chapter14 oscillation

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14-1

Conceptual Questions

14.1. The period of a block oscillating on a spring is 2 /T m k= . We are told that 1 2 0 sT = . . (a) In this case the mass is doubled: 2 12m m= .

22 2 1

1 1 11

2 / 2 22 /

m kT m mT m mm k

= = = =

So 2 12 2(2 0 s) 2 8 sT T= = . = . . (b) In this case the spring constant is doubled: 2 12k k= .

22 1 1

1 2 11

2 / 122 / 2

m kT k kT k km k

= = = =

So 2 1/ 2 (2 0 s)/ 2 1 4 sT T= = . = . . (c) The formula for the period does not contain the amplitude; that is, the period is independent of the amplitude. Changing (in particular, doubling) the amplitude does not affect the period, so the new period is still 2.0 s. It is equally important to understand what doesnt appear in a formula. It is quite startling, really, the first time you realize it, that the amplitude doesnt affect the period. But this is crucial to the idea of simple harmonic motion. Of course, if the spring is stretched too far, out of its linear region, then the amplitude would matter.

14.2. The period of a simple pendulum is 2 /T L g= . We are told that 1 2 0 sT = . . (a) In this case the mass is doubled: 2 12m m= . However, the mass does not appear in the formula for the period of a pendulum; that is, the period does not depend on the mass. Therefore the period is still 2.0 s. (b) In this case the length is doubled: 2 12L L= .

22 2 1

1 1 11

2 / 2 22 /

L gT L LT L LL g

= = = =

So 2 12 2(2 0 s) 2 8 sT T= = . = . . (c) The formula for the period of a simple small-angle pendulum does not contain the amplitude; that is, the period is independent of the amplitude. Changing (in particular, doubling) the amplitude, as long as it is still small, does not affect the period, so the new period is still 2.0 s. It is equally important to understand what doesnt appear in a formula. It is quite startling, really, the first time you realize it, that the amplitude max( ) doesnt affect the period. But this is crucial to the idea of simple harmonic motion. Of course, if the pendulum is swung too far, out of its linear region, then the amplitude would matter. The amplitude does appear in the formula for a pendulum not restricted to small angles because the small-angle approximation is not valid; but then the motion is not simple harmonic motion.

OSCILLATIONS

14

14-2 Chapter 14


14.3. (a) The amplitude is the maximum displacement obtained. From the graph, 10 cmA = .

(b) From the graph, the period 2 0 sT = . . The angular frequency is thus 2 3 1 rad/sT

= = . .

(c) The phase constant specifies at what point the cosine function starts. From the graph, the starting point looks like 12

A. So at 0,t = 1cos2

A A= means 0 ( 60 )3= . Since the oscillator is moving to the left at 0,t = it is in the

upper half of the circular-motion diagram, and we choose 0 3 = + .

14.4. (a) A position-vs-time graph plots 0( ) cos( )x t A wt = + . The graph of x(t) starts at 12 A and is increasing. So

at 0,t = 0 01 cos2 3

A A= = . We choose 0 3= since the particle is moving to the right, indicating that it is

in the bottom half of the circular-motion diagram. (b) The phase at each point can be determined in the same manner as for part (a). For points 1 and 3, the amplitude is

again 12

A. At point 1, the particle is moving to the right so 1 3 = . At point 3, the particle is moving left, so

3 .3 = + At point 2, the amplitude is A, so 2 2cos 1 0= = .

14.5. (a) A velocity-vs-time graph is a plot of 0( ) sin ( )v t A t= + . At 0,t = the amplitude is max1 12 2 =v A

0sin , = A so 10 1 7sin or 2 6 6

= = .

Since ( 0 s)v t = is increasing, we need 0sin ( )t + increasing at

0,t = so we choose 076

= .

(b) The phase is 0wt= + . At points 1 and 3, max1( ) ,2v t v= so 1 1 5sin or

2 6 6

= = . 2 At point 1, we need

1sin , increasing, so 1 56 = . At point 3, 3sin is decreasing, so 3 6

= . At point 2, the amplitude is max ,v so

2sin 1, = thus 2 .2 =

14.6. Energy is transferred back and forth between all potential energy at the extremes 212

kA and all kinetic

energy at the equilibrium point(s) 2max12

mv . The equation does not say that the particle ever has amplitude A and speed maxv at the same time. The equation relates expressions for the energy at two different times.

14.7. (a) The equilibrium length is 20 cm since that is where the potential energy 21 ( ) 02 e

U k x x= = .

(b) The turning points occur where the potential energy is equal to the total energy. These points are at 14 cmx = and 26 cm. (c) The maximum kinetic energy of 7 J is obtained when the potential energy is minimal, at the equilibrium point. (d) The total energy will be around 14 J. If a horizontal line is drawn at that energy, it intersects the potential energy curve at the turning points 12 cmx = and 28 cm.

Oscillations 14-3


14.8. One expression for the total energy is 21 ,2

E kA= which occurs at the turning points when ( ) 0v t = . If the total

energy is doubled, 2 21 12 ( ) ( 2 )

2 2E k A k A= =

Thus the new amplitude is 2 2(20 cm) 28 cmA A = = = .

14.9. One expression for the total energy is 2max1 ,2

E mv= which occurs at the equilibrium point when 0U = . If the

total energy is doubled, 2 2

max max1 12 ( ) ( 2 )2 2

E m v m v= =

Thus the new maximum speed is max max2 2(20 cm/s) 28 cm/sv v = = =

14.10. The time constant mb

= decreases as b increases, meaning energy is removed from the oscillator more

quickly since /0( ) .tE t E e =

(a) The medium is more resistive. (b) The oscillations damp out more quickly. (c) The time constant is decreased.

14.11. (a) T, the period, is the time for each cycle of the motion, the time required for the motion to repeat itself. , the damping time constant, is the time required for the amplitude of a damped oscillator to decrease to 1 37%e

of its original value. (b) , the damping time constant, is the time required for the amplitude of a damped oscillator to decrease to

1 37%e of its original value, while 1/2,t the half-life, is the time required for the amplitude of a damped oscillator to decrease to 50% of its original value.

14.12. Natural frequency is the frequency that an oscillator will oscillate at on its own. You may drive an oscillator at a frequency other than its natural frequency. For example, if you are pushing a child in a swing, you build up the amplitude of the oscillation by driving the oscillator at its natural frequency. You can achieve resonance by driving an oscillator at its natural frequency.

Exercises and Problems

Section 14.1 Simple Harmonic Motion

14.1. Solve: The frequency generated by a guitar string is 440 Hz. The period is the inverse of the frequency, hence 31 1 2 27 10 s 2 27 ms

440 HzT

f

= = = . = .

14.2. Model: The air-track glider oscillating on a spring is in simple harmonic motion. Solve: The glider completes 10 oscillations in 33 s, and it oscillates between the 10 cm mark and the 60 cm mark.

(a) 33 s 3 3 s/oscillation 3 3 s10 oscillations

T = = . = .

(b) 1 1 0 303 Hz 0 30 Hz3 3 s

fT

= = = . .

.

(c) 2 2 0 303 Hz 1 90 rad/sf= = ( . ) = .

14-4 Chapter 14


(d) The oscillation from one side to the other is equal to 60 cm 10 cm 50 cm 0 50 m = = . . Thus, the amplitude is 12 (0 50 m) 0 25 mA = . = . .

(e) The maximum speed is

max2 (1 90 rad/s)(0 25 m) 0 48 m/sv A AT

= = = . . = .

14.3. Model: The air-track glider attached to a spring is in simple harmonic motion. Visualize: The position of the glider can be represented as ( ) cosx t A t= . Solve: The glider is pulled to the right and released from rest at 0 st = . It then oscillates with a period 2 0 sT = . and a maximum speed max 40 cm/s 0 40 m/sv = = . .

(a) maxmax2 2 0 40 m/s and rad/s 0 127 m 12 7 cm 13 cm

2 0 s rad/svv A A

T.

= = = = = = = . = . =.

(b) The gliders position at 0 25 st = . is 0 25 s (0 127 m)cos[( rad/s)(0 25 s)] 0 090 m 9 0 cmx . = . . = . = .

Section 14.2 Simple Harmonic Motion and Circular Motion

14.4. Model: The oscillation is the result of simple harmonic motion. Solve: (a) The amplitude 20 cmA = . (b) The period 4 0 s,T = . thus

1 1 0 25 Hz4 0 s

fT

= = = .

