2 20

2

1

3

22

2

2

&& &q q q

mml2 && &q q q+ ÊËÁ

ˆ¯

+ ÊËÁ

ˆ¯

=3 3

40

c

m

k

m �

(b) Substituting q l= e t into the differential equation obtained in (a), we obtain the characteristic equation,

l l2 3 3

40+ Ê

ËÁˆ¯

+ =c

m

k

m

and obtain the roots l =( ) ( )- -3 3 2 3

2

cm

cm

kmm

The critical damping coefficient, cc , is the value of c, for which the radicand is zero.

Thus, 3 32c

m

k

mcÊ

ËÁˆ¯

= ckm

c =3

�

205


PROBLEM 19.138

A 2-kg uniform rod is supported by a pin at O and a spring at A, and is connected to a dashpot at B. Determine (a) the differential equation of motion for small oscillations, (b) the angle that the rod will form with the horizontal 5 s after end B has been pushed 25 mm down and released.

SOLUTION

Small angles: sin ,

. .

. .

.

q q qd q q

d q q

d q

ª ª cos 1

y

y

y

A

C

B

= ( ) =

= ( ) =

= ( ) =

0 2 0 2

0 2 0 2

0 6

m

m

m 00 6. q

(a) Newton’s Law: S SM M0 0= ( )eff

+ - -0 2 0 2 2 9 81 0 6. . ( )( . ) .m m m( ) + ( ) ( )F Fs D

= ( )I mata + 0 2. m (1)

F k y k

F c y c

I ml

s A A A

D B

= + = +( )= =

= =

( ( ) ) . ( )

.

d d q d

d qST ST0 2

0 6

1

12

1

122

& &

mm( . )0 84

752 =

m

Kinematics: a q a q= = ( ) =&& &&, . . mat 0 2 0 2

Thus, from Eq. (1), 4

750 2 0 6 0 2 0 2 0 2 22 2m

m c k A+ÈÎÍ

˘˚

+ + ( ) -( . ) ( . ) . ( . ) ( . )(&& &q q q + (dST ))( . )9 81 0= (2)

But in equilibrium, SM0 0=

+ k kA A( ) ( )( . )( . ) , . ( ) ( . )( )( . )d dST ST(0.2) - 2 9 81 0 2 0 0 2 0 2 2 9 81= =

206



Equation (2) becomes

7

75

9

25

1

250

7

75

7

752 0

ÊËÁ

ˆ¯

+ ÊËÁ

ˆ¯

ÊËÁ

ˆ¯

=

= ÊËÁ

ˆ¯

=

m c k

m

&& &q q q

( ) ..

( ) .

18667

9

25

9

258 2 88

25

50

252

c

k

= ÊËÁ

ˆ¯

=

= = 0 18667 2 88 2 0. .&& &q q q+ + = �

(b) Substituting e tl into the above differential equation,

0 18667 2 88 2 0

0 7289 14 699

2. .

. , .

l ll

+ + == -- i

Since the roots are negative and real, the system is heavily damped. (Eq. 19.46):

q

q

= +

= - --

c e c e

c e c

t t

t

10 7289

114 699

10 7289

20 7289 14 699

- -. .

.( . ) ( .& )) .e t-14 699 (3)

Initial conditions. q

q

01

0

1 2

1

25

6000 04168

0

0 04168

0 0 07289 14 6

= =

== += - -

-sin .

.

. .

rad

c c

c 999 2c

Solving for c1 and c2: c c1 230 04385 2 1747 10= = - -. , . ¥

At t = 5, q = -=

-0 04385 2 1747 10

0 1146

0 7289 5 3. .

.

( . )( )e e- ¥= 0.657∞rad

207


PROBLEM 19.139

A 500-kg machine element is supported by two springs, each of constant 50 kN/m. A periodic force of 150 N amplitude is applied to the element with a frequency of 2.8 Hz. Knowing that the coefficient of damping is 1 8. kN s/m,◊ determine the amplitude of the steady-state vibration of the element.

SOLUTION

Eq. (19.52): xP

k m cm

m

f f

=( ) +- w w2 2 2( )

Total spring constant: k =

=

( )(2 503

kN/m)

100 N/m¥10

w p p pf f

m

m

f

m

P

c

x

= = =

=== ◊

=

2 2 2 8 5 6

500

150

1800

150

( . ) .

[

rad/s

N

N s/m

kg

1100 000 500 5 6 1800 5 6

150

2 9982 10 1 0028

2 2 2

9

, ( )( . ) ] [ ( . ) ]

. .

- p p+

=¥ + ¥¥

= ¥ -10

2 3714 10

9

3. m 2.3714 mm= xm = 2 37. mm �

208


PROBLEM 19.140

In Problem 19.139, determine the required value of the constant of each spring if the amplitude of the steady-state vibration is to be 1.25 mm.

SOLUTION

w p p pf f

m

f

P

mW

g

c

= = =

=

= =

= ◊

2 2 2 8 5 6

150

500

1800

( )( . ) . rad/s

N

N s/m

kg

xxm = ¥ -1 25 3. mm 1.25 10 m¥

Eq. (19.52): xP

k m cm

m

f f

=+( ) ( )- w w2 2 2

Solve for k. ( ) ( )k m cP

xf fm

m- w w2 2 2

2

+ = ÊËÁ

ˆ¯

k mP

xcf

m

mf= +

ÊËÁ

ˆ¯

= + ÊËÁ

ˆ¯-

w w

p

22

2

23

2

500 5 6150

1 25 10

-

¥

( )

( )( . ).

--

-

[( )( . )]

. . .

.

1800 5 6

154 755 10 1 44 10 1 00281 10

154 75

2

3 10 9

p

= ¥ + ¥ ¥

= 55 10 115 746 10

270 501 10

3 3

3

¥ + ¥

= ¥

.

. N/m

The above calculated value is the equivalent spring constant for two springs.

For each spring, the spring constant is k

2.

k

2135 3= . kN/m �

209


PROBLEM 19.141

In the case of the forced vibration of a system, determine the range of values of the damping factor c cc/ for which the magnification factor will always decrease as the frequency ratio w wf n/ increases.

SOLUTION

From Eq. ( . ) :19 53 ¢

Magnification factor: xmPk c

c

m

f

n c

f

n

=

- ( )ÈÎÍ

˘˚

+ ( )( )ÈÎÍ

˘˚

1

1 22 2 2w

www

Find value of c

cc

for which there is no maximum for xmPkm

as ww

f

n

increases.

d

d

x cc

mPmk

f

n

f

n c

f

n

( )( )

=- - ( )Ê

ËÁˆ¯

-( ) +È

ÎÍ

˘

˚˙

-

2

2

22 1 1 4

1

2

2

ww

ww

ww(( )È

ÎÍ˘˚

+ ( )( )ÈÎÍ

˘˚

ÏÌÔ

ÓÔ

¸˝Ô

Ô

=

- + ÊËÁ

ˆ¯

2 2 22

2

0

2 2

ccc

f

n

f

n

ww

ww

22 2

2 2

2

4 0

1 2

2+ =

ÊËÁ

ˆ¯ = -

c

c

c

c

c

f

nc

ww

For c

cc

2

2

1

2� , there is no maximum for

xmPkm( ) and the magnification factor will decrease as

ww

f

n

increases.

c

cc

�1

2

c

cc

� 0 707. �

210


PROBLEM 19.142

Show that for a small value of the damping factor c cc/ , the maximum amplitude of a forced vibration occurs when w wf nª and that the corresponding value of the magnification factor is 1

2 ( ).c cc/

SOLUTION

From Eq. (19.53' ):

Magnification factor = =

- ( )ÈÎÍ

˘˚

+ ( )( )È

xmPk c

c

m

f

n c

f

n

1

1 22 2

ww

wwÎÎÍ

˘˚

2

Find value of ww

f

n

for which xmPkm

is a maximum.

