Vibrations and Waves€¦ · Vibrations and Waves CLICKER QUESTIONS Question G1.01 Description:...

13Vibrations and Waves

CLICKER QUESTIONS

Question G1.01

Description: Understanding harmonic oscillation frequency.

Question

Two identical springs are attached to two identical masses. Both are resting on a frictionless horizontal surface. One of the springs is displaced a distance d from its equilibrium position, while the other is displaced a distance 2d.

d

Xeq

(A)

(B)

2d

If both masses are released at the same instant, which of the springs, A or B, returns to its unstretched length fi rst?

1. Spring A 2. Spring B 3. It’s a tie 4. Not enough information

Commentary

Purpose: To develop your understanding of what a harmonic oscillator’s frequency means.

Discussion: The period of an oscillating object of mass m attached to a spring of spring constant k is

2π m k . Note that the period does not depend on the amplitude; it takes the same amount of time to complete one large oscillation as one small oscillation. In both cases shown, the block is undergoing one quarter of a complete oscillation, so the time required is one quarter of a period. If the springs are of equal stiffness and the blocks have the same mass, both will reach their equilibrium point at the same time.

Although case B travels a larger distance, it has a larger average velocity and covers that distance more quickly. To stretch the spring twice as far, four times the potential energy is required. Block B will thus have four times as much kinetic energy as block A when they reach their equilibrium points, which means it will be moving twice as fast. Also, the initial spring force on B will be twice as large as on A, so it will accelerate twice as quickly when the blocks are released.

625

56157_13_ch13_p625-656.indd 62556157_13_ch13_p625-656.indd 625 1/4/08 1:48:57 PM1/4/08 1:48:57 PM

626 Chapter 13

Key Points:

• The period and frequency of a simple harmonic oscillator are independent of amplitude.

For Instructors Only

This aspect of simple harmonic oscillator behavior is counterintuitive to most students; they will expect case B to take longer, though not twice as long. A demonstration may help.

You should stress that this is only true for ideal oscillators; real physical springs, pendula, and other oscillators deviate increasingly from ideal behavior as amplitude is increased. Nevertheless, the approximation is good enough for small oscillations that pendulum-based clocks depend upon it for keeping time.

Question G1.02

Description: Understanding what harmonic oscillator frequency depends upon.

Question

Three systems are constructed from identical springs and masses as shown.

21 3

Which statement is true regarding the frequencies of these systems?

1. ω ω ω1 2 3> > 2. ω ω ω1 2 3> < 3. ω ω ω1 2 3< = 4. ω ω ω1 2 3< < 5. ω ω ω1 2 3= < 6. ω ω ω1 2 3= = 7. None of the above. 8. Cannot be determined.


Vibrations and Waves 627

Commentary

Purpose: To develop your understanding of what the frequency of an oscillator depends upon.

Discussion: Consider a plot of net force (along the direction of motion of the block) vs. block position for case 1.

F

0L0

L

The equilibrium point, about which oscillation occurs, is at the spring’s resting length. When the spring is stretched ( L > L

0 ), it pulls back; when compressed ( L < L

0 ), it pushes back; this is a “restoring force” that

causes oscillations about the equilibrium point. The angular frequency of oscillations is related to the slope of the force vs. position plot (−k , where k is the spring’s force constant) and to the block mass m, and has a

value of ω = k m .

Now consider case 2. The net force on the block is due to the spring force and gravity, and a plot of the net force vs. block position (in terms of the spring length, where downward is positive) looks the same as for case 1, except that the entire line has been shifted upward by a constant amount mg.

F

0L0

L

(In the plot, the dotted green line represents the spring force alone, the dotted red line represents the weight, and the solid blue line represents the net force.) We can see that the block’s equilibrium point (where the net force is zero) is different than in case 1, but the slope of the line is the same. This means that gravity will stretch the spring and cause oscillations to occur about a larger length value, but the frequency of

oscillations will be the same: ω = k m.