.

(c) The position of an object undergoing simple harmonic motion is 0( ) cos( )x t A t= + . At 00 s, 10 cmt x= = . Thus, 1 1

0 010 cm 110 cm (20 cm)cos cos cos rad 60 .20 cm 2 3

= = = = = Because the object is moving to the right at 0 s,t = it is in the lower half of the circular motion diagram and thus

must have a phase constant between and 2 radians. Therefore, 0 rad 603 = = .

14.5. Model: The oscillation is the result of simple harmonic motion. Solve: (a) The amplitude 10 cmA = . (b) The time to complete one cycle is the period, hence 2 0 sT = . and

1 1 0 50 Hz2 0 s

fT

= = = .

.

(c) The position of an object undergoing simple harmonic motion is 0( ) cos( )x t A t= + . 0At 0 s, 5 cm,t x= = thus 0

10 0

5 cm (10 cm)cos[ (0 s) ]

5 cm 1 1 2cos cos rad or 12010 cm 2 2 3

= +

= = = =

Since the oscillation is originally moving to the left, 0 120 = + .

14.6. Visualize: The phase constant 23 has a plus sign, which implies that the object undergoing simple harmonic motion is in the second quadrant of the circular motion diagram. That is, the object is moving to the left. Solve: The position of the object is

20 0 3( ) cos( ) cos(2 ) (4 0 cm)cos[(4 rad/s) rad]x t A t A ft t = + = + = . +

The amplitude is 4 cmA = and the period is 1/ 0 50 sT f= = . . A phase constant 0 2 /3 rad 120 = = (second quadrant) means that x starts at 12 A and is moving to the left (getting more negative).

Oscillations 14-5


Assess: We can see from the graph that the object starts out moving to the left.

14.7. Visualize: A phase constant of 2

implies that the object that undergoes simple harmonic motion is in the

lower half of the circular motion diagram. That is, the object is moving to the right. Solve: The position of the object is given by the equation

0 0( ) cos( ) cos(2 ) (8 0 cm)cos rad/s rad2 2x t A t A ft t = + = + = .

The amplitude is 8 0 cmA = . and the period is 1/ 4 0 sT f= = . . With 0 /2 rad, = x starts at 0 cm and is moving to the right (getting more positive).

Assess: As we see from the graph, the object starts out moving to the right.

14.8. Solve: The position of the object is given by the equation

0 0( ) cos( ) cos(2 )x t A t A ft= + = + We can find the phase constant 0 from the initial condition:

1 10 0 0 20 cm (4 0 cm)cos cos 0 cos (0) rad

= . = = = Since the object is moving to the right, the object is in the lower half of the circular motion diagram. Hence,

10 2 rad = . The final result, with 4 0 Hz,f = . is

12( ) (4 0 cm)cos[(8 0 rad/s) rad]x t t = . .

14-6 Chapter 14


14.9. Solve: The position of the object is given by the equation

0( ) cos( )x t A t= + The amplitude is 8 0 cmA = . . The angular frequency 2 2 0 50 Hz rad/sf = = ( . ) = . Since at 0t = it has its most negative position, it must be about to move to the right, so 0 = . Thus

( ) 8 0 cm cos[( rad/s) rad]x t t= ( . )

14.10. Model: The air-track glider is in simple harmonic motion. Solve: (a) We can find the phase constant from the initial conditions for position and velocity:

0 0 0 0cos sinxx A v A= = Dividing the second by the first, we see that

0 00

0 0

sin tancos

xvx

= =

The glider starts to the left 0( 5 00 cm)x = . and is moving to the right 0( 36 3 cm/s)xv = + . . With a period of 321 5 s s,. =

the angular frequency is 432 / rad/sT= = . Thus

1 1 20 3 3

36 3 cm/stan rad (60 ) or rad ( 120 )(4 /3 rad/s)( 5 00 cm)

.= =

.

The tangent function repeats every 180, so there are always two possible values when evaluating the arctan function. We can distinguish between them because an object with a negative position but moving to the right is in the third quadrant of the corresponding circular motion. Thus 20 3 rad, = or 120 . (b) At time t, the phase is 4 20 3 3( rad/s) radt t = + = . This gives 23 rad,= 0 rad, 23 rad, and 43 rad at, respectively, 0 s,t = 0.5 s, 1.0 s, and 1.5 s. This is one period of the motion.

Section 14.3 Energy in Simple Harmonic Motion

Section 14.4 The Dynamics of Simple Harmonic Motion

14.11. Model: The block attached to the spring is in simple harmonic motion. Solve: The period of an object attached to a spring is

02 2 0 smT Tk

= = = .

where m is the mass and k is the spring constant. (a) For mass 2 ,m=

022 ( 2) 2 8 smT Tk

= = = .

(b) For mass 12 ,m

12

02 / 2 1 41 sm

T Tk

= = = .

(c) The period is independent of amplitude. Thus 0 2 0 sT T= = . (d) For a spring constant 2 ,k=

02 / 2 1 41 s2mT Tk

= = = .

Oscillations 14-7


14.12. Model: The air-track glider attached to a spring is in simple harmonic motion. Solve: Experimentally, the period is (12 0 s)/(10 oscillations) 1 20 sT = . = . . Using the formula for the period,

2 22 22 (0 200 kg) 5 48 N/m1 20 s

mT k mk T

= = = . = .

.

14.13. Model: The mass attached to the spring oscillates in simple harmonic motion. Solve: (a) The period 1/ 1/2 0 Hz 0 50 sT f= = . = . .

(b) The angular frequency 2 2 2 0 Hz 4 rad/sf= = ( . ) = . (c) Using energy conservation

2 2 21 1 10 02 2 2 xkA kx mv= +

Using 0 5 0 cm,x = . 0 30 cm/sxv = and 2 2(0 200 kg)(4 rad/s) ,k m= = . we get 5 54 cmA = . .

(d) To calculate the phase constant 0, 0 0

10

cos 5 0 cm5 0 cmcos 0 45 rad

5 54 cm

A x

= = .

. = = . .

(e) The maximum speed is max (4 rad/s)(5 54 cm) 70 cm/sv A= = . = .

(f) The maximum acceleration is 2 2

max (4 rad/s)(70 cm/s) 8 8 m/sa A A= = ( ) = = . (g) The total energy is 2 21 1max2 2 (0 200 kg)(0 70 m/s) 0 049 JE mv= = . . = . .

(h) The position at 0 40 st = . is

0 4 s (5 54 cm)cos[(4 rad/s)(0 40 s) 0 45 rad] 3 8 cmx . = . . + . = + .

14.14. Model: The oscillating mass is in simple harmonic motion. Solve: (a) The amplitude 2 0 cmA = . . (b) The period is calculated as follows:

2 210 rad/s 0 63 s10 rad/s

TT

= = = = . (c) The spring constant is calculated as follows:

2 2(0 050 kg)(10 rad/s) 5 0 N/mk k mm

= = = . = .

(d) The phase constant 10 4 rad= . (e) The initial conditions are obtained from the equations

1 14 4( ) (2 0 cm)cos(10 ) and ( ) (20 0 cm/s)sin (10 )xx t t v t t= . = .

At 0 s,t = these equations become 1 1

0 04 4(2 0 cm)cos( ) 1 41 cm and (20 cm/s)sin ( ) 14 1cm/sxx v= . = . = = .

In other words, the mass is at 1 41 cm+ . and moving to the right with a velocity of 14.1 cm/s. (f) The maximum speed is max (2 0 cm)(10 rad/s) 20 cm/sv A= = . = .

(g) The total energy 2 2 31 12 2 (5 0 N/m)(0 020 m) 1 00 10 JE kA

= = . . = . .

(h) At 0 41 s,t = . the velocity is 1

0 4(20 cm/s)sin[(10 rad/s)(0 40 s) ] 1 46 cm/sxv = . = .