0

2 1 1 4

1

2

2

2 2

2

=( )( )

= -- ( )È

ÎÍ˘˚

-( ) + ( )È

ÎÍ

˘

˚˙

-

d

d

x cc

mPmk

f

n

f

n c

ww

ww

www

ww

ww

f

n c

f

n

cc

f

n

( )ÈÎÍ

˘˚

+ ( )( )ÈÎÍ

˘˚

ÏÌÔ

ÓÔ

¸˝Ô

Ô

- +ÊËÁ

ˆ

2 2 22

2

2 2¯

+ÊËÁ

ˆ¯

=2 2

4 0c

cc

For small c

cc

, ww

w wf

nf nª ª1

For ww

f

n

= 1

xmPk c

c

m

c

=-[ ] + ( )È

Î˘˚

1

1 1 2 12 2

x c

cm

Pk

c

m( ) = 1

2 �

211


PROBLEM 19.143

A 50-kg motor is directly supported by a light horizontal beam, which has a static deflection of 6 mm due to the mass of the motor. The unbalance of the rotor is equivalent to a mass of 100 g located 75 mm from the axis of rotation. Knowing that the amplitude of the vibration of the motor is 0.8 mm at a speed of 400 rpm, determine (a) the damping factor c cc/ , (b) the coefficient of damping c.

SOLUTION

Spring constant: km mg= = =

¥= ¥ ◊-d dST ST

N m( )( . )

.50 9 81

6 1081 75 10

33

Natural undamped circular frequency: wnk

m= = ¥ =81 75 10

5040 435

3.. rad/s

Unbalance: ¢ = == =

m g

r

100 0 100

75 0 075

.

.

kg

mm m

Forcing frequency: w f = =400 41 888rpm rad/s.

Unbalance force: P m rm f= ¢ = =w2 20 100 0 075 41 888 13 1595( . )( . )( . ) . N

Static deflection: dST m= =¥

= ¥ -P

km 13 1595

81 75 100 16097 10

33.

..

Amplitude: xm = = ¥ -0 8 0 8 10 3. .mm m

Frequency ratio: ww

f

n

= =41 888

40 4351 0359

.

..

Eq. (19.53): xm

cc

f

n c

f

n

=

- ( )ÈÎÍ

˘˚

+ ( )( )ÈÎÍ

˘˚

d

ww

ww

ST

1 22 2 2

1 22 2 2

-ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

+ÊËÁ

ˆ¯

ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

=ww

ww

df

n c

f

n m

c

c xSTÊÊ

ËÁˆ¯

- +ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

=

2

2 22

1 1 0359 2 1 03590 16097

[ ( . ) ] ( . ).c

cc

¥¥¥

È

ÎÍ

˘

˚˙

+ÊËÁ

ˆ¯

=

-

-10

0 8 10

0 0053523 4 2924 0 040486

3

3

2

2

.

. . .c

c

c

c

c

cc

ÊËÁ

ˆ¯

=2

0 008185.

212



(a) Damping factor. c

cc

= 0 090472. c

cc

= 0 0905. �

Critical damping factor. c kmc =

= ¥

= ¥ ◊

2

2 81 75 10 50

4 0435 10

3

3

( . )( )

. N s/m

(b) Coefficient of damping. cc

cc

cc=

ÊËÁ

ˆ¯

= ¥( . )( . )0 090472 4 0435 103 c = ◊366 N s/m �

213


PROBLEM 19.144

A 15-kg motor is supported by four springs, each of constant 45 kN/m. The unbalance of the motor is equivalent to a mass of 20 g located 125 mm from the axis of rotation. Knowing that the motor is constrained to move vertically, determine the amplitude of the steady-state vibration of the motor at a speed of 1500 rpm, assuming (a) that no damping is present, (b) that the damping factor c cc/ is equal to 1.3.

SOLUTION

Mass of motor: m = 15 kg

Equivalent spring constant: k = ¥

= ¥

( )( )4 45 10

180 10

3

3 lb/ft

Undamped natural circular frequency: wnk

m= = ¥

=

180 10

15109 545

3

. rad/s


Frequency ratio: ww

f

n

= =157 08

109 5451 43394

.

..

Unbalance: ¢ = = ¥= =

-m g

r

20 20 10

125 0 125

3 kg

mm m.

Unbalance force: P m rm f= ¢ = ¥ =-w2 3 220 10 0 125 157 08 61 685( )( . )( . ) . N

Static deflection: dST

m

= =¥

= ¥ -

P

km 61 685

180 10

0 3427 10

3

3

.

.

Eq. (19.53): xm

cc

f

n c

f

n

=

- ( )ÈÎÍ

˘˚

+ ( )( )ÈÎÍ

˘˚

d

ww

ww

ST

1 22 2 2

(a) No damping. 1 1 1 43394 1 115222 2

2 2- ( )ÈÎÍ

˘˚

= - =ww

f

n

[ ( . ) ] .

xm =

= ¥

= ¥

-

-

dST

m

1 11522

0 3427 10

1 05618

0 324 10

3

3

.

.

.

. xm = 0 324. mm �

214



(b) c

cc

cc

f

n c

f

n= - ( )È

ÎÍ˘˚

+ ( )( )ÈÎÍ

˘˚

1 3 1 22 2 2

. ww

ww

= + =

= = ¥

1 11522 2 1 3 1 43394 15 015

15 015

0 3427 10

2. [( )( . )( . )] .

.

.xm

dST--

-= ¥

3

3

3 8749

0 0884 10

.

. m xm = 0 0884. mm �

215


PROBLEM 19.145

A 100-kg motor is supported by four springs, each of constant 90 kN/m, and is connected to the ground by a dashpot having a coefficient of damping c = ◊6500 N s/m. The motor is constrained to move vertically, and the amplitude of its motion is observed to be 2.1 mm at a speed of 1200 rpm. Knowing that the mass of the rotor is 15 kg, determine the distance between the mass center of the rotor and the axis of the shaft.

SOLUTION

Equation (19.52): Rotor

xP

k m cm

m

f f

f

f

=-( ) +

= ÈÎÍ

˘˚

= -

w w

w p

w

2 2 2

22

2 2

1200 2

60

15 791

( )

( )( )

, s

kk

P m e

e

m f

= ¥

= ¥

= ¢

==

-

4 90 10

360 10

15 15 791

236 8

3

3

2

2

( )

( ) ( , )

,

N/m

kg s

w

665e

2 1 10236 865

360 10 100 15 791 6500 15 79

3

3 2 2.