Case 3 is the same as case 2, except that only one component of the gravitational force mg cosθ( ) affects the blocks’ motion, so the equilibrium point does not shift as much as it does for case 2. The slope of the net force vs. position line is still the same, so the frequency must be the same also.

In other words, for any orientation of the spring, the restoring force can be written as F kd= − , where d is the displacement of the mass from its equilibrium position (not necessarily from the point at which the spring is relaxed). Since the restoring force has the same proportionality constant k, oscillations must have the same frequency.


628 Chapter 13

Key Points:

• The oscillation frequency of an object on a spring does not depend on the orientation of the spring, only on the spring’s stiffness and the object’s mass.

• Changing the orientation of a mass on a spring changes the equilibrium point about which oscillations occur.

• For an object oscillating due to a restoring force of strength F kd= − , where d is the object’s

displacement from equilibrium, the angular frequency of oscillation will be ω = k m .


Intuitively, many students have diffi culty believing that orientation makes no difference. Sometimes, students convince themselves that cases 1 and 2 have the same frequency, but case 3 differs. They want to see a factorof cosθ worked in somewhere. A demonstration, if available, may be useful.

We have not derived the relationship between restoring force and angular frequency of oscillation here; how that is handled depends upon the level of your course and the mathematical sophistication of your students. This question is intended for use after the simple harmonic oscillator, case 1, has been introduced and its angular frequency presented.

Question G3.01

Description: Distinguishing interference from related concepts and phenomena.

Question

The phenomenon of interference is the result of:

1. Conservation of energy 2. Superposition 3. The Fourier theorem 4. Conservation of mass 5. All of the above

Commentary

Purpose: To hone the concept of “interference,” distinguishing it from other concepts and situations it is frequently related to.

Discussion: The principle of superposition states that if two or more traveling waves are moving through a medium and combine at a given point, the resultant displacement of the medium at that point is the sum of the displacements of the individual waves.

Interference is the combination of separate waves in a region of space to produce a resultant wave. It can occur due to the convergence of waves from multiple sources, or from one source via multiple paths. Of the four listed answers, it depends only on superposition—answer (2).

“Diffraction” is not a different phenomenon from interference, but a particular case in which many waves (from every point in a fi nite-width slit, for example, or from every gap in a grating) interfere.



Key Points:

• Interference is a consequence of superposition.

• A wide variety of phenomena, including “diffraction,” are instances of interference.


While discussing interference, be careful to distinguish between amplitude and intensity; students frequently confuse these. This is especially true in many demonstrations of interference. The intensity is usually observed, though it is superposition of amplitudes that is discussed.

Question G3.02

Description: Honing the concept of interference.

Question

Two sound waves can interfere:

1. Only when traveling in the same direction 2. Only when the frequencies are the same 3. Only when both are sinusoidal 4. Only when the phase difference is constant 5. None of the above

Commentary

Purpose: To distinguish the essential phenomenon of “interference” from many of the contextual features it is frequently associated with.

Discussion: Any time two waves overlap in a linear medium (one that obeys the principle of superposition), interference between the waves occurs. The principle of superposition states that the total excitation of the medium (displacement in a vibrational wave, fi eld strength in an electromagnetic wave, etc.) is equal to the sum of the excitations due to each individual wave acting alone.

Interference occurs regardless of the waves’ directions, frequencies, waveforms, or phase differences. If the waves overlap, some kind of interference will occur. (To get specifi c kinds of interference — for example, standing waves, stationary nodes and antinodes, and the like — additional conditions are necessary.) Thus, “None of the above” is the appropriate answer here.

Key Points:

• Interference occurs any time two waves overlap in a linear medium (a medium obeying the principle of superposition).

• Interference does not necessarily result in a stable pattern of nodes and antinodes, standing waves, or anything else. Those are special cases.


630 Chapter 13


This question serves well as a follow-up to Question G3.01, reinforcing the ideas developed there.