14-8 Chapter 14


14.15. Model: The block attached to the spring is in simple harmonic motion. Visualize:

Solve: (a) The conservation of mechanical energy equation f sf i siK U K U+ = + is 2 2 2 2 21 1 1 1 11 0 02 2 2 2 2

0

( ) 0 J 0 J 0 J

1 0 kg (0 40 m/s) 0 10 m 10 cm16 N/m

mv k x mv kA mv

mA vk

+ = + + = +. = = . = . =

(b) We have to find the velocity at a point where /2x A= . The conservation of mechanical energy equation 2 s2 i siK U K U+ = + is

22 2 2 2 2 2 2 22 0 2 0 0 0 0

2 0

1 1 1 1 1 1 1 1 1 1 3 10 J2 2 2 2 2 2 4 2 2 4 2 4 2

3 3 (0 40 m/s) 0 346 m/s4 4

Amv k mv mv mv kA mv mv mv

v v

+ = + = = =

= = . = .

The velocity is 35 cm/s.

Section 14.5 Vertical Oscillations

14.16. Model: The vertical oscillations constitute simple harmonic motion. Visualize:

Oscillations 14-9


Solve: (a) At equilibrium, Newtons first law applied to the physics book is sp

2

( ) 0 N 0 N

/ (0 500 kg)(9 8 m/s )/( 0 20 m) 24 5 N/m 25 N/m

yF mg k y mg

k mg y

= = = = . . . = .

(b) To calculate the period: 24 5 N/m 2 2 rad7 0 rad/s and 0 90 s0 500 kg 7 0 rad/s

k Tm

.

= = = . = = = .

. .

(c) The maximum speed is max 0 10 m 7 0 rad/s 0 70 m/sv A= = ( . )( . ) = .

Maximum speed occurs as the book passes through the equilibrium position.


Solve: The period and angular frequency are 20 s 2 20 6667 s and 9 425 rad/s

30 oscillations 0 6667 sT

T= = . = = = .

.

(a) The mass can be found as follows:

2 215 N/m 0 169 kg 0 17 kg

(9 425 rad/s)k kmm

= = = = . ..

(b) The maximum speed max (9 425 rad/s)(0 060 m) 0 57 m/sv A= = . . = . .

14.18. Model: The vertical oscillations constitute simple harmonic motion. Solve: To find the oscillation frequency using 2 / ,f k m = = we first need to find the spring constant k. In equilibrium, the weight mg of the block and the spring force k L are equal and opposite. That is, /mg k L k mg L= = . The frequency of oscillation f is thus given as

21 1 / 1 1 9 8 m/s 3 5 Hz2 2 2 2 0 020 m

k mg L gfm m L

.= = = = = .

.

Section 14.6 The Pendulum

14.19. Model: Assume a small angle of oscillation so there is simple harmonic motion. Solve: The period of the pendulum is

00 2 4 0 s

LTg

= = .

14-10 Chapter 14


(a) The period is independent of the mass and depends only on the length. Thus 0 4 0 sT T= = . . (b) For a new length 02 ,L L=

00

22 2 5 7 sLT Tg

= = = .

(c) For a new length 0/2,L L=

00

/2 12 2 8 s2

LT Tg

= = = .

(d) The period is independent of the amplitude as long as there is simple harmonic motion. Thus 4 0 sT = . .

14.20. Model: Assume the small-angle approximation so there is simple harmonic motion. Solve: The period is 12 s/10 oscillations 1 20 sT = = . and is given by the formula

2 221 20 s2 (9 8 m/s ) 36 cm

2 2L TT L gg

. = = = . =

14.21. Model: Assume a small angle of oscillation so there is simple harmonic motion. Solve: (a) On the earth the period is

earth 21 0 m2 2 2 0 s

9 80 m/sLTg

.

= = = .

.

(b) On Venus the acceleration due to gravity is 11 2 2 24

2VenusVenus 2 6 2

Venus

Venus 2Venus

(6 67 10 N m /kg )(4 88 10 kg) 8 86 m/s(6 06 10 m)

1 0 m2 2 2 1 s8 86 m/s

GMgR

LTg

. . = = = .

.

. = = = ..

14.22. Model: Assume the pendulum to have small-angle oscillations. In this case, the pendulum undergoes simple harmonic motion. Solve: Using the formula 2/ ,g GM R= the periods of the pendulums on the moon and on the earth are

2 2earth earth moon moon

earth moonearth moon

2 2 and 2L L R L RT Tg GM GM

= = =

Because earth moon ,T T= 22 2

earth earth moon moon moon earthmoon earth

earth moon earth moon222 6

24 6

2 2

7 36 10 kg 6 37 10 m (2 0 m) 33 cm5 98 10 kg 1 74 10 m

L R L R M RL LGM GM M R

= =

. .

= . = . .

14.23. Model: Assume a small angle of oscillation so that the pendulum has simple harmonic motion. Solve: The time periods of the pendulums on the earth and on Mars are

earth Marsearth Mars

2 and 2L LT Tg g

= =

Dividing these two equations, 2 2

2 2earth Mars earthMars earth

Mars earth Mars

1 50 s(9 8 m/s ) 3 67 m/s2 45 s

T g Tg gT g T

. = = = . = .

.

Oscillations 14-11


14.24. Model: This is not a simple pendulum, but is a physical pendulum. Model the bar as thin enough to use 21

3I ML= where L is the full length of the rod.

Visualize: For a small-angle physical pendulum 2 IMglT = where l is the distance from the pivot to the center of

mass. Note that = /2.l L We seek L. Solve: Substitute in for I and then solve for L.

2 21 23 3 22 2 2 2

/2 3ML LI LT

Mgl MgL gL g= = = =

2 22 2

2 23 3 3 (9.8 m/s ) (1.2 s) 0.54 m 54 cm

2 2 8 8T g gL T

= = = = =

Assess: It seems reasonable that a uniform bar 54 cm long would have a period of 1.2 s.

Section 14.7 Damped Oscillations

Section 14.8 Driven Oscillations and Resonance

14.25. Model: The spider is in simple harmonic motion. Solve: Your tapping is a driving frequency. Largest amplitude at ext 1 0 Hzf = . means that this is the resonance frequency, so 0 ext 1 0 Hzf f= = . . That is, the spiders natural frequency of oscillation 0f is 1.0 Hz and 0 02 f= =

2 rad/s. We have

2 20 0 (0 0020 kg)(2 rad/s) 0 079 N/m

k k mm

= = = . = .

14.26. Model: The motion is a damped oscillation. Solve: The amplitude of the oscillation at time t is given by Equation 14.58: /20( ) ,

tA t A e= where /m b= is the time constant. Using 0 368 Ax = . and 10 0 s,t = . we get

10.0 s/2 10 s 10.0 s0.368 ln(0.368) 5.00 s2 2ln(0.368)

A Ae

= = = =

14.27. Model: The motion is a damped oscillation. Solve: The position of a damped oscillator is ( /2 ) 0( ) cos( )

tx t Ae t= + . The frequency is 1.0 Hz and the damping time constant is 4.0 s. Let us assume 0 0 = rad and 1A = with arbitrary units. Thus,

t/(8 0 s) 0 125 t( ) cos[2 1 0 Hz ] ( ) cos(2 )x t e t x t e t . .= ( . ) = where t is in s. Values of x(t) at selected values of t are displayed in the following table:

t(s) x(t) t(s) x(t) t(s) x(t) 0 1 2.00 0.779 6.00 0.472 0.25 0 2.50 0.732 6.50 0.444 0.50 0.939 3.00 0.687 7.00 0.417 0.75 0 3.50 0.646 7.50 0.392 1.00 0.882 4.00 0.607 8.00 0.368 1.25 0 4.50 0.570 8.50 0.346 1.50 0.829 5.00 0.535 9.00 0.325 1.75 0 5.50 0.503 9.50 0.305

10.00 0.286

14-12 Chapter 14


14.28. Model: The pendulum is a damped oscillator. Solve: The period of the pendulum and the number of oscillations in 4 hours are calculated as follows:

osc215 0 m 4(3600 s)2 2 7 773 s 1850

7 773 s9 8 m/sLT Ng

.

= = = . = =..

The amplitude of the pendulum as a function of time is /2( ) bt mA t Ae= . The exponent of this expression can be calculated to be

(0.010 kg/s)(4 3600 s) 0.65452 2(110 kg)

= =

btm

We have 0 6545( ) 1 50 m 0 78 mA t e .= ( . ) = . . 14.29. Model: Assume the eye is a simple driven oscillator. Visualize: Given the mass and the resonant frequency, we can determine the effective spring constant using the relationship 2 /f k m= = . Solve: Solving the above expression for the spring constant, obtain

2 2 3(2 ) [2 29 Hz ] (7 5 10 kg) 249 N/m 250 N/mk f m = = ( ) . = Assess: As spring constants go, this is a fairly large value, however the musculature holding the eyeball in the socket is strong and hence will have a large effective spring constant.