,

[( ) ( )( , )] ( ) ( ,¥ =

¥ - +- m

e

11

0 16141

13 01 10 3

)

.

.

=

= ¥ -

e

e m e = 13 01. mm �

216


PROBLEM 19.146

A counter-rotating eccentric mass exciter consisting of two rotating 400-g masses describing circles of 150-mm radius at the same speed, but in opposite senses, is placed on a machine element to induce a steady-state vibration of the element and to determine some of the dynamic characteristics of the element. At a speed of 1200 rpm, a stroboscope shows the eccentric masses to be exactly under their respective axes of rotation and the element to be passing through its position of static equilibrium. Knowing that the amplitude of the motion of the element at that speed is 15 mm, and that the total mass of the system is 140 kg, determine (a) the combined spring constant k, (b) the damping factor c cc/ .

SOLUTION


Unbalance of one mass: m g

r

= == =

400 0 4

150 0 15

.

.

kg

mm m

Shaking force: P mr t

t

f f

f

=

=

= ¥

2

2 0 4 0 15 125 664

1 89497 10

2

2

3

w w

w

sin

( )( . )( . )( . ) sin

. siin

.

w f

m

t

P = ¥1 89497 103 N

Total mass: M = 140 kg

By Eqs. (19.48) and (19.52), the vibratory response of the system is

x x tm f= -sin( )w j

where xP

k M cm

m

f f

=-( ) +w w2 2 2( )

(1)

and tanjw

w=

-

c

k M

f

f2

(2)

Since j p= ∞ =902

, tan .j w- =and k M f2 0

(a) Combined spring constant.

k M f=

=

= ¥

w2

2

6

140 125 664

2 2108 10

( )( . )

. N/m k = 2 21. Mn/m �

217



The observed amplitude is xm = =15 0 015mm m.

From Eq. (1): cP

xk M

P

xf

m

mf

m

f m

=ÊËÁ

ˆ¯

- - =

= ¥

1

1 89497 10

125 664 0 01

22

3

ww

w( )

.

( . )( . 55

1 00531 103

)

.= ¥ ◊N s/m

Critical damping coefficient: c kMc =

= ¥

= ¥ ◊

2

2 2 2108 10 140

35 186 10

6

3

( . )( )

. N s/m

(b) Damping factor. c

cc

= ¥¥

1 00531 10

35 186 10

3

3

.

.

c

cc

= 0 0286. �

218


PROBLEM 19.147

A simplified model of a washing machine is shown. A bundle of wet clothes forms a mass mb of 10 kg in the machine and causes a rotating unbalance. The rotating mass is 20 kg (including mb) and the radius of the washer basket e is 25 cm. Knowing the washer has an equivalent spring constant k = 1000 N/m and damping ratio z = c/ cc = 0.05 and during the spin cycle the drum rotates at 250 rpm, determine the amplitude of the motion and the magnitude of the force transmitted to the sides of the washing machine.

SOLUTION

Forced circular frequency: w pf = =( )( )

.2 250

6026 18 rad/s

System mass: m = 20 kg

Spring constant: k = 1000 N/m


m= = =1000

207 0711. rad/s

Critical damping constant: c kmc = = = ◊2 2 1000 20 282 84( )( ) . N s/m

Damping constant: cc

cc

c

=ÊËÁ

ˆ¯

= = ◊( . )( . ) .0 05 141 42 14 1421 N s/m

Unbalance force: P m e

P

m b f

m

=

= =

w2

210 0 25 26 18 1713 48( )( . )( . ) .kg m rad/s N

The differential equation of motion is

mx cx kx P tm f&& &+ + = sinwThe steady state response is

x x tm f= -sin( )w j &x x tf m f= -w w jcos( )

where xP

k m cm

m

f f

=-( ) +

=- +

w w2 2 2

2 2

1713 48

1000 20 26 18 141421

( )

.

[ ( )( . ) ] [( ))( . )]

.

( , . ) ( . )

.

, .

26 18

1713 48

12 707 8 370 24

1713 48

12 713 2

2

2 2=

- += = 00 13 478. , m

xm = 134 8. mm �

219



(a) Amplitude of vibration. x t

x t

f

f

= -

= -

=

0 13478

26 18 0 13478

3 5285

. sin( )

( . )( . )cos( )

. cos

w j

w j&(( )w jf t -

Spring force: kx t

t

f

f

= -

= -

( )( . )sin( )

. sin( )

1000 0 13478

134 78

w j

w j

Damping force: cx t

t

f

f

& = -

= -

( . )( . )cos( )

. cos( )

14 1421 3 5285

49 901

w j

w j

(b) Total force: F t tf f= - + -134 78 49 901. sin( ) . cos( )w j w j

Let F F t F t

F t

m f m f

m f

= - + -

= - +

cos sin( ) sin sin( )

sin( )

y w j j w j

w j y

Maximum force. F F Fm m m2 2 2 2 2= +cos siny y

= +=

( . ) ( . )

,

134 78 49 901

20 656

2 2

Fm = 143 7. N �

220


PROBLEM 19.148

A machine element is supported by springs and is connected to a dashpot as shown. Show that if a periodic force of magnitude P = Pm sin wf t is applied to the element, the amplitude of the fluctuating force transmitted to the foundation is

F Pm m

cc

cc

c

f

n

f

n c

f

n

=+ ( )( )È

ÎÍ˘˚

- ( )ÈÎÍ

˘˚

+ ( )( )ÈÎ

1 2

1 2

2

2 2 2

ww

ww

wwÍÍ

˘˚

2

SOLUTION

From Equation (19.48), the motion of the machine is x x tm f= -sin( )w f

The force transmitted to the foundation is

Springs: F kx kx ts m f= = -sin( )w f

Dashpot: F cx cx t

F x k c t

D m f f

T m f f f

= = -

= - + -

& w w f

w f w w f

cos( )

[ sin( ) cos( )]

or recalling the identity,

A y B y A B y

B

A BA

A B

F x k cT m

sin cos sin( )

sin

cos

(

+ = + +

=+

=+

= +

2 2

2 2

2 2

2

f

y

y

w ff f t) sin( )2ÈÎ

˘˚ - +w f y

Thus, the amplitude of FT is F x k cm m f= +2 2( )w (1)

From Equation (19.53): xm

Pk

cc

m

f

n c

f

n

=

- ( )ÈÎÍ

˘˚

+ ( )ÈÎÍ

˘˚1 2

2 2 2ww

ww

Substituting for xm in Equation (1), FP

k

m

m

mc

k

cc

n

f

f

n c

f

n

=+ ( )

- ( )ÈÎÍ

˘˚

+ ( )( )ÈÎÍ

˘˚

=

1

1 2

2

2 2

2

2

w

ww

ww

w

(2)

221



and Equation (19.41), c m

mc

c

k

c

m

c

c

c n

n

f f

n c

f

n

=

=

= =ÊËÁ

ˆ¯

ÊËÁ

ˆ¯

2

2

22

ww

w w

w

ww

Substituting in Eq. (2), FP

m

mcc

cc

c

f

n

f

n c

f

n

=+ ( )( )È

ÎÍ˘˚

- ( )ÈÎÍ

˘˚

+ ( )( )ÈÎÍ

1 2

1 2

2

2 2

ww

ww

ww

˘˚

2 Q.E.D. �

222


PROBLEM 19.149

A 100-kg machine element supported by four springs, each of constant k = 200 N/m, is subjected to a periodic force of frequency 0.8 Hz and amplitude 100 N. Determine the amplitude of the fluctuating force transmitted to the foundation if (a) a dashpot with a coefficient of damping c = 420 N · s/m is connected to the machine element and to the ground, (b) the dashpot is removed.