The question’s real purpose is to instigate a discussion about what students associate the word “interference” with, so that they can learn to distinguish the general phenomenon and its preconditions from its consequences under special cases. Too frequently, students see interference only through a small set of these special cases, and do not learn to generalize the phenomenon.

QUICK QUIZZES

1. (d). To complete a full cycle of oscillation, the object must travel distance 2 A to position x A= − and then travel an additional distance 2 A returning to the original position at x A= + .

2. (c). The force producing harmonic oscillation is always directed toward the equilibrium position, and hence, directed opposite to the displacement from equilibrium. The acceleration is in the direction of the force. Thus, it is also always directed opposite to the displacement from equilibrium.

3. (b). In simple harmonic motion, the force (and hence, the acceleration) is directly proportional to the displacement from equilibrium. Therefore, force and acceleration are both at a maximum when the displacement is a maximum.

4. (a). The period of an object-spring system is T m k= 2π . Thus, increasing the mass by a factor of 4 will double the period of oscillation.

5. (c). The total energy of the oscillating system is equal to 12

2kA , where A is the amplitude of oscillation. Since the object starts from rest at displacement A in both cases, it has the same amplitude of oscillation in both cases.

6. (d). The expressions for the total energy, maximum speed, and maximum acceleration are E kA A k m a A k m= = = ( )1

22, ,max max and v where A is the amplitude. Thus, all are changed

by a change in amplitude. The period of oscillation is T m k= 2π and is unchanged by altering the amplitude.

7. (c), (b). An accelerating elevator is equivalent to a gravitational fi eld. Thus, if the elevator is accelerating upward, this is equivalent to an increased effective gravitational fi eld magnitude g, and the period will decrease. Similarly, if the elevator is accelerating downward, the effective value of g is reduced and the period increases. If the elevator moves with constant velocity, the period of the pendulum is the same as that in the stationary elevator.

8. (a). The clock will run slow. With a longer length, the period of the pendulum will increase. Thus, it will take longer to execute each swing, so that each second according to the clock will take longer than an actual second.

9. (b). Greater. The value of g on the Moon is about one-sixth the value of g on Earth, so the period of the pendulum on the moon will be greater than the period on Earth.



ANSWERS TO MULTIPLE CHOICE QUESTIONS

1. The wavelength of a wave is the distance from crest to the following crest. Thus, the distance between a crest and the following trough is a half wavelength, giving λ = ( ) =2 2 4 m m. The speed of the wave is then v = = ( )( ) =λ f 4 2 8m Hz m s, and (c) is the correct choice.

2. When an object undergoes simple harmonic motion, the position as a function of time may be written as x A t A ft= = ( )cos cosω π2 . Comparing this to the given relation, we see that the frequency of vibration is f = 3 Hz, and the period is T f= =1 1 3 s, so the correct answer is (c).

3. In this spring-mass system, the total energy equals the elastic potential energy at the moment the mass is temporarily at rest at x A= = 6 cm (i.e., at the extreme ends of the simple harmonic motion) Thus, E kA= 2 2 and we see that as long as the spring constant k and the amplitude A remain unchanged, the total energy is unchanged. Hence, the energy is still 12 J and (a) is the correct choice.

4. The energy given the vibratory system equals the elastic potential energy at the extremes of the motion, x A= ± . Thus, E k A= 2 2 and this energy will all be in the form of kinetic energy as the body passes through the equilibrium position, giving m kAvmax

2 22 2= , or

vmax ..

..= = ( ) =A

k

m0 10

80 0

0 401 4m

N m

kgm s

and (b) is the correct choice.

5. The frequency of vibration is

fk

m= =ω

π π2

1

2

Thus, increasing the mass by a factor of 9 will decrease the frequency to 1 3 of its original value, and the correct answer is (b).

6. When the object is at its maximum displacement, the magnitude of the force exerted on it by

the spring is F k xs = = ( )( ) =max . . .8 0 0 10 0 80 N m m N. This force will give the mass an

acceleration of a F ms= = =0 80 2 0. . N 0.40 kg m s2, making (d) the correct choice.