14.30. Solve: The position and the velocity of a particle in simple harmonic motion are 0 0 max 0( ) cos( ) and ( ) sin ( ) sin ( )xx t A t v t A t v t = + = + = +

From the graph, 12 sT = and the angular frequency is 2 2 rad/s

12 s 6T= = =

(a) Because max 60 cm/s,v A= = we have 60 cm/s 60 cm/s 115 cm

/6 rad/sA = = =

(b) At 0 s,t =

0 0 0

1 510 6 6

sin 30 cm/s 60 cm/s sin 30 cm/s

sin (0 5 rad) rad (30 ) or rad (150 )

xv A

= = ( ) = = . =

Because the velocity at 0 st = is negative and the particle is slowing down, the particle is in the second quadrant of the circular motion diagram. Thus 50 6 rad = . (c) At 0 s,t = 50 6(115 cm)cos( rad) 100 cmx = = .

Oscillations 14-13


14.31. Solve: The position and the velocity of a particle in simple harmonic motion are 0 0 max 0( ) cos( ) and ( ) sin( ) sin ( )xx t A t v t A t v t= + = + = +

(a) At 0 s,t = the equation for x yields 1 2

0 0 3( 5 0 cm) (10 0 cm)cos( ) cos ( 0 5) rad

. = . = . = Because the particle is moving to the left at 0 s,t = it is in the upper half of the circular motion diagram, and the phase constant is between 0 and radians. Thus, 20 3 rad= . (b) The period is 4.0 s. At 0 s,t =

0 02 2sin (10 0 cm) sin 13 6 cm/s

3xv A

T

= = . = . (c) The maximum speed is

max2 (10 0 cm) 15 7 cm/s

4 0 sv A = = . = .

.

Assess: The negative velocity at 0 st = is consistent with the position-vs-time graph and the positive sign of the phase constant.

14.32. Model: The vertical mass/spring systems are in simple harmonic motion. Solve: Spring/mass A undergoes three oscillations in 12 s, giving it a period A 4 0 sT = . . Spring/mass B undergoes 2 oscillations in 12 s, giving it a period B 6 0 sT = . . We have

A B A A BA B

A B B B A

4 0 s 22 and 26 0 s 3

m m T m kT Tk k T m k

.= = = = =

.

If A B,m m= then

B A

A B

4 9 2 259 4

k kk k

= = = .

14.33. Solve: The objects position as a function of time is 0( ) cos( )x t A t= + . Letting 0 mx = at 0 s,t = gives 1

0 0 20 cosA= = Since the object is traveling to the right, it is in the lower half of the circular motion diagram, giving a phase constant between and 0 radians. Thus, 10 2= and

1 12 2( ) cos( ) ( ) sin (0 10 m)sin ( )x t A t x t A t t= = = .

where we have used 0 10 mA = . and 2 2 rad rad/s

4 0 s 2T= = =

.

Let us now find t where 0 060 m:= .x 12 0 060 m0 060 m (0 10 m)sin sin 0 41 s

2 0 10 mt t . . = . = = .

.

Assess: The answer is reasonable because it is approximately 18 of the period.

14.34. Model: The block attached to the spring is in simple harmonic motion. Visualize: The position and the velocity of the block are given by the equations

0 0( ) cos( ) and ( ) sin ( )xx t A t v t A t = + = + Solve: (a) To graph x(t) we need to determine , 0, and A. These quantities will be found by using the initial ( 0 s)t = conditions on x(t) and ( )xv t . The period is

1 0 kg 2 2 rad2 2 1 405 s 1 4 s 4 472 rad/s20 N/m 1 405 s

mTk T

.

= = = . . = = = . .

14-14 Chapter 14


At 0 s,t = 0 0 0 0cos and sinxx A v A= = . Dividing these equations, 0

0 00

( 1 0 m s)tan 1 1181 0 841 rad(4 472 rad/s)(0 20 m)

xv /x

.

= = = . = .. .

From the initial conditions, 2 2

2 200

1 0 m/s(0 20 m) 0 30 m4 472 rad/s

xvA x . = + = . + = . .

14.35. Model: The astronaut attached to the spring is in simple harmonic motion. Solve: (a) From the graph, 3 0 s,T = . so we have

2 23 0 s2 (240 N/m) 55 kg2 2

m TT m kk

. = = = =

(b) Oscillations occur about an equilibrium position of 1.0 m. From the graph, 1 02 (0 80 m) 0 40 m, 0 rad, andA = . = . = 2 2 2 1rad/s

3 0 sT= = = .

.

The equation for the position of the astronaut is ( ) cos 1 0 m (0 4 m) cos[(2 1 rad/s) ] 1 0 m

1 2 m (0 4 m)cos[(2 1 rad/s) ] 1 0 m cos[(2 1 rad/s) ] 0 5 0 50 sx t A t t

t t t= + . = . . + .

. = . . + . . = . = .

The equation for the velocity of the astronaut is

0 5 s

( ) sin ( )(0 4 m)(2 1 rad/s)sin[(2 1 rad/s)(0 50 s)] 0 73 m/s

xv t A tv

.

=

= . . . . = .

Thus her speed is 0.73 m/s.

14.36. Model: The particle is in simple harmonic motion. Solve: The equation for the velocity of the particle is

( ) (25 cm)(10 rad/s)sin (10 )= xv t t Substituting into 2K U= gives

2 2 2 2

2 22 2

2 2

2 2 2 1

1 1 1( ) 2 ( ) [ (250 cm/s)sin (10 )] [(25 cm)cos(10 )]2 2 2

sin (10 ) (25 cm) 12 2 s100cos (10 ) (250 cm/s)

1 1tan (10 ) 2(10 rad/s) s 2.0 tan 2.0 0.09100 10

= =

= = = = = =

xmv t kx t m t k t

t kmt

t t 6 s

14.37. Model: The spring undergoes simple harmonic motion. Solve: (a) Total energy is 212E kA= . When the displacement is

12 ,x A= the potential energy is

( ) ( )22 2 31 1 1 1 1 12 2 2 4 2 4 4U kx k A kA E K E U E= = = = = = One quarter of the energy is potential and three-quarters is kinetic. (b) To have 12U E= requires

( )2 21 1 1 12 2 2 2 2AU kx E kA x= = = = 14.38. Solve: Average speed is avg /v x t= . During half a period 12( ),t T = the particle moves from x A= to

( 2 )x A x A= + = . Thus

avg max max avg2 4 4 2 2( )/2 2 / 2

x A A Av A v v vt T T

= = = = = = =

Oscillations 14-15


14.39. Model: The ball attached to a spring is in simple harmonic motion. Solve: (a) Let 0 st = be the instant when 0 5 0 cmx = . and 0 20 cm/sv = . The oscillation frequency is

2 5 N/m 5 0 rad/s0 100 kg

km

.

= = = .

.

Solving Equation 14.26 for the amplitude gives 22

2 200

20 cm/s( 5 0 cm) 6 4 cm5 0 rad/s

vA x

= + = . + = . .

(b) The maximum acceleration is 2 2max 160 cm/sa A= = . (c) For an oscillator, the acceleration is most positive max( )a a= when the displacement is most negative

max( )x x A= = . So the acceleration is maximum when 6 4 cmx = . . (d) We can use the conservation of energy between 0 5 0 cmx = . and 1 3 0 cm:x = .

2 2 2 2 2 2 21 1 1 10 0 1 1 1 0 0 12 2 2 2 ( ) 0 283 m/s

kmv kx mv kx v v x xm

+ = + = + = . The speed is 28 cm/s. Because k is known in SI units of N/m, the energy calculation must be done using SI units of m, m/s, and kg.

14.40. Model: The block on a spring is in simple harmonic motion. Solve: (a) The position of the block is given by 0( ) cos( )x t A t= + . Because ( )x t A= at 0 s,t = we have

0 0 rad,= and the position equation becomes ( ) cosx t A t= . At 0 685 s,t = . 3 00 cm cos(0 685 )A. = . and at 0 886 s,t = . 3 00 cm cos(0 886 )A . = . . These two equations give

cos(0 685 ) cos(0 886 ) cos( 0 886 )0 685 0 886 2 00 rad/s

. = . = .