SOLUTION

Forcing frequency: w p p pf ff= = =2 2 0 8 1 6( )( . ) . rad/s

Exciting force amplitude: Pm = 100 N

Mass: m = 100 kg

Equivalent spring constant: k = =( )( )4 200 800N/m N/m

Natural frequency: wnk

m=

=

800

1002 8284.

Frequency ratio: ww

pf

n

=

=

1 6

2 8284

1 77715

.

.

.

Critical damping coefficient: c kmc =

== ◊

2

2 800 100

565 685

( )( )

. N s/m

From the derivation given in Problem 19.148, the amplitude of the force transmitted to the foundation is

FP

m

mcc

cc

c

f

n

f

n c

f

n

=+ ( )( )È

ÎÍ˘˚

- ( )ÈÎÍ

˘˚

+ ( )( )ÈÎÍ

1 2

1 2

2

2 2

ww

ww

ww

˘˚

-ÊËÁ

ˆ¯

= - = -

2

221 1 1 77715 2 1583

ww

f

n

( . ) .

(1)

223



(a) Fm when c = ◊420 N s/m: c

c

c

c

c

c

f

n

=

=

ÊËÁ

ˆ¯

ÊËÁ

ˆ¯

=

420

565 685

0 74263

2 2 0 74263 1

.

.

( )( . )( .ww

777715

2 63894

)

.=

From Eq. (1): Fm =+

- +

100 1 2 63894

2 1583 2 63894

2

2 2

( . )

( . ) ( . )

= 100 7 964

11 6223

.

. Fm = 82 8. N �

(b) Fm when c = 0: FP

mm

f

n

=- ( )

=1

100

2 15832ww

. Fm = 46 3. N �

224


PROBLEM 19.150*

For a steady-state vibration with damping under a harmonic force, show that the mechanical energy dissipated per cycle by the dashpot is E cxm f= p w2 , where c is the coefficient of damping, xm is the amplitude of the motion, and w f is the circular frequency of the harmonic force.

SOLUTION

Energy is dissipated by the dashpot.

From Equation (19.48), the deflection of the system is

x x tm f= -sin( )w j

The force on the dashpot. F cx

F cx tD

D m f f

== -&

w w jcos( )

The work done in a complete cycle with

c

E F dx

dx x t

ff

d

m f f

f

=

= ¥

= -

Ú

2

0

2

pw

w w

p w/ i.e., force distance( )

cos( jj

w w j

w jw

p w

)

cos ( )

cos ( )[ cos(

dt

E cx t dt

tt

m f f

Df

f= -

- =- -

Ú 2 2 2

0

2

2 1 2

/

ff

ww f

w w

p w

)]

cos( )

sin(

21 2

2

2

2

2 2

0

2

2 2

E cxt

dt

Ecx

tt

m ff

m f f

f=- -

= --

Ú/

ffw

p w)

f

fÈ

ÎÍÍ

˘

˚˙˙0

2 /

Ecxm f

f f

= - - -È

ÎÍÍ

˘

˚˙˙

2 2

2

2 22

w pw w

p j j(sin( ) sin )0

E cxm f= p w2 Q.E.D. �

225


PROBLEM 19.151*

The suspension of an automobile can be approximated by the simplified spring-and-dashpot system shown. (a) Write the differential equation defining the vertical displacement of the mass m when the system moves at a speed v over a road with a sinusoidal cross section of amplitude dm and wavelength L. (b) Derive an expression for the amplitude of the vertical displacement of the mass m.

SOLUTION

(a)

+ SF ma W k x cdx

dt

d

dtm

d x

dt= - + - - -Ê

ËÁˆ¯

=: ( )d d dST

2

2

Recalling that W k= dST, we write

md x

dtc

dx

dtkx k c

d

dt

2

2+ + = +d d

(1)

Motion of wheel is a sine curve, d d w= m f tsin . The interval of time needed to travel a distance L at a

speed v is tL

v= , which is the period of the road surface.

Thus, w pt

p pf

fLv

v

L= = =2 2 2

and d d w d d pw= =m f

mLv

ftd

dttsin cos

2

Thus, Equation (1) is

md x

dtc

dx

dtkx k t c tf f f m

2

+ + = +( sin cos )w w w d �

226


PROBLEM 19.151* (Continued)

(b) From the identity A y B y A B y

B

A BA

A B

sin cos sin( )

sin

cos

+ = + +

=+

=+

2 2

2 2

2 2

y

j

j

We can write the differential equation

md x

dtc

dx

dtkx k c t

c

k

m f f

f

2

22 2

1

+ + = + +

= -

d w w y

yw

( ) sin( )

tan

The solution to this equation is analogous to Equations 19.47 and 19.48, with

P k cm m f= +d w2 2( )

x x tm f= - +sin( )w j y (where analogous to Equations (19.52)) �

xk c

k m cm

f

f f

=+

-( ) +

d w

w w

2 2

2 2 2

( )

( ) �

tanjw

w=

-

c

k m

f

f2

�

tanyw

=c

kf

�

227


PROBLEM 19.152*

Two blocks A and B, each of mass m, are supported as shown by three springs of the same constant k. Blocks A and B are connected by a dashpot, and block B is connected to the ground by two dashpots, each dashpot having the same coefficient of damping c. Block A is subjected to a force of magnitude P = P tm fsin .w Write the differential equations defining the displacements xA and xB of the two blocks from their equilibrium positions.

SOLUTION

Since the origins of coordinates are chosen from the equilibrium position, we may omit the initial spring compressions and the effect of gravity

For load A,

+ SF ma P t k x x c x x mxA m f B A B A A= + - + - =: sin ( ) ( )w 2 & & && (1)

For load B,

+ SF ma k x x c x x kx cx mxB B A B A B B B= - - - - - - =: 2 2( ) ( )& & & && (2)

Rearranging Equations (1) and (2), we find: mx c x x k x x P tA A B A B m f&& & &+ - + - =( ) ( ) sin2 w �

mx cx cx kx kxB B A B A&& & &+ - + - =3 3 2 0 �

228


PROBLEM 19.153

Express in terms of L, C, and E the range of values of the resistance R for which oscillations will take place in the circuit shown when switch S is closed.

SOLUTION

For a mechanical system, oscillations take place if c cc� (lightly damped).

