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Vector Mechanics for Engineers: Dynamics

Tenth Edition

Free Vibrations of Particles. Simple Harmonic Motion

19 - 2

•  If a particle is displaced through a distance xm from its equilibrium position and released with no velocity, the particle will undergo simple harmonic motion,

( )0=+

−=+−==

kxxmkxxkWFma st

!!

δ

•  General solution is the sum of two particular solutions,

( ) ( )tCtC

tmkCt

mkCx

nn ωω cossin

cossin

21

21

+=

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛=

•  x is a periodic function and ωn is the natural circular frequency of the motion.

•  C1 and C2 are determined by the initial conditions:

( ) ( )tCtCx nn ωω cossin 21 += 02 xC =

nvC ω01 =( ) ( )tCtCxv nnnn ωωωω sincos 21 −== !

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Vector Mechanics for Engineers: Dynamics

Tenth Edition

Free Vibrations of Particles. Simple Harmonic Motion

19 - 3

( )φω += txx nm sin

==n

n ωπ

τ2 period

===πω

τ 21 n

nnf natural frequency

( ) =+= 20

20 xvx nm ω amplitude

( ) == −nxv ωφ 00

1tan phase angle

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Vector Mechanics for Engineers: Dynamics

Tenth Edition

Free Vibrations of Particles. Simple Harmonic Motion

19 - 4

( )φω += txx nm sin

•  Velocity-time and acceleration-time curves can be represented by sine curves of the same period as the displacement-time curve but different phase angles.

( )( )2sin

cosπφωω

φωω

++=

+=

=

txtx

xv

nnm

nnm

!

( )

( )πφωω

φωω

++=

+−=

=

tx

tx

xa

nnm

nnm

sin

sin2

2

!!

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Vector Mechanics for Engineers: Dynamics

Tenth Edition

Simple Pendulum (Approximate Solution)

19 - 5

•  Results obtained for the spring-mass system can be applied whenever the resultant force on a particle is proportional to the displacement and directed towards the equilibrium position.

for small angles,

( )

gl

tlg

nn

nm

πωπ

τ

φωθθ

θθ

22

sin

0

==

+=

=+!!

:tt maF =∑

•  Consider tangential components of acceleration and force for a simple pendulum,

0sin

sin

=+

=−

θθ

θθ

lg

mlW

!!

!!

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Vector Mechanics for Engineers: Dynamics

Tenth Edition

Simple Pendulum (Exact Solution)

19 - 6

0sin =+ θθlg!! An exact solution for

leads to ( )

∫−

=2

022 sin2sin1

4π

φθ

φτ

mn

dgl

which requires numerical solution.

⎟⎟⎠

⎞⎜⎜⎝

⎛=

glK

n ππ

τ 22

(1)

, with K being integral in (1)

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Vector Mechanics for Engineers: Dynamics

Tenth Edition

Concept Question

19 - 7

The amplitude of a vibrating system is shown to the right. Which of the following statements is true (choose one)?

a)   The amplitude of the acceleration equals the amplitude of the displacement

b)  The amplitude of the velocity is always opposite (negative to) the amplitude of the displacement

c)   The maximum displacement occurs when the acceleration amplitude is a minimum

d)  The phase angle of the vibration shown is zero

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Vector Mechanics for Engineers: Dynamics

Tenth Edition

Sample Problem 19.1

19 - 8

A 50-kg block moves between vertical guides as shown. The block is pulled 40mm down from its equilibrium position and released.

For each spring arrangement, determine a) the period of the vibration, b) the maximum velocity of the block, and c) the maximum acceleration of the block.

SOLUTION:

•  For each spring arrangement, determine the spring constant for a single equivalent spring.

•  Apply the approximate relations for the harmonic motion of a spring-mass system.