. = . = .

(b) Substituting into the position equation, 3 00 cm3 00 cm cos((2 00 rad/s)(0 685 s)) cos(1 370) 0 20 15 0 cm

0 20A A A A .. = . . = . = . = = .

.

14.41. Model: The oscillator is in simple harmonic motion. Energy is conserved. Solve: The energy conservation equation 1 2E E= is

2 2 2 21 1 1 11 1 2 22 2 2 2

2 2 2 21 1 1 1(0 30 kg)(0 954 m/s) (0 030 m) (0 30 kg)(0 714 m/s) (0 060 m)2 2 2 2

44 48 N/m

mv kx mv kx

k k

k

+ = +

. . + . = . . + .

= .

The total energy of the oscillator is 2 2 2 2

total 1 11 1 1 1(0 30 kg)(0 954 m/s) (44 48 N/m)(0 030 m) 0 1565 J2 2 2 2

E mv kx= + = . . + . . = .

Because 21total max2 ,E mv=

2max max

10 1565 J (0 300 kg) 1 02 m/s2

v v. = . = . Assess: A maximum speed of 1.02 m/s is reasonable.

14.42. Model: The transducer undergoes simple harmonic motion. Solve: Newtons second law for the transducer is

3 8 2restoring max max max40,000 N (0 10 10 kg) 4 0 10 m/sF ma a a

= = . = .

14-16 Chapter 14


Because 2max ,a A= 8 2

5max2 6 2

4 0 10 m/s 1 01 10 m 10 1 m[2 1 0 10 Hz ]

aA . = = = . = .( . )

(b) The maximum velocity is 6 5

max 2 1 0 10 Hz 1 01 10 m 64 m/sv A

= = ( . )( . ) =

14.43. Model: The block attached to the spring is in simple harmonic motion. Solve: (a) The frequency is

1 1 2000 N/m 3 183 Hz2 2 5 0 kg

kfm

= = = .

.

The frequency is 3.2 Hz. (b) From energy conservation,

2 22 200

1 0 m/s(0 050 m) 0 0707 m2 3 183 Hz

vA x . = + = . + = . .

The amplitude is 7.1 cm. (c) The total mechanical energy is

2 21 12 2 (2000 N/m)(0 0707 m) 5 0 JE kA= = . = .

14.44. Model: Model this situation as a simple harmonic oscillator without damping. Visualize: The period is independent of the amplitude. If the oscillator is undamped the amplitude will be constant; the amplitude is just the arbitrary distance from the equilibrium from which the mass was released. The column of amplitude data is not needed.

The period is related to the spring constant however: 2 .mkT =

Solve: The equation 2 24 mkT = leads us to believe that a graph of 2T vs. m would give a straight line graph whose

slope is 24 /k . Note that the period is 1/10 of the times given in the table.

From the spreadsheet we see that the linear fit is very good and that the slope is 26.0997s /kg. 2

22

4slope 4 / 6.5 N/m6.0997s /kg

k k= = =

Assess: Check the units carefully in the last equation. 6.5 N/m is a reasonable spring constant.

Oscillations 14-17


14.45. Model: The block undergoes simple harmonic motion. Visualize:

Solve: (a) The frequency of oscillation is 1 1 10 N/m 1 125 Hz

2 2 0 20 kgkfm

= = = .

.

The frequency is 1.1 Hz. (b) Using conservation of energy, 2 2 2 21 1 1 11 1 0 02 2 2 2 ,mv kx mv kx+ = + we find

2 2 2 2 2 21 0 0 1

0 20 kg( ) ( 0 20 m) ((1 00 m/s) 0 50 m/s )10 N/m

0 2345 m or 23 cm

mx x v vk

.

= + = . + . ( . )= .

(c) At time t, the displacement is 0cos( )x A t= + . The angular frequency is 2 7 071 rad/sf= = . . The amplitude is 2 2

2 200

1 00 m/s( 0 20 m) 0 245 m7 071 rad/s

vA x . = + = . + = . .

The phase constant is 1 10

00 200 mcos cos 2 526 rad or 145

0 245 mxA

. = = = .

. A negative displacement (below the equilibrium point) and positive velocity (upward motion) indicate that the corresponding circular motion is in the third quadrant, so 0 2 526 rad= . . Thus at 1 0 s,t = .

(0 245 m)cos((7 071 rad/s)(1 0 s) 2 526 rad) 0 0409 m 4 09 cmx = . . . . = . = . The block is 4.1 cm below the equilibrium point.

14.46. Model: The mass is in simple harmonic motion. Visualize:

The high point of the oscillation is at the point of release. This conclusion is based on energy conservation. Gravitational potential energy is converted to the springs elastic potential energy as the mass falls and stretches the spring, then the elastic potential energy is converted 100% back into gravitational potential energy as the mass rises, bringing the mass back to exactly its starting height. The total displacement of the oscillationhigh point to low pointis 20 cm. Because the

14-18 Chapter 14


oscillations are symmetrical about the equilibrium point, we can deduce that the equilibrium point of the spring is 10 cm below the point where the mass is released. The mass oscillates about this equilibrium point with an amplitude of 10 cm, that is, the mass oscillates between 10 cm above and 10 cm below the equilibrium point. Solve: The equilibrium point is the point where the mass would hang at rest, with sp GF F mg= = . At the equilibrium

point, the spring is stretched by 10 cm 0 10 my = = . . Hookes law is sp ,F k y= so the equilibrium condition is 2

2sp G

9 8 m/s[ ] [ ] 98 s0 10 m

k gF k y F mgm y

.

= = = = = = .

2

The ratio /k m is all we need to find the oscillation frequency:

21 1 98 s 1 58 Hz 1 6 Hz2 2

kfm

= = = . .

14.47. Model: The spring is ideal, so the apples undergo simple harmonic motion. Solve: The spring constant of the scale can be found by considering how far the pan goes down when the apples are added.

20 N 222 N/m0 090 m

mg mgL kk L

= = = = .

The frequency of oscillation is

21 1 222 N/m 1 66 Hz 1 7 Hz

2 2 (20 N/9 8 m/s )kfm

= = = . .

.

Assess: An oscillation of fewer than twice per second is reasonable.

14.48. Model: The compact car is in simple harmonic motion. Solve: (a) The mass on each spring is (1200 kg)/4 300 kg= . The spring constant can be calculated as follows:

2 2 2 2 4(2 ) (300 kg)[2 (2 0 Hz)] 4 74 10 N/mk k m m fm

= = = = . = . The spring constant is 44 7 10 N/m. . (b) The car carrying four persons means that each spring has, on the average, an additional mass of 70 kg. That is,

300 kg 70 kg 370 kgm = + = . Thus, 41 1 4 74 10 N/m 1 8 Hz

2 2 2 370 kgkfm

. = = = = .

Assess: A small frequency change from the additional mass is reasonable because frequency is inversely proportional to the square root of the mass.

14.49. Model: Assume simple harmonic motion for the two-block system without the upper block slipping. We will also use the model of static friction between the two blocks. Visualize:

Solve: The net force on the upper block 1m is the force of static friction due to the lower block 2m . The two blocks ride together as long as the static friction doesnt exceed its maximum possible value. The model of static friction gives the maximum force of static friction as

s max s s 1 1 max max s2 22

max max maxs 2

( ) ( )

2 2 0 40 m 0 721 5 s 9 8 m/s

f n m g m a a g

a A Ag g T g

= = = = . = = = = = .

. .

Assess: Because the period is given, we did not need to use the block masses or the spring constant in our calculation.

Oscillations 14-19


14.50. Model: The DNA and cantilever undergo simple harmonic motion. Solve: The cantilever has the same spring constant with and without the DNA molecule. The frequency of oscillation without the DNA is

1 13

kM

=

With the DNA, the frequency of oscillation is

2 13

kM m

=

+

where m is the mass of the DNA. Divide the two equations, and express 2 1 ,= where 2 2 (50 Hz)f = = .