But from Equation (19.41), c mk

mkmc = =2 2

Therefore, c km� 2 (1)

From Table 19.2: c ® R

m ® L

k ® 1

C (2)

Substituting in Eq. (1) the analogous electrical values in Eq. (2), we find that oscillations will take place if

RC

L� 21Ê

ËÁˆ¯

( ) RL

C� 2 �

229


PROBLEM 19.154

Consider the circuit of Problem 19.153 when the capacitor C is removed. If switch S is closed at time t = 0, determine (a) the final value of the current in the circuit, (b) the time t at which the current will have reached ( )1 1- /e times its final value. (The desired value of t is known as the time constant of the circuit.)

SOLUTION

Electrical system Mechanical system

The mechanical analogue of closing a switch S is the sudden application of a constant force of magnitude P to the mass.

(a) Final value of the current corresponds to the final velocity of the mass, and since the capacitance is zero, the spring constant is also zero

+ SF ma P Cdx

dtm

d x

dt= - =:

2

2 (1)

Final velocity occurs when d x

dt

P Cdx

dt

dx

dtv

2

20

0

=

- = =final final

final

vP

Cfinal =

From Table 19.2: v ® L, P ® E, C ® R

Thus, LE

Rfinal = �

230



(b) Rearranging Equation (1), we have

md x

dtC

dx

dtP

2

2+ =

Substitute dx

dtAe

P

C

d x

dtA et t= + = -- -l ll;

2

m A e C AeP

CPt t-ÈÎ ˘ + +È

ÎÍ˘˚

=- -l l l

- + = =m Cc

ml l0

Thus, dx

dtAe

p

cc m t= +-( )/

At t = 0, dx

dtA

P

CA

P

C= = + = -0 0

vdx

dt

p

ce c m t= = -ÈÎ ˘-1 ( )/

From Table 19.2: v ® c, p ® E, c ® R, m ® L

LE

Re R L t= -ÈÎ ˘-1 ( )/

For LE

R e= Ê

ËÁˆ¯

-ÊËÁ

ˆ¯

11

, R

LtÊ

ËÁˆ¯

= 1 tL

R= �

231


PROBLEM 19.155

Draw the electrical analogue of the mechanical system shown. (Hint: Draw the loops corresponding to the free bodies m and A.)

SOLUTION

We note that both the spring and the dashpot affect the motion of Point A. Thus, one loop in the electrical circuit should consist of a capacitor k cfi( )1 and a resistance ( ).c Rfi

The other loop consists of ( sinP tm fw ®E tm fsin ),w an inductor (m®L) and the resistor (c ®R).

Since the resistor is common to both loops, the circuit is

�

232


PROBLEM 19.156

Draw the electrical analogue of the mechanical system shown. (Hint: Draw the loops corresponding to the free bodies m and A.)

SOLUTION

Loop 1 (Point A): k1 ® 1

12c

k, ® 1

21c

c, ® R1

Loop 2 (Mass m): k2 ® 1

2cm, ® L c, 2 ® R2

With k2 ® 12c common to both loops,

�

233


PROBLEM 19.157

Write the differential equations defining (a) the displacements of the mass m and of the Point A, (b) the charges on the capacitors of the electrical analogue.

SOLUTION

(a) Mechanical system.

Point A: + SF = 0: cd

dtx x kxA m A( )- + = 0 �

Mass m:

+ SF ma cd

dtx x P t m

d x

dtm A m f

m= - - = -: ( ) sinw2

2

md x

dtc

d

dtx x P tm

m A m f

2

2+ - =( ) sinw �

(b) Electrical analogue.

From Table 19.2:

m ® L

c ® R

k ® 1

Cx ® q

P ® E

234



Substituting into the results from Part (a), the analogous electrical characteristics,

Rd

dtq q

CqA m n( )- + Ê

ËÁˆ¯

=10 �

Ld q

dtR

d

dtq q E tm

m A m f

2

2+ - =( ) sinw �

Note: These equations can also be obtained by summing the voltage drops around the loops in the circuit of Problem 19.155.

235


PROBLEM 19.158

Write the differential equations defining (a) the displacements of the mass m and of the Point A, (b) the charges on the capacitors of the electrical analogue.

SOLUTION

(a) Mechanical system.

+ SF = 0

Point A: k x cdx

dtk x xA

AA m1 1 2 0+ + - =( ) c

dx

dtk k x k xA

A m1 1 2 2 0+ + - =( ) �

Mass m: + SF ma k x x cdx

dtm

d x

dtA m

m m= - - =: ( )2 2

2

2 m

d x

dtc

dx

dtk x xm m

m A

2

2 2 2 0+ + - =( ) �

(b) Electrical analogue.

Substituting into the results from Part (a) using the analogous electrical characteristics from Table 19.2 (see left),

Rdq

dt C Cq

CqA

A m11 2 2

1 1 10+ +

ÊËÁ

ˆ¯

- = �

Ld q

dtR

dq

dt Cq qm m

m A

2

2 22

10+ + - =( ) �

236


PROBLEM 19.159

A thin square plate of side a can oscillate about an axis AB located at a distance b from its mass center G. (a) Determine the period of small oscillations if b a= 1

2 . (b) Determine a second value of b for which the period of small oscillations is the same as that found in Part a.

SOLUTION

Let the plate be rotated through angle q about the axis as shown.

+ S SM M mgb I ma bAB AB t= = - -( ) sin ( )( )eff : q a

Kinematics: a q= &&

a b bt = =ª

a qq

&&sin 0

Moment of inertia: I ma= 1

122

Then ( )I mb mgb

a b gb

gb

a b

+ + =

+ÊËÁ

ˆ¯

+ =

++

=

2

2 2

2 2

0

1

120

12

120

&&

&&

&&

q q

q q

q q

Natural circular frequency: wngb

a b=

+12

122 2

(a) b a= 1

2: wn

ga

a a

g

a=

+=16

3

3

22 2

Period of vibration: t pwn

n

= 2 t pn

a

g= 2

2

3 �

237



(b) Another value of b giving the same period:

wngb

a b

g

a

b

a b a

ab a b

=+

=

+=

= +

12

12

3

2

12

12

3

2

24 3 36

2 2

2 2

2 2

36 24 3 0

0 5 0 16667

2 2b ab a

b a b a

- + == =. .and b a= 0 1667. �

238


PROBLEM 19.160

A 150-kg electromagnet is at rest and is holding 100 kg of scrap steel when the current is turned off and the steel is dropped. Knowing that the cable and the supporting crane have a total stiffness equivalent to a spring of constant 200 kN/m, determine (a) the frequency, the amplitude, and the maximum velocity of the resulting motion, (b) the minimum tension which will occur in the cable during the motion, (c) the velocity of the magnet 0.03 s after the current is turned off.