7. The car will continue to compress the spring until all of the car’s original kinetic energy has been

converted into elastic potential energy within the spring, i.e., until kx m i2 22 2= v , or

xm

ki= = ( ) ××

=v 2 03 0 10

100 77

5

..

.m skg

2.0 N mm6

The correct choice is seen to be (a).

8. The period of a simple pendulum is T g= 2π � , and its frequency is f T g= = ( )1 1 2π �.

Thus, if the length is doubled so ′ =� �2 , the new frequency is

′ =′

= =⎛⎝⎜

⎞⎠⎟

=fg g g f1

2

1

2 2

1

2

1

2 2π π π� � �

and we see that (d) is the correct response.


632 Chapter 13

9. The period of a simple pendulum is T g= 2π � . If the length is changed to ′ =� �4 , the newperiod will be

′ = ′ = =⎛⎝⎜

⎞⎠⎟

=Tg g g

T2 24

2 2 2π π π� � �

or the period will be doubled. The correct choice is (e).

10. For a particle executing simple harmonic motion about an equilibrium point x0, its position as a function of time is given by x x A t− = ( )0 cos ω , and the turning points (i.e., the extremes of the position) are at x x A= ±0 . That is, the equilibrium position is midway between the turning points, so the correct response is choice (c).

11. The only false statement among the listed choices is choice (d). At the equilibrium position, x = 0,

the elastic potential energy PE kxs =( )2 2 is a minimum and the kinetic energy is a maximum.

12. In a vertical mass-spring system, the equilibrium position is the point at which the mass will hang at rest on the lower end of the spring. If the mass is raised distance A above this position and released from rest, it will undergo simple harmonic motion, with amplitude A, about the equilibrium position. The upper turning point of the motion is at the point of release, and the lower turning point is distance A below the equilibrium position or distance 2 A below the release point. Thus, if the release point is 15 cm above the equilibrium position, the mass drops 30 cm before stopping momentarily and reversing direction. The correct answer is choice (c).

ANSWERS TO EVEN-NUMBERED CONCEPTUAL QUESTIONS

2. Each half-spring will have twice the spring constant of the full spring, as shown by the following argument. The force exerted by a spring is proportional to the separation of the coils as the spring is extended. Imagine that we extend a spring by a given distance and measure the distance between coils. We then cut the spring in half. If one of the half-springs is now extended by the same distance, the coils will be twice as far apart as they were for the complete spring. Thus, it takes twice as much force to stretch the half-spring, from which we conclude that the half-spring has a spring constant which is twice that of the complete spring.

4. Friction. This includes both air resistance and damping within the spring.

6. No. The period of vibration is T L g= 2π and g is smaller at high altitude. Therefore, the period is longer on the mountain top and the clock will run slower.

8. Shorten the pendulum to decrease the period between ticks.

10. The speed of the pulse is v = F µ , so increasing the tension F in the hose increases the speed of the pulse. Filling the hose with water increases the mass per unit length µ, and will decrease the speed of the pulse.

12. The speed of a wave on a string is given by v = F µ . This says the speed is independent of the frequency of the wave. Thus, doubling the frequency leaves the speed unaffected.



PROBLEM SOLUTIONS

13.1 (a) Taking to the right as positive, the spring force acting on the block at the instant of release is

F kxs i= − = − ( ) +( ) = −130 0 13 17N m m N or. 17 N to the left

(b) At this instant, the acceleration is

aF

ms= = − = −17

28N

0.60 kgm s2 or a = 28 m s to the left2

13.2 When the object comes to equilibrium (at distance y0 below the unstretched position of the end of

the spring), ΣF k y mgy = − −( ) − =0 0 and the force constant is

k

mg

y= =

( )( )×

= ×−0

4 25 9 80

101 59

. ..

kg m s

2.62 m

2

2 110 1 593 N kN= .