( )( )

1 131 1 1 3

12 2 1 3

13

kM M mf fkf f f M

M m

+= = = =

+

Thus

( ) ( )2 21 11 13 32 2 2

1 1 11 13 2 3 2

1 1

( )

( ( ) ) 1( ) ( )

f M f f M m

f f f fm M Mf f f f

= +

= =

Since 1,f f (50 Hz 12 MHz),2

2 2 21 1 1

1 1( ) 1 1 2f ff f f f

f f

= + .

Thus

1 23 3

1 11 2 1f fm M M

f f

= + =

The mass of the cantilever 3 9 9 16(2300 kg/m )(4000 10 m)(100 10 m) 3 68 10 kgM = = .

Thus the mass of the DNA molecule is

16 216

2 50 Hz(3 68 10 kg) 1 02 10 kg3 12 10 Hz

m = . = .

Assess: The mass of the DNA molecule is about 56 2 10. atomic mass units, which is reasonable for such a large molecule.

14.51. Model: Assume that the swinging lamp makes a small angle with the vertical so that there is simple harmonic motion. Visualize:

14-20 Chapter 14


Solve: (a) Using the formula for the period of a pendulum, 2 2

2 5 5 s2 (9 8 m/s ) 7 5 m2 2

L TT L gg

. = = = . = .

(b) The conservation of mechanical energy equation 0 g0 1 g1K U K U+ = + for the swinging lamp is

2 2 21 1 10 0 1 1 max2 2 2

max

2

0 J 0 J

2 2 ( cos3 )

2(9 8 m/s )(7 5 m)(1 cos3 ) 0 45 m/s

mv mgy mv mgy mgh mv

v gh g L L

+ = + + = + = =

= . . = .

14.52. Model: Assume that the angle with the vertical that the pendulum makes is small enough so that there is simple harmonic motion. Solve: The angle made by the string with the vertical as a function of time is

max 0( ) cos( )t t= + The pendulum starts from maximum displacement, thus 0 0= . Thus, max( ) cost t= . To find the time t when the pendulum reaches 4.0 on the opposite side:

1( 4 0 ) 8 0 cos cos ( 0 5) 2 094 radt t . = ( . ) = . = . Using the formula for the angular frequency,

29 8 m/s 2 0944 rad 2 094 rad3 130 rad/s 0 669 s1 0 m 3 130 rad/s

g tL

. . .

= = = . = = = .. .

The time 0 67 st = . . Assess: Because 2 / 2 0 s,T = = . a value of 0.67 s for the pendulum to cover a little less than half the oscillation is reasonable.

14.53. Model: Assume the orangutan is a simple small-angle pendulum. Visualize: One swing of the arm would be half a period of the oscillatory motion. The horizontal distance traveled in that time would be 2(0 90m)sin (20 ) 0 616m. = . from analysis of a right triangle. Solve:

22 2 0 90 m 0 952s

2 2 2 9 8 m/sT L

g.

= = = .

.

dist 0 616 mspeed 0 65 m/stime 0 952 s

.

= = = .

.

Assess: This isnt very fast, but isnt out of the reasonable range.

14.54. Model: The mass is a particle and the string is massless. Solve: Equation 14.51 is

MglI

=

The moment of inertia of the mass on a string is 2,I Ml= where l is the length of the string. Thus

2Mgl g

lMl= =

This is Equation 14.48 with L l= . Assess: Equation 14.48 is really a specific case of the more general physical pendulum described by Equation 14.51.

14.55. Model: The rod is thin and uniform with moment of inertia described in Table 12.2. The clay ball is a particle located at the end of the rod. The ball and rod together form a physical pendulum. The oscillations are small.

Oscillations 14-21


Visualize:

Solve: The moment of inertia of the composite pendulum formed by the rod and clay ball is 2 2

rod ball rod ball rod ball

2 2 3 2

13

1 (0 200 kg)(0 15 m) 0 020 kg 0 15 m 1 95 10 kg m3

I I I m L m L+

= + = +

= . . + ( . )( . ) = .

The center of mass of the rod and ball is located at a distance from the pivot point of

2cm

0 15(0 200 kg) m 0 020 kg 0 15 m2 8 18 10 m

(0 200 kg 0 020 kg)y

. . + ( . )( . )

= = . . + .

The frequency of oscillation of a physical pendulum is 2 2

3 21 1 (0 220 kg)(9 8 m/s )(8 18 10 m) 1 51 Hz

2 2 1 95 10 kg mMglf

I

. . . = = = .

.

The period of oscillation 1Tf

= = 0.66 s.

14.56. Model: Model the rod as a small-angle physical pendulum without damping. Visualize: Figure 14.22 in the chapter shows that l is the distance from the pivot to the center of mass; in this case

/4.l L= Solve: We first need to compute the moment of inertia of a thin rod about an axis 1/4 of the way from one end. Use the parallel-axis theorem to do so:

22 2 2 2

c.m.1 1 1 7

12 4 12 16 48LI I Md ML M ML ML = + = + = + =

For the physical pendulum

( )427

48

1 1 1 12 1 3= =2 2 2 7 7

LMgMgl g gfI L LML

= =

Assess: The period 1/T f= is a little shorter than swinging the rod from the end, which is what we expect.

14.57. Model: Model the sphere as a small-angle physical pendulum without damping. Visualize: Figure 14.22 in the chapter shows that l is the distance from the pivot to the center of mass; in this case l = R. Solve: We first need to compute the moment of inertia of a sphere about an axis tangent to the edge. Use the parallel-axis theorem to do so:

2 2 2 2c.m.

2 75 5

I I Md MR MR MR= + = + =

For the physical pendulum

275

1 1 1 52 2 2 7

Mgl MgR gfI RMR

= = =

Assess: The frequency is less than for a simple pendulum of length R as we expect.

14-22 Chapter 14


14.58. Model: Model this situation as a small-angle physical pendulum without damping. The moment of inertia for a uniform meter stick pivoted on one end is 213 .ML

Visualize: The frequency and length of a physical pendulum are related by 2 ,Mgl

If 1

= where /2.l L=

Solve: Square both sides of the equation above and substitute in for I.

2 22 2 2 2 21

3

1 1 3 3 14 4 8 8

LMgMgl g gfI L LML

= = = =

This leads us to believe that a graph of 2f vs. 1/L would produce a straight line whose slope is 23

8.g

From the spreadsheet we see that the linear fit is excellent and that the slope is 20.3707m/s .

2 22

23 (8 )(0.3707 m/s )slope 9.76 m/s

38g g= = =

Assess: If the uncertainties in our measurements were small enough then our answer would show that g varies slightly from place to place on the earth.

14.59. Model: Treat the lower leg as a physical pendulum. Visualize: We can determine the moment of inertia by combining 2 /T I mgL= and 1/T f= .

Solve: Combining the above expressions and solving for the moment of inertia we obtain 2 2 2 2 2/(2 ) (5 0 kg)(9 80 m/s )(0 18 m)/[2 (1 6 Hz)] 8 7 10 kg mI mgL f = = . . . . = .

Assess: NASA determines the moment of inertia of the shuttle in a similar manner. It is suspended from a heavy cable, allowed to oscillate about its vertical axis of symmetry with a very small amplitude, and from the period of oscillation one may determine the moment of inertia. This arrangement is called a torsion pendulum.

14.60. Model: A completely inelastic collision between the two gliders resulting in simple harmonic motion.

Oscillations 14-23


Visualize:

Let us denote the 250 g and 500 g masses as 1m and 2,m which have initial velocities i1v and i2v . After 1m collides with and sticks to 2,m the two masses move together with velocity fv . Solve: The momentum conservation equation f ip p= for the completely inelastic collision is 1 2 f 1 i1 2 i2( ) .m m v m v m v+ = + Substituting the given values,

f f(0 750 kg) (0 250 kg)(1 20 m/s) (0 500 kg)(0 m/s) 0 400 m/sv v. = . . + . = . We now use the conservation of mechanical energy equation:

2 21 1s compressed s equilibrium 1 2 f2 2

1 2f

( ) 0 J ( ) 0 J

0 750 kg (0 400 m/s) 0 11 m10 N/m

K U K U kA m m v

m mA vk

+ = ( + ) + = + ++ . = = . = .

The period is

1 2 0 750 kg2 2 1 7 s10 N/m

m mTk+ .

= = = .