SOLUTION

Data: m m k1 23150 100 200 10= = = ¥ kg kg N/m

From the first two sketches, T kx m m gm0 1 2+ = +( ) (1)

T m g0 1= (2)

Subtracting Eq. (2) from Eq. (1), kx m gm = 2

xm g

km = =¥

= ¥ =-23

3100 9 81

200 104 905 10

( )( . ). m 4.91 mm


m= = ¥ =

1

3200 10

15036 515. rad/s

Natural frequency: f fnn

n= = =w

p p2

36 515

25 81

.. Hz

Maximum velocity: v xm n m= = ¥ =-w ( . )( . ) .36 515 4 905 10 0 17913 m/s

(a) Resulting motion: amplitude mmxm = 4 91. �

frequency Hzfn = 5 81. �

maximum velocity m/svm = 0 1791. �

239



(b) Minimum value of tension occurs when x xm= - .

T T kx

m g m g

m m g

mmin

( )

( )( . )

= -= -= -=

0

1 2

1 2

50 9 81 Tmin = 491 N �

The motion is given by x x t

x x tm n

n m n

= += +

sin( )

cos( )

w jw w j&

Initally, x x

x

x x t

m

n m n

0

0

1

0

2

2

= - = -= =

= -

= -ÊËÁ

ˆ¯

or sin

or cos 0

jj

j p

w w p

&

& cos

(c) Velocity at s.t = 0 03.

ww j

w j

n

n

n

t

t

t

= =- = -

-

( . )( . ) .

.

)

36 515 0 03 1 09545

0 47535

rad

rad

cos( == 0 88913.

&x = ¥ -( . )( . )( . )36 515 4 905 10 0 889133 &x = 0 1592. m/s� �

240


PROBLEM 19.161

Disks A and B weigh 15 kg and 6 kg, respectively, and a small 2.5-kg block C is attached to the rim of disk B. Assuming that no slipping occurs between the disks, determine the period of small oscillations of the system.

SOLUTION

Small oscillations: h rr

B mB B= - ª( cos )12

2

qq

Positionj r rB B A A& &q q=

T m r I Ir

r

Im r

Im r

C B m B m AB

Am

BB B

AA

12 2

2

2

1

2

1

2

1

2

2

= + +ÊËÁ

ˆ¯

=

=

( )& & &q q q

AA

C BB B A A B

AmT m r

m r m r r

r

T

2

12

2 2 22

2

1

2 2 2= + + Ê

ËÁˆ¯

ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙&q

112 2

1

1

2 2 20

= + +ÊËÁ

ˆ¯

ÈÎÍ

˘˚

=

mm m

r

V

CB A

B m&q

241



Positionk T e

V m gh

m r

C

C m

2

2

2

0

2

==

=&q


T V T V

mm m

rm

m n m

CB A

B n m

1 1 2 2

2 2 21

2 2 20 0

+ = +

=

+ +ÊËÁ

ˆ¯

ÈÎÍ

˘˚

+ = +

&q w q

w q CC B m

nC

CB A B

n

gr

m

mm m

g

r

q

w

w

2

2

2

2

2

2 5 9 81

2 515

=+ +È

ÎÍ˘˚

=+ +

( )

( . )( . )

.( 66

20 15

12 57692 2

)( . )

.

ÈÎÍ

˘˚

= -wn s

Period of small oscillations. t pw

pn

n

= =2 2

12 5769. t n = 1 772. s �

242


PROBLEM 19.162

A period of 6.00 s is observed for the angular oscillations of a 120-g gyroscope rotor suspended from a wire as shown. Knowing that a period of 3.80 s is obtained when a 30 mm-diameter steel sphere is suspended in the same fashion, determine the centroidal radius of gyration of the rotor. (Density of steel = 7800 kg/m3.)

SOLUTION

+ S SM M K I= - =( ) :eff q q&&

&&q q

w

t p

+ =

=

=

K

IK

I

I

K

n

0

2

2

(1)

KI= 4 2

2

pt

(2)

IK= t

p

2

24 (3)

For the sphere, rd= = =2

30

215 mm = 0.015 m

Volume: V rs = =

= ¥ -

4

3

4

30 015

1 41372 10

3 3

5

p p ( . )

. m3

Mass: m SVs s=

= ¥=

-( )( . )

.

7800 1 41372 10

0 11027

5kg/m

kg

3

Moment of inertia: I m rs= =

= ¥ -

2

5

2

50 11027 0 015

9 9243 10

2 2

6

( . )( . )

. kgm2

Period: t s = 3 80. s

243



From Eq. (2): K =¥

= ¥ ◊

-

-

4 9 9243 10

3 80

2 7133 10

2 6

2

5

p ( . )

( . )

. N m/rad

For the rotor, m = ==

120 0 12

6 00

g kg

s

.

.t

From Eq. (3): I = ¥

= ¥ ◊

-

-

( . )( . )

.

2 7133 106 00

4

2 4742 10

52

2

5

pkg m2

Radius of gyration. I mk= 2

kI

m=

= ¥

=

-2 4742 10

0 120 014359

5.

.. m k = 14 36. mm �

244


PROBLEM 19.163

A 1.5-kg block B is connected by a cord to a 2-kg block A, which is suspended from a spring of constant 3 kN/m. Knowing that the system is at rest when the cord is cut, determine (a) the frequency, the amplitude, and the maximum velocity of the resulting motion, (b) the minimum tension that will occur in the spring during the motion, (c) the velocity of block A 0.3 s after the cord has been cut.

SOLUTION

Before the cord is cut, the tension in the spring is

T m m gA B0

2 1 5 9 81

34 335

= += +=

( )

( . )( . )

. N

The elongation of the spring is eT

k00

3

3

34 335

3 10

11 445 10

=¢

=¥

= ¥ -

.

. m

After the cord is cut, the tension in the equilibrium position is

¢ ===

T m gA0

2 0 9 81

19 62

( . )( . )

. N

The corresponding elongation is ¢ =¢

=¥

= ¥ -

eT

k00

3

3

19 62

3 10

6 54 10

.

. m

Let x be measured downward from the equilibrium position.

T T kx= ¢ +0

(a) Newton’s Second Law after the cord is cut.mx m g T

m x m g kx T kx

m x kx

k

m

A

A A

A

nA

&&&&

&&

= -+ - - = -

+ =

= = ¥ =

0

3

0

3 10

238 72w . 998

2

38 7298

2

rad/s

fnn= =

wp p

. fn = 6 16. Hz �

245



The resulting motion is x x t

x xm n

n m t

= += +

sin ( )

cos( )

w jw w j&

Initial condition: x e e

x

x

xm

n m

0 0 0

3

0

3

4 905 10

0

0 75375 10

0

= - ¢

= ¥=

¥ ==

-

-

.

. sin

cos

m

&j

w j

j p=

=24 905xm . m xm = 4 91. mm �

x t

x t

n

n

= ¥ +ÊËÁ

ˆ¯

= +ÊËÁ

ˆ¯

-4 905 102

0 189972

3. sin ( )

. cos

w p

w p

m

(m/& ss) &xm = 0 1900. m/s �

(b) Minimum tension occurs when x is minimum.

T T kxmmin

. ( )( . )

= ¢ -

= - ¥ ¥ -0

3 319 62 3 10 4 905 10 Tmin .= 4 91 N �

(c) Velocity when t = 0.3 s.

w j pnt

x

- = +

==

( . )( . )

.