13.3 (a) Since the collision is perfectly elastic, the ball will rebound to the height of 4.00 m before coming to rest momentarily. It will then repeat this motion over and over again with a regular period.

(b) From ∆y t a ty y= +v012

2, with v0 0y = , the time required for the ball to reach the ground is

ty

ay

=( ) =

−( )−

=2 2 4 00

9 800 904

∆ .

..

m

m ss2

This is one-half of the time for a complete cycle of the motion. Thus, the period is

T = 1 81. s .

(c) No . The net force acting on the object is a constant given by F mg= − (except when it is contact with the ground). This is not in the form of Hooke’s law.

13.4 (a) The spring contant is

kF

x

mg

xs= = =

×= ×−

501 0 103N

5.0 10 mN m2 .

F F kxs= = = ×( )( ) = ×1 0 10 0 11 1 103. . .N m m 1 N2

(b) The graph will be a straight line passing through the origin with a slope equal to

k = ×1 0 103. N m.

13.5 When the system is in equilibrium, the tension in the spring F k x= must equal the weight of the object. Thus, k x mg= , giving

mk x

g= =

( ) ×( )=

−47 5 5 00 10

9 800 242

2. .

..

N m

m sk2 gg


634 Chapter 13

13.6 (a) The free-body diagram of the point in the center of the string is given at the right. From this, we see that

ΣF F Tx = ⇒ − =0 2 35 0 0°sin .

or

TF=

°=

°=

2 35 0

375

2 35 0327

sin . sin .

NN

(b) Since the bow requires an applied horizontal force of 375 N to hold the string at 35.0° from the vertical, the tension in the spring must be 375 N when the spring is stretched 30.0 cm. Thus, the spring constant is

kF

x= = = ×375

0 3001 25 103N

mN m

..

13.7 (a) When the block comes to equilibrium, ΣF ky mgy = − − =0 0, giving

ymg

k0

10 0 9 80

4750 206= − = −

( )( )= −

. ..

kg m s

N mm

2

or the equilibrium position is 0 206. m below the unstretched position of the lower end of the spring.

(b) When the elevator (and everything in it) has an upward acceleration of a = 2 00. m s2, applying Newton’s second law to the block gives

ΣF k y y mg may y= − +( ) − =0 or ΣF ky mg ky may y= − −( ) − =0

where y = 0 at the equilibrium position of the block. Since − − =ky mg0 0 [see part (a)],this becomes − =ky ma and the new position of the block is

yma

ky=

−=

( ) +( )−

= − ×10 0 2 00

4 21 1. .

.kg m s

475 N m

2

00 4 212− = −m cm.

or 4.21 cm below the equilibrium position.

(c) When the cable breaks, the elevator and its contents will be in free fall with a gy = − . The new “equilibrium” position of the block is found from ΣF ky mg m gy = − ′ − = −( )0 , which yields ′ =y0 0. When the cable snapped, the block was at rest relative to the elevator at distance y y0 0 206 0 042 1 0 248+ = + =. . .m m m below the new “equilibrium” position.

Thus, while the elevator is in free fall, the block will oscillate with amplitude m= 0 248.

about the new “equilibrium” position, which is the unstretched position of the spring’s lower end.

13.8 (a) When the gun is fi red, the elastic potential energy initially stored in the spring is transformed into kinetic energy of the projectile. Thus, it is necessary to have

1

2

1

202

02k x m= v or k

m

x= =

×( )( )×

−

−

v02

02

3 2

2

3 00 10 45 0

8 00 10

. .

.

kg m s

mN m

( )=2 949

(b) The magnitude of the force required to compress the spring 8.00 cm and load the gun is

F k xs = = ( ) ×( ) =−949 8 00 10 75 92N m m N. .

→F

→T

→T

35.0°

35.0°

→F

→T

→T

35.0°

35.0°



13.9 (a) Assume the rubber bands obey Hooke’s law. Then, the force constant of each band is

kF

xs= =

×= ×−

151 5 103N

1.0 10 mN m2 .