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Vector Mechanics for Engineers: Dynamics

Tenth Edition

Sample Problem 19.1

19 - 9

mkN6mkN4 21 == kk SOLUTION: •  Springs in parallel:

-  determine the spring constant for equivalent spring

mN10mkN10 4

21

21

==

+==

+=

kkPk

kkP

δ

δδ

-  apply the approximate relations for the harmonic motion of a spring-mass system

nn

n mk

ωπ

τ

ω

2

srad14.14kg20N/m104

=

===

s 444.0=nτ

( )( )srad 4.141m 040.0=

= nmm xv ω

sm566.0=mv

2sm00.8=ma( )( )2

2

srad 4.141m 040.0=

= nmm axa

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Vector Mechanics for Engineers: Dynamics

Tenth Edition

Sample Problem 19.1

19 - 10

mkN6mkN4 21 == kk •  Springs in series: -  determine the spring constant for equivalent spring

-  apply the approximate relations for the harmonic motion of a spring-mass system

nn

n mk

ωπ

τ

ω

2

srad93.6kg20

400N/m2

=

===

s 907.0=nτ

( )( )srad .936m 040.0=

= nmm xv ω

sm277.0=mv

2sm920.1=ma( )( )2

2

srad .936m 040.0=

= nmm axa1 2

1 2

1 2

1 2

1 2

1 2

( )

2.4kN m 2400 N m

P Pk k

k k Pk kk kPkk k

= + = +

+=

= = =+

d d d

d

d

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Vector Mechanics for Engineers: Dynamics

Tenth Edition

Free Vibrations of Rigid Bodies

19 - 11

•  If an equation of motion takes the form 0or0 22 =+=+ θωθω nn xx !!!!

the corresponding motion may be considered as simple harmonic motion.

•  Analysis objective is to determine ωn.

( ) ( )[ ] mgWmbbbmI ==+= ,22but 23222

121

053sin

53

=+≅+ θθθθbg

bg !!!!

gb

bg

nnn 3

522,53then π

ωπ

τω ===

•  For an equivalent simple pendulum, 35bl =

•  Consider the oscillations of a square plate ( ) ( ) θθθ !!!! ImbbW +=− sin

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Vector Mechanics for Engineers: Dynamics

Tenth Edition

Sample Problem 19.2

19 - 12

k

A cylinder of weight W is suspended as shown.

Determine the period and natural frequency of vibrations of the cylinder.

SOLUTION:

•  From the kinematics of the system, relate the linear displacement and acceleration to the rotation of the cylinder.

•  Based on a free-body-diagram equation for the equivalence of the external and effective forces, write the equation of motion.

•  Substitute the kinematic relations to arrive at an equation involving only the angular displacement and acceleration.

© 2013 The McGraw-Hill Companies, Inc. All rights reserved.

Vector Mechanics for Engineers: Dynamics

Tenth Edition

Sample Problem 19.2

19 - 13

SOLUTION: •  From the kinematics of the system, relate the linear

displacement and acceleration to the rotation of the cylinder. θrx = θδ rx 22 ==

θα !!rra ==θα !!"= θ!!

"ra =

•  Based on a free-body-diagram equation for the equivalence of the external and effective forces, write the equation of motion.

( ) :∑∑ = effAA MM ( ) αIramrTWr +=− 22( )θδ rkWkTT 2but 2

102 +=+=

•  Substitute the kinematic relations to arrive at an equation involving only the angular displacement and acceleration.

( )( ) ( )

038

22 221

21

=+

+=+−

θθ

θθθ

mk

mrrrmrkrWWr

!!

!!

mk

n 38

=ωkm

nn 8

322π

ωπ

τ ==mkf n

n 38

21

2 ππω

==

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Vector Mechanics for Engineers: Dynamics

Tenth Edition

Sample Problem 19.3

19 - 14

n

10 kg1.13s

m ==t s93.1=nτ

The disk and gear undergo torsional vibration with the periods shown. Assume that the moment exerted by the wire is proportional to the twist angle.

Determine a) the wire torsional spring constant, b) the centroidal moment of inertia of the gear, and c) the maximum angular velocity of the gear if rotated through 90o and released.

SOLUTION:

•  Using the free-body-diagram equation for the equivalence of the external and effective moments, write the equation of motion for the disk/gear and wire.

•  With the natural frequency and moment of inertia for the disk known, calculate the torsional spring constant.

•  With natural frequency and spring constant known, calculate the moment of inertia for the gear.

•  Apply the relations for simple harmonic motion to calculate the maximum gear velocity.

© 2013 The McGraw-Hill Companies, Inc. All rights reserved.

Vector Mechanics for Engineers: Dynamics

Tenth Edition

Sample Problem 19.3

19 - 15

s13.1lb20

n =

=

τ

Ws93.1=nτ

SOLUTION: •  Using the free-body-diagram equation for the equivalence

of the external and effective moments, write the equation of motion for the disk/gear and wire.