( . )cos( .

38 7298 0 32

13 1897

0 18997 13 1

radians

& 8897

0 18997 0 8119

)

( . )( . )= &x = 0 1542. m/s ¯ �

246


PROBLEM 19.164

Two rods, each of mass m and length L, are welded together to form the assembly shown. Determine (a) the distance b for which the frequency of small oscillations of the assembly is maximum, (b) the corresponding maximum frequency.

SOLUTION

Positionj V1 0=

T mv mv I I

v b

vL

CD AB CD m AB m

CD m

AB

12 2 2 21

2

2

= + + +ÈÎ ˘

=

= ÊËÁ

ˆ¯

& &

&

&

q q

q

qmm

CD AB

m

I I

mL

T m bL

L L

=

=

= + ÊËÁ

ˆ¯ + +

È

ÎÍÍ

˘

˚˙˙

1

12

1

2 2

1

12

1

12

2

12

22 2 2&q

== +Ê

ËÁˆ

¯m

bL

m2

5

122

22&q

247



Positionk V mgb mgL

m m2 12

1= - + -( cos ) ( cos )q q

Small angles: 1 22 2

2 20

22

2

2

2

- = ÊËÁ

ˆ¯ ª

= +ÊËÁ

ˆ¯

=

cos sinq q aq

q

mm m

mV mg bL

T


T V T V

m b Lmg

bL

m m

m

1 1 2 2

2 2 2 21

2

5

120 0

2 2

+ = +

+ÈÎÍ

˘˚

+ = + +ÈÎÍ

˘˚

=

&

&q q

q wnn m

n

Lg b

b L

q

w2 22 5

122

=+( )

+( ) (1)

(a) Distance b for maximum frequency.

Maximum whenww

nnd

db2

2

0=

d

db

b L g g b b

b L

b Lb

nLw2 2 5

122

2

2 512

2 2

2

20

5

12

=+( ) - +( )

+( )=

- - + ÊËÁ

ˆ¯

( )

˜ =

=- + ( )

=

L

bL L L

L L

2

2 2012

2

0

20 316 1 317

m. , . b L= 0 316. �

(b) Corresponding maximum frequency.

From Eq. (1) and the answer to Part (a):

w

wp p

n

nn

g

L

g

L

fg

L

22 5

12

0 316 0 5

0 316

1 580

2

1 580

2

= ++ÈÎ ˘

=

= =

[ . . ]

( . )

.

.

== 0 200.g

LHz �

248


PROBLEM 19.165

As the rotating speed of a spring-supported motor is slowly increased from 200 to 500 rpm, the amplitude of the vibration due to the unbalance of the rotor is observed to decrease steadily from 8 mm to 2.5 mm. Determine (a) the speed at which resonance would occur, (b) the amplitude of the steady-state vibration at a speed of 100 rpm.

SOLUTION

Since the amplitude decreases with increasing speed over the range w1 200= rpm to w2 500= rpm, the motion is out of phase with the force.

xm

Pk

mr

k

mrM

m

f

n

f

f

n

f

n

f

n

=- ( )

=- ( )

=( )( )

- ( )1 1 12 2

2

2

2

ww

w

ww

ww

ww

| |xm

mrM

f

n

f

n

=( )( )( ) -

ww

ww

2

21

Let un

=ww

1 , where w1 200 20 944= =rpm rad/s. .

| |xu

um

mrM

1

2

2 1=

( )-

(1)

At w w

ww

f

f

n

u u

= =

= =

2 500

500

2002 5

rpm

.

| |( . )

.x

u

um

mrM

2

2

2

6 25

6 25 1=

( )-

(2)

Dividing Eq. (1) by Eq. (2),

| |

| |

( . )

( )( . ).

.

x

x

u u

u u

u u

m

m

1

2

2 2

2 2

4

6 25 1

1 6 25

83 2

6 25

= --

= =

-

mm

2.5 mm22 4 2

4 2

20 20

13 75 19 019

13 751 17551

= -

- = = =

u u

u u u..

.

(a) Natural frequency. ww

n u= = =1 20 944

1 1755117 817

.

.. rad/s wn = 170 1. rpm �

249



At 200 rpm, | | .x um 138 8 10 1 17551= = ¥ =-mm m,

From Eq. (1): 8 101 17551

1 17551 1

19

5 25

2 2105 10

32

2¥ =

( )-

=

= ¥

-

-

mrM

mr

Mmr

M

( . )

( . )

.

. 33 m

(b) Amplitude at 100 rpm. w w

ww

f n

f

n

=

= =

10 472

10 472

17 8170 58775

.

.

..

rad/s �

The motion is in phase with the force.

x

x

m

mrM

m

f

n

f

n

=( )( )

- ( )= ¥

-

-

ww

ww

2

2

3 2

1

2 2105 10 0 58775

1 0 587

( . )( . )

( . 775

1 167 10

2

3

)

.= ¥ - m xm = 1 167. mm �

250


PROBLEM 19.166

The compressor shown has a mass of 250 kg and operates at 2000 rpm. At this operating condition, undesirable vibration occurs when the compressor is attached directly to the ground. To reduce the vibration of the concrete floor that is resting on clay soil, it is proposed to isolate the compressor by mounting it on a square concrete block separated from the rest of the floor as shown. The density of concrete is 2400 kg/m3 and the spring constant for the soil is found to be 80 106¥ N/m. The geometry of the compressor leads to choosing a block that is 1.5 m by 1.5 m. Determine the depth h that will reduce the force transmitted to the ground by 75%.

SOLUTION

Forced circular frequency corresponding to 2000 rpm:

w pf = =( )( )

.2 2000

60209 44 rad/s

Natural circular frequency for compressor attached directly to the ground:

w11

680 10565 68= =

¥=k

m

N/m

250 kgrad/s.

Let P be the force due to the compressor unbalance and F1 be the force transmitted to the ground.

F kx kP

F

P

Pk

f

n

f

n

f

n

1 1 2 2

12 209 44

565

1 1

1

1

1

1

= =- ( )

=- ( )

=- ( )

=-

ww

ww

ww

.

.668

21 15951

( )= .

251



Modify the system by making the mass much larger so that the natural frequency of modified system is much less than the forcing frequency. Let w2 be the new natural circular frequency. For the ground force to be reduced by 75%,

F

P

F

P f

n

f

2 12

2

2

0 25 0 25 1 15951 0 289881

1= = = =

( ) -

ÊËÁ

ˆ¯

=

. ( . )( . ) .ww

ww

111

0 289884 4497 2 109

4 4497

209 44

4 4497

2

22

2 2

+ = =

= = =

.. .

.

( . )

.

ww

ww

f

f 99 8579 10

80 10

3 2

22

2

222

6

. ( )

(

¥

=

= =¥¥

rad/s

N/m

9.8579 10 rad/3

w

w

k

m

mk

ss)kg

2= 8115

Required properties of attached concrete block:

mass kg

volumemass

density

kg

kg/

= - = - =

= =

m m2 1 8115 250 7865

7865

2400 mmm

area m m m

depthvolume

area

m

33

2

3

=

= ¥ =

= =

3 277

1 5 1 5 2 25

3 277

2

.