14.61. Model: The block attached to the spring is oscillating in simple harmonic motion. Solve: (a) Because the frequency of an object in simple harmonic motion is independent of the amplitude and/or the maximum velocity, the new frequency is equal to the old frequency of 2.0 Hz. (b) The speed 0v of the block just before it is given a blow can be obtained by using the conservation of mechanical energy equation as follows:

2 2 21 1 1max 02 2 2

0 (2 ) (2 )(2 0 Hz)(0 02 m) 0 25 m/s

kA mv mv

kv A A f Am

= =

= = = = . . = .

The blow to the block provides an impulse that changes the velocity of the block: f 0

3f f( 20 N)(1 0 10 s) (0 200 kg) (0 200 kg)(0 25 m/s) 0 150 m/s

x xJ F t p mv mv

v v= = =

. = . . . = . Since fv is the new maximum velocity of the block at the equilibrium position, it is equal to A . Thus,

0 150 m/s 0 150 m/s 0 012 m 1 19 cm 1 2 cm 2 (2 0 Hz)

A . . = = = . = . ..

Assess: Because fv is positive, the block continues to move to the right even after the blow.

14-24 Chapter 14


14.62. Model: Assume the small-angle approximation. Visualize:

Solve: The tension in the two strings pulls downward at angle . Thus Newtons second law is 2 siny yF T ma = =

From the geometry of the figure we can see that

2 2sin y

L y=

+

If the oscillation is small, then y L and we can approximate sin /y L . Since /y L is tan , this approximation is equivalent to the small-angle approximation sin tan if 1 rad. With this approximation, Newtons second law becomes

2 2

2 22 22 sin yT d y d y TT y ma m yL mLdt dt

= = = This is the equation of motion for simple harmonic motion (see Equations 14.32 and 14.46). The constants 2T/mL are equivalent to k/m in the spring equation or g/L in the pendulum equation. Thus the oscillation frequency is

1 22

TfmL

=

14.63. Solve: The potential energy curve of a simple harmonic oscillator is described by 212 ( ) ,U k x= where

0x x x = is the displacement from equilibrium. From the graph, we see that the equilibrium bond length is 0 0.13 nm.x = We can find the bonds spring constant by reading the value of the potential energy U at a displacement x and using the potential energy formula to calculate k.

x (nm) x (nm) U (J) k (N/m)

0.11 0.02 190 8 10 J. 400 0.10 0.03 191 9 10 J. 422 0.09 0.04 193 4 10 J. 425

The three values of k are all very similar, as they should be, with an average value of 416 N/m. Knowing the spring constant, we can now calculate the oscillation frequency of a hydrogen atom on this spring to be

1327

1 1 416 N/m 7 9 10 Hz2 2 1 67 10 kg

kfm

= = = . .

Oscillations 14-25


14.64. Model: Assume that the size of the ice cube is much less than R and that is a small angle. Visualize:

Solve: The ice cube is like an object on an inclined plane. The net force on the ice cube in the tangential direction is 2 2

G 2 2sin sind dF ma mR mR mg mRdt dt

( ) = = = = where is the angular acceleration. With the small-angle approximation sin , this becomes

22

2d g

Rdt= =

This is the equation of motion of an object in simple harmonic motion with a period of 2 2 RT

g= =

14.65. Visualize:

Solve: (a) Newtons second law applied to the penny along the y-axis is

net yF n mg ma= =

netFG

is upward at the bottom of the cycle (positive ),ya so n mg> . The speed is maximum when passing through

equilibrium, but 0ya = so n mg= . The critical point is the highest point. netFG

points down and ya is negative. If

ya becomes sufficiently negative, n drops to zero and the penny is no longer in contact with the surface.

(b) When the penny loses contact ( 0),n = the equation for Newtons law becomes maxa g= . For simple harmonic motion,

22

max9 8 m/s 15 65 rad0 040 m

15 65 rad/s 2 5 Hz2 2

ga AA

f

.

= = = = ..

. = = = .

14-26 Chapter 14



Solve: At the equilibrium position, the net force on mass m on Planet X is: X

net X 0 Nk gF k L mgm L

= = =

For simple harmonic motion 2/ ,k m = thus 2 2

2 2X XX

2 2 2 (0 312 m) 5 86 m/s14 5 s/10

g g g LL L T T

= = = = = . = . .

Assess: Because ordinary tasks seemed easier than on earth we expected an answer less than 29.8 m/s .

14.67. Model: The dolls head is in simple harmonic motion and is damped. Solve: (a) The oscillation frequency is

2 2 21 (2 ) (0 015 kg)(2 ) (4 0 Hz) 9 475 N/m2

kf k m fm

= = = . . = .

The spring constant is 9.5 N/m. (b) Using /20( ) ,

bt mA t A e= we get (4 0 s)/(2 0 015 kg) (133 3 s/kg)(0 5 cm) (2 0 cm) 0 25

133 33 s/kg ln 0 25 0 0104 kg/s 0 010 kg/s

b be eb b

. . .. = . . = ( . ) = . = . .

14.68. Model: The oscillator is in simple harmonic motion. Solve: The maximum displacement at time t of a damped oscillator is

/2 maxmax

( )( ) ln2

t t x tx t AeA

= =

Using max 0 98x A= . at 0 50 s,t = . we can find the time constant to be 0 50 s 12 375 s

2ln(0 98).

= = .

.

25 oscillations will be completed at 25 12 5 st T= = . . At that time, the amplitude will be 12 5 s/(2)(12 375 s)

max, 12 5 s (10 cm) 6 0 cmx e . .

.

= = .

14.69. Model: The vertical oscillations are damped and follow simple harmonic motion. Solve: The position of the ball is given by ( /2 ) 0( ) cos( )

tx t Ae t= + . The amplitude ( /2 )( ) tA t Ae= is a function of time. The angular frequency is

(15 0 N/m) 25 477 rad/s 1 147 s0 500 kg

k Tm

.

= = = . = = ..

Oscillations 14-27


Because the balls amplitude decreases to 3.0 cm from 6.0 cm after 30 oscillations, that is, after 30 1 147 s 34.41 s, . = we have

(34 414 s/2 ) (34 41 s/2 ) 34 41 s3 0 cm (6 0 cm) 0 50 ln(0 50) 25 s 2

e e . . .. = . . = . = =

14.70. Model: The motion is a damped oscillation. Solve: The position of the air-track glider is ( /2 ) 0( ) cos( ),

tx t Ae t= + where /m b= and 2

24k bm m

=

Using 0 20 m,A = . 0 0 rad,= and 0 015 kg/s,b = . 2

42

4 0 N/m (0 015 kg/s) 16 9 10 rad/s 4 0 rad/s0 250 kg 4(0 250 kg)

. .

= = = . . .

Thus the period is 2 2 rad 1 57 s

4 0 rad/sT = = = .

.

The amplitude at 0 st = is 0x A= and the amplitude will be equal to 1e A at a time given by

( /2 )1 2 2 33 3 st mA Ae te b

= = = = . The number of oscillations in a time of 33.3 s is (33 3 s)/(1 57 s) 21. . = .

14.71. Model: The oscillator is in simple harmonic motion. Solve: The maximum displacement, or amplitude, of a damped oscillator decreases as /2max ( ) ,

tx t Ae= where is the time constant. We know max/ 0 60x A = . at 50 s,t = so we can find as follows:

max ( ) 50 sln 48 9 s2 2ln(0 60)t x t

A

= = = . .