Thus, when both bands are stretched 0.20 m, the total elastic potential energy is

PE k xs = ⎛⎝

⎞⎠ = ×( )( ) =2

1

21 5 10 0 20 602 3 2. .N m m J

(b) Conservation of mechanical energy gives KE PE KE PEs f s i+( ) = +( ) , or

1

20 0 602mv + = + J, so v = ( )

×=−

2 6049

J

50 10 kgm s3

13.10 (a) kF

x= = =max

max

N

0.400 mN m

230575

(b) work done PE k xs N m= = = ( )( ) =1

2

1

2575 0 400 46 02 2. . J

13.11 From conservation of mechanical energy,

KE PE PE KE PE PEg s f g s i+ +( ) = + +( ) or 0 0 0 0

1

22+ + = + +mgh kxf i

giving

kmgh

xf

i

= =( )( )( )2 2 0 100 9 80 0 600

22

. . .

.

kg m s m2

000 102 94 10

2 23

×( )= ×

− mN m.

13.12 Conservation of mechanical energy, ( ) ( )KE PE PE KE PE PEg s f g s i+ + = + + , gives 1

22 1

220 0 0 0m k xi fv + + = + + , or

vi i

k

mx= = × ×( ) =−5 00 10

10003 16 10 2 2

62.

. .N m

kgm 33 m s

13.13 An unknown quantity of mechanical energy is converted into internal energy during the collision. Thus, we apply conservation of momentum from just before to just after the collision and obtain m M M m Viv + ( ) = +( )0 , or the speed of the block and embedded bullet just after collision is

Vm

M m i=+

⎛⎝

⎞⎠ = ×⎛

⎝⎜⎞⎠⎟

−

v10 0 10

3003. kg

2.01 kgm ss m s( ) = 1 49.

Now, we use conservation of mechanical energy from just after collision until the block comes to rest. This gives 0 1

22 1

22+ = +( )k x M m Vf , or

x VM m

kf = + = ( ) =1 492 01

0 477..

.m skg

19.6 N mm

13.14 (a) At either of the turning points, x A= ± , the constant total energy of the system is momentarily

stored as elastic potential energy in the spring. Thus, E k A= 2 2 .

(b) When the object is distance x from the equilibrium position, the elastic potential energy is PE k xs = 2 2 and the kinetic energy is KE m= v2 2. At the position where KE PEs= 2 , is it necessary that

1

22

1

22 2m kxv = ⎛

⎝⎞⎠ or

1

22 2m kxv =

continued on next page


636 Chapter 13

(c) When KE PEs= 2 , conservation of energy gives E KE PE PE PE PEs s s s= + = ( ) + =2 3 , or

1

23

1

2

2

3 22 2

2

kA kx xk A

k= ⎛

⎝⎞⎠ ⇒ = ± or x

A= ±3

13.15 (a) At maximum displacement from equilibrium, all of the energy is in the form of elastic

potential energy, giving E kx= max2 2, and

kE

x= = ( )

( )= ×2 2 47 0

0 2401 63 102 2

3

max

.

..

J

mN m

(b) At the equilibrium position x =( )0 , the spring is momentarily in its relaxed state and PEs = 0, so all of the energy is in the form of kinetic energy. This gives

KE m Ex = = = =021

247 0vmax . J

(c) If, at the equilibrium position, v v= =max .3 45 m s, the mass of the block is

mE= = ( )

( )=2 2 47 0

3 457 902 2vmax

.

..