. . .

.

.2251 456

mm

2= . h = 1 456. m �

252


PROBLEM 19.167

If either a simple or a compound pendulum is used to determine experimentally the acceleration of gravity g, difficulties are encountered. In the case of the simple pendulum, the string is not truly weightless, while in the case of the compound pendulum, the exact location of the mass center is difficult to establish. In the case of a compound pendulum, the difficulty can be eliminated by using a reversible, or Kater, pendulum. Two knife edges A and B are placed so that they are obviously not at the same distance from the mass center G, and the distance l is measured with great precision. The position of a counterweight D is then adjusted so that the period of oscillation t is the same when either knife edge is used. Show that the period t obtained is equal to that of a true simple pendulum of length l and that g = 4p 2l/t 2.

SOLUTION

From Problem 19.52, the length of an equivalent simple pendulum is:

l rk

rA = +2

and l Rk

RB = +2

But t t

p p

A B

A Bl

g

l

g

=

=2 2

Thus, l lA B=

For l l

rk

rR

k

R

r R k R r R k r

r R r R k r R

r R

A B=

+ = +

+ = +

- = -- =

2 2

2 2 2 2

2

0

[ ] [ ]

( )

Thus, r R k= 2

or rk

R

Rk

r

=

=

2

2

253



Thus, AG GA BG GB= ¢ = ¢and

That is, A A B B= ¢ = ¢and

Noting that l l l

l

g

A B= =

=t p2

or gl= 4 2

2

pt

254


PROBLEM 19.168

A 400-kg motor supported by four springs, each of constant 150 kN/m, is constrained to move vertically. Knowing that the unbalance of the rotor is equivalent to a 23-g mass located at a distance of 100 mm from the axis of rotation, determine for a speed of 800 rpm (a) the amplitude of the fluctuating force transmitted to the foundation, (b) the amplitude of the vertical motion of the motor.

SOLUTION


Unbalance: ¢ = == =

m

r

23 0 023

100 0 100

g kg

mm m

.

.

Forcing frequency: w f =

=

800

83 776

rpm

rad/s.

Spring constant: k = ¥

= ¥

( )( )4 150 10

600 10

3

3

N/m

N/m


m=

= ¥

=

600 10

40038 730

3

. rad/s

Frequency ratio: ww

f

n

= 2 1631.

Unbalance force: P m rm f= ¢

==

w2

20 023 0 100 83 776

16 1424

( . )( . )( . )

. N

Static deflection: P

km =

¥= ¥ -

16 1424

600 10

26 904 10

3

6

.

. m

255



Amplitude of vibration. From Eq. (19.33):

xm

Pkm

f

n

=- ( )

= ¥-

= - ¥

-

-

1

26 904 10

1 2 1631

7 313 10

2

6

2

6

ww

.

( . )

. m

(a) Transmitted force. F kxm m= - = - ¥ - ¥ -( )( . )600 10 7 313 103 6 Fm = 4 39. N �

(b) Amplitude of motion. | | .xm = ¥ -7 313 10 6 m | | .xm = 0 00731 mm �

256


PROBLEM 19.169

Solve Problem 19.168, assuming that a dashpot of constant c = ◊6500 N s/m is introduced between the motor and the ground.

PROBLEM 19.168 A 400-kg motor supported by four springs, each of constant 150 kN/m, is constrained to move vertically. Knowing that the unbalance of the rotor is equivalent to a 23-g mass located at a distance of 100 mm from the axis of rotation, determine for a speed of 800 rpm (a) the amplitude of the fluctuating force transmitted to the foundation, (b) the amplitude of the vertical motion of the motor.

SOLUTION


Unbalance: m

r

= == =

23 0 023

100 0 100

g kg

mm m

.

.

Forcing frequency: w f =

=

800

83 776

rpm

rad/s.

Spring constant: ( )( )4 150 10 600 103 3¥ = ¥N/m N/m


m= = ¥

=

600 10

40038 730

3

. rad/s

Frequency ratio: ww

f

n

= 2 1631.

Viscous damping coefficient: c = ◊6500 N s/m

Critical damping coefficient: c kMc = = ¥= ◊

2 2 600 10 400

30 984

3( )( )

, N s/m

Damping factor: c

cc

= 0 20978.

Unbalance force: P mrm f= =

=

w2 20 023 0 100 83 776

16 1424

( . )( . )( . )

. N

Static deflection: dST

m

= =¥

= ¥ -

P

km 16 1424

600 10

26 904 10

3

6

.

.

257



Amplitude of vibration. Use Eq. (19.53).

xm

Pk

cc

m

f

n c

f

n

=

- ( )ÈÎÍ

˘˚

+ ( )( )ÈÎÍ

˘˚

1 1 22 2 2

ww

ww

where 1 1 2 1631 3 6792

2-ÊËÁ

ˆ¯

= - = -ww

f

n

( . ) .

and 2 2 0 20978 2 1631 0 90755

26 904

c

c

x

c

f

n

m

ÊËÁ

ˆ¯

ÊËÁ

ˆ¯

= =

=

ww

( )( . )( . ) .

. ¥¥

- +

= ¥

-

-

10

3 679 0 90755

7 1000 10

3

2 2

6

( . ) ( . )

. m

Resulting motion: x x t

x x t

m f

f m f

= -

= -

sin ( )

cos( )

w j

w w j&Spring force: F kx kx t ts m f f= = - = -sin ( ) .w j w j4 26 sin ( )

Damping force: F cx c x t td f m f f= = - = -& w w j w jcos( ) . cos( )3 8663

Let F F t F F ts m f d m f= - = -cos sin ( ) sin cos( )y w j y w jand

Total force: F F t F t

F t

m f m f

m f

= - + -

= - +

cos sin ( ) sin cos( )

sin ( )

y w j y w j

w j y

(a) Force amplitude. F F F

F kx c x

m m m

m m f m

= +

= +

( cos ) ( sin )

( ) ( )

y y

w

2 2

2 2

= +( . ) ( . )4 26 3 86632 2 Fm = 5 75. N �

(b) Amplitude of vibration. xm = 0 00710. mm �

258


PROBLEM 19.170

A small ball of mass m attached at the midpoint of a tightly stretched elastic cord of length l can slide on a horizontal plane. The ball is given a small displacement in a direction perpendicular to the cord and released. Assuming the tension T in the cord to remain constant, (a) write the differential equation of motion of the ball, (b) determine the period of vibration.

SOLUTION

(a) Differential equation of motion.

+ SF ma T mx= = -: sin2 q &&

For small x, sin tanq qª = ( ) =x x

ll2

2

mx Tx

l&&+ Ê

ËÁˆ¯ =( )2

20 �

mxT

lx&&+ Ê

ËÁˆ¯

=40

Natural circular frequency. w

w

n

n

T

ml

T

ml

2 4

2

=

=

(b) Period of vibration. t pwn

n

= 2 t pn

ml

T= �

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