Now we can find the time 30t at which max/ 0 30:x A = .

max30

( )2 ln 2(48 9 s)ln(0 30) 118 sx ttA

= = . . =

The undamped oscillator has a frequency 2 Hz 2f = = oscillations per second. Damping changes the oscillation frequency slightly, but the text notes that the change is negligible for light damping. Damping by air, which allows the oscillations to continue for well over 100 s, is certainly light damping, so we will use 2 0 Hzf = . . Then the number of oscillations before the spring decays to 30% of its initial amplitude is

30 (2 oscillations/s) (118 s) 236 oscillationsN f t= = =

14.72. Solve: The solution of the equation 2

2 0d x b dx k x

m dt mdt+ + =

is /2 0( ) cos( ).bt mx t Ae t = + The first and second derivatives of x(t) are

/2 /20 0

2 22 /2

0 02 2

cos( ) sin ( )2

cos( ) sin ( )4

bt m bt m

bt m

dx Ab e t A e tdt m

d x Ab AbA t t emdt m

= + +

= + + +

14-28 Chapter 14


Substituting these expressions into the differential equation, the terms involving 0sin ( )t + cancel and we obtain the simplified result

22

02 cos( ) 04b k t

mm + + =

Because 0cos( )t + is not equal to zero in general, 2 2

22 204 4

b k k bm mm m

+ = =

14.73. Model: The two springs obey Hookes law. Visualize:

Solve: There are two restoring forces on the block. If the blocks displacement x is positive, both restoring forcesone pushing, the other pullingare directed to the left and have negative values:

net sp 1 sp 2 1 2 1 2 eff( ) ( ) ( ) ( )x x xF F F k x k x k k x k x= + = = + =

where eff 1 2k k k= + is the effective spring constant. This means the oscillatory motion of the block under the influence of the two springs will be the same as if the block were attached to a single spring with spring constant effk . The frequency of the blocks, therefore, is

2 2eff 1 2 1 21 22 2

1 12 2 4 4

k k k k kf f fm m m m

+= = = + = +

14.74. Model: The two springs obey Hookes law. Assume massless springs. Visualize: Each spring is shown separately. Note that 1 2x x x = + .

Solve: Only spring 2 touches the mass, so the net force on the mass is 2 on m mF F= . Newtons third law tells us that

2 on on 2m mF F= and that 2 on 1 1 on 2F F= . From net ,F ma= the net force on a massless spring is zero. Thus w on 1F

2 on 1 1 1= = F k x and on 2 1 on 2 2 2mF F k x= = . Combining these pieces of information,

1 1 2 2mF k x k x= =

The net displacement of the mass is 1 2,x x x = + so

1 21 2

1 2 1 2 1 2

1 1m mm m

F F k kx x x F Fk k k k k k

+ = + = + = + =

Oscillations 14-29


Turning this around, the net force on the mass is

1 2 1 2eff eff

1 2 1 2 where m

k k k kF x k x kk k k k

= = =+ +

eff ,k the proportionality constant between the force on the mass and the masss displacement, is the effective spring constant. Thus the masss angular frequency of oscillation is

eff 1 2

1 2

1k k km m k k

= =

+

Using 21 1/k m= and 22 2/k m= for the angular frequencies of either spring acting alone on m, we have

2 21 2 1 2

2 21 2 1 2

( / )( / )( / ) ( / )

k m k mk m k m

= =

+ +

Since the actual frequency f is simply a multiple of , this same relationship holds for :f

2 21 2

2 21 2

f fff f

=

+

14.75. Model: The blocks undergo simple harmonic motion. Visualize:

The length of the stretched spring due to a block of mass m is 1L . In the case of the two-block system, the spring is further stretched by an amount 2L . Solve: The equilibrium equations from Newtons second law for the single-block and double-block systems are

1 1 2( ) and ( ) (2 )L k mg L L k m g = + =

Using 2 5 0 cm,L = . and subtracting these two equations, gives us

1 2 1( ) (2 ) (0 05 m)L L k L k m g mg k mg + = . = 2

29 8 m/s 196 s0 05 m

km

. = =.

With both blocks attached, giving total mass 2m, the angular frequency of oscillation is

21 1196 s 9 90 rad/s2 2 2k km m

= = = = .

Thus the oscillation frequency is /2 1 6 Hzf = = . .

14-30 Chapter 14


14.76. Model: A completely inelastic collision between the bullet and the block resulting in simple harmonic motion. Visualize:

Solve: (a) The equation for conservation of energy after the collision is

2 2b B f f

b B

1 1 2500 N/m( ) (0 10 m) 5 0 m/s2 2 1 010 kg

kkA m m v v Am m

= + = = . = . + .

The momentum conservation equation for the perfectly inelastic collision after beforep p= is

b B f b b B B2

b b

( )

(1 010 kg)(5 0 m/s) (0 010 kg) 1 00 kg 0 m/s 5 0 10 m/s

m m v m v m v

v v

+ = +

. . = . + ( . )( ) = .

(b) No. The oscillation frequency b B/( )k m m+ depends on the masses but not on the speeds.

14.77. Model: The block undergoes simple harmonic motion after sticking to the spring. Energy is conserved throughout the motion. Visualize:

Its essential to carefully visualize the motion. At the highest point of the oscillation the spring is stretched upward. Solve: Weve placed the origin of the coordinate system at the equilibrium position, where the block would sit on the spring at rest. The spring is compressed by L at this point. Balancing the forces requires k L mg = . The

Oscillations 14-31


angular frequency is 2 / / ,w k m g L= = so we can find the oscillation frequency by finding L . The block hits the spring (1) with kinetic energy. At the lowest point (3), kinetic energy and gravitational potential energy have been transformed into the springs elastic energy. Equate the energies at these points:

2 21 11 1g 3s 3g 12 2 ( ) ( )K U U U mv mg L k L A mg A+ = + + = + +

Weve used 1y L= as the block hits and 3y A= at the bottom. The spring has been compressed by y L A = + .

Speed 1v is the speed after falling distance h, which from free-fall kinematics is 21 2v gh= . Substitute this expression

for 21v and /mg L for k, giving

2( ) ( )2( )

mgmgh mg L L A mg AL

+ = + +

The mg term cancels, and the equation can be rearranged into the quadratic equation 2 2( ) 2 ( ) 0L h L A + =

The positive solution is 2 2 2 2(0 030 m) 0 100 m 0 030 m 0 0744 mL h A h = + = . + ( . ) . = .

Now that L is known, we can find 29 80 m/s 11 48 rad/s 1 8 Hz

0 0744 m 2g fL

.

= = = . = = . .

14.78. Model: Assume that the pendulum is restricted to small angles so sin . Visualize: Because the string is massless the center of mass is at the center of the sphere, so l L= in the formula for

a physical pendulum: 2 .IMglT =

Solve: (a) First we use the parallel axis theorem to find I for the sphere around the pivot point at the top (L away):

2 2 2cm .I I Md MR ML25= + = +

2 2 2 22 25 52 2 2

MR ML R LITMgl MgL gL

+ += = =

This reduces to the traditional simple pendulum formula 2 LgT = in the limit L R . (b)

2 225 2 2 2 22 2

5 5real

simple

2 (0.010 m) (1.0 m)1.00002

1.0 m2 Lg

R LR LgLT

T L

+

+ += = = =

Assess: Making the simple pendulum approximation introduces an error of only 0.002% with these numbers.

14-32 Chapter 14


14.79. Model: Assume a simple system with a massless spring: 2 .mkT = Visualize: The simplest way to find T is to consider T as a function of m and use calculus to take the differential

dT and consider that a good stand-in for .T Also note that 222 4 .m mk TT k = =

Solve:

(a) Re-write 2 mkT = as a function of m.

1/22( )T m mk

=

Now take the differential of T with respect to m using the exponent rule and chain rule.

1/22 12

dT m dmk

=

Now cancel the 2s and plug in 224 .m

Tk =

22

1 124 m

T

TdT dm dm dmmk m m

= = =

Or, in the notation given in the problem:

2T mT

m =

(b) The new period, ,T is given by

1 11 (2.000 s) 1 (0.001) 2.001s2 2 2T m mT T T T T

m m

= + = + = + = + =

Assess: We did not expect a 0.1% change in mass to change the period by much.

14.80. Model: The rod is thin and uniform. Visualize:

Solve: We must derive our own equation for this combination of a pendulum and spring. For small oscillations, sFG

remains horizontal. The net torque around the pivot point is

Gcos sin2net sLI F L F = =

With 2

2 ,ddt

=

G ,F mg= s sin ,F k x kL= = and 2

1 ,3

I mL=

2

23 3sin cos sin

2d k g

m Ldt=

Oscillations 14-33


We can use 1sin cos sin 22

= . For small angles, sin and sin 2 2 . So 2

23 3

2d k g

m Ldt

= +

This is the same as Equations 14.32 and 14.46 with 3 3

2k g

m L= +

The frequency of oscillation is thus 21 3(3 0 N/m) 3(9 8 m/s ) 1 73 Hz

2 (0 200 kg) 2(0 20 m)f . .= + = .

. .

The period 1Tf

= = 0.58 s.

Assess: Fewer than two oscillations per second is reasonable. The rods angle from the vertical must be small enough that sin 2 2 . This is more restrictive than other examples, which only require that sin .

chapter14 oscillation

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