( ) :∑∑ = effOO MM

0=+

−=+

θθ

θθ

IK

IK

!!

!!

KI

IK

nnn π

ωπ

τω 22===

•  With the natural frequency and moment of inertia for the disk known, calculate the torsional spring constant.

( ) ( )21 2 22

1 10 kg 0.2 m 0.2 kg m2

I mr= = = ◊

0.21.13 2K

= p 6.18N m radK = ◊

© 2013 The McGraw-Hill Companies, Inc. All rights reserved.

Vector Mechanics for Engineers: Dynamics

Tenth Edition

Sample Problem 19.3

19 - 16

s13.1lb20

n =

=

τ

Ws93.1=nτ

K = 6.18 N ⋅m rad

KI

IK

nnn π

ωπ

τω 22===

•  With natural frequency and spring constant known, calculate the moment of inertia for the gear.

1.93 26.183I= p Igear = 0.583 kg ⋅m2

•  Apply the relations for simple harmonic motion to calculate the maximum gear velocity.

nmmnnmnm tt ωθωωωθωωθθ === sinsin

rad 571.190 =°=mθ

( ) ⎟⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

s 93.12rad 571.12 π

τπ

θωn

mm

srad11.5=mω

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Vector Mechanics for Engineers: Dynamics

Tenth Edition

Group Problem Solving

19 - 17

A uniform disk of radius 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 period of the resulting oscillation.

SOLUTION:

•  Using the free-body and kinetic diagrams, write the equation of motion for the pendulum.

•  Determine the natural frequency and moment of inertia for the disk (use the small angle approximation).

•  Calculate the period.

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Vector Mechanics for Engineers: Dynamics

Tenth Edition

Group Problem Solving

19 - 18

Draw the FBD and KD of the pendulum (mbar ~ 0).

Bn Bt

man mat

Iαmg

Determine the equation of motion.

B BM I αΣ =

( )2sinmgl I mlθ α− = + sin tmgl I lmaθ α− = +

*Note that you could also do this by using the “moment” from at, and that at = lα

θ l r

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Vector Mechanics for Engineers: Dynamics

Tenth Edition

Group Problem Solving

19 - 19

Find I, set up equation of motion using small angle approximation

Determine the natural frequency

( )2sinmgl I mlθ α− = +

21 , sin2

I mr θ θ= ≈

12mr2 +ml2

!

"#

$

%& θ +mglθ = 0

( )2

22

2

2 212

(9.81)(0.650)(0.250) (0.650)

14.0533.7487 rad/s

n r

n

gll

ω

ω

=+

=+

=

=

Calculate the period

2 1.676 snn

πτ

ω= =

1.676 snτ =

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Vector Mechanics for Engineers: Dynamics

Tenth Edition

Damped Free Vibrations

19 - 20

•  With viscous damping due to fluid friction, :maF =∑ ( )

0=++

=−+−

kxxcxmxmxcxkW st

!!!

!!!δ

•  Substituting x = eλt and dividing through by eλt yields the characteristic equation,

mk

mc

mckcm −⎟

⎠⎞

⎜⎝⎛±−==++

22

220 λλλ

•  Define the critical damping coefficient such that

ncc m

mkmc

mk

mc

ω2202

2===−⎟

⎠

⎞⎜⎝

⎛

•  All vibrations are damped to some degree by forces due to dry friction, fluid friction, or internal friction.

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Vector Mechanics for Engineers: Dynamics

Tenth Edition

Damped Free Vibrations

19 - 21

•  Characteristic equation,

mk

mc

mckcm −⎟

⎠⎞

⎜⎝⎛±−==++

22

220 λλλ

== nc mc ω2 critical damping coefficient

•  Heavy damping: c > cc tt eCeCx 21 21

λλ += - negative roots - nonvibratory motion

•  Critical damping: c = cc ( ) tnetCCx ω−+= 21 - double roots

- nonvibratory motion

•  Light damping: c < cc ( ) ( )tCtCex dd

tmc ωω cossin 212 += −

=⎟⎟⎠

⎞⎜⎜⎝

⎛−=

21

cnd c

cωω damped frequency

Free  Vibra)ons  

Free  Vibra)ons  Solu)on  

Damped  vibra)ons  

Damped  vibra)ons  solu)on