J

m skg

(d) At any position, the constant total energy is, E KE PE m k xs= + = +v2 22 2, so at x = 0 160. m

v = − =( ) − ×( )( )2 2 47 0 1 63 10 0 1602 3 2

E kx

m

. . .J N m m

77 902 57

..

kgm s=

(e) At x = 0 160. m, where v = 2 57. m s, the kinetic energy is

KE m= = ( )( ) =1

2

1

27 90 2 57 26 12 2v . . .kg m s J

(f ) At x = 0 160. m, where KE = 26 1. J, the elastic potential energy is

PE E KEs = − = − =47 0 26 1 20 9. . .J J J

or alternately,

PE k xs = = ×( )( ) =1

2

1

21 63 10 0 160 20 92 3 2. . .N m m J

(g) At the fi rst turning point (for which x < 0 since the block started from rest at x = +0 240. m and has passed through the equilibrium at x = 0) all of the remaining energy is in the form of elastic potential energy, so

1

247 0 14 0 33 02k x E E= − = − =loss J J J. . .

and

xk

= − ( )= − ( )

×= −2 33 0 2 33 0

1 63 100 2013

. .

..

J J

N mmm



13.16 (a) F k x= = ( ) ×( ) =−83 8 5 46 10 4 582. . .N m m N

(b) E PE k xs= = = ( ) ×( ) =−1

2

1

283 8 5 46 10 0 1252 2 2

. . .N m m J

(c) While the block was held stationary at x = 5 46. cm, ΣF F Fx s= − + = 0, or the spring force was equal in magnitude and oppositely directed to the applied force. When the applied force is suddenly removed, there is a net force Fs = 4 58. N directed toward the equilibrium position acting on the block. This gives the block an acceleration having magnitude

aF

ms= = =4 58

0 25018 3

.

..

N

kgm s2

(d) At the equilibrium position, PEs = 0, so the block has kinetic energy KE E= = 0 125. J and speed

v = = ( )=2 2 0 125

0 2501 00

E

m

.

..

J

kgm s

(e) If the surface was rough, the block would spend energy overcoming a retarding friction force as it moved toward the equilibrium position, causing it to arrive at that position with

a lower speed than that computed above. Computing a number value for this lower speed

requires knowledge of the coefficient of friction between the block and surface.

13.17 From conservation of mechanical energy,

KE PE PE KE PE PEg s f g s i+ +( ) = + +( )

we have 12

2 12

2 12

20 0 0m k x k Av + + = + + , or

v = −( )k

mA x2 2

(a) The speed is a maximum at the equilibrium position, x = 0.

vmax

N m

kgm= =

( )( ) ( ) =k

mA2 219 6

0 400 040 0 28

.

.. . m s

(b) When x = − 0 015. m,

v =( )( ) ( ) − −( )⎡19 6

0 400 040 0 0152 2.

.. .

N m

kgm m⎣⎣ ⎤⎦ = 0 26. m s

(c) When x = + 0 015. m,

v =( )( ) ( ) − +( )⎡19 6

0 400 040 0 0152 2.

.. .

N m

kgm m⎣⎣ ⎤⎦ = 0 26. m s

(d) If v v= 12 max, then

k

mA x

k

mA2 2 21

2−( ) =

This gives A x A2 2 2 4− = , or x A= = ( ) =3 2 4 0 3 2 3 5. . cm cm .


638 Chapter 13

13.18 (a) KE x A= =0 at , so E KE PE k As= + = +0 12

2, or the total energy is

E k A= = ( )( ) =1

2

1

2250 0 035 0 152 2N m m J. .

(b) The maximum speed occurs at the equilibrium position where PEs = 0. Thus, E m= 1

22vmax, or

vmax mN m

kgm s= = = ( ) =2

0 035250

0 500 78

E

mA

k

m.

..

(c) The acceleration is a F m kx m= = −Σ . Thus, a a= max at x x A= − = −max .

ak A

m

k

mAmax

N m

kg= − −( )

= ⎛⎝

⎞⎠ =

⎛⎝⎜

⎞⎠⎟

250

0 500 0

.. 335 18m m s2( ) =

13.19 The maximum speed occurs at the equilibrium position and is

vmax = k

mA

Thus

mkA= =

( )( )( )

=2

2

2

2

16 0 0 200

0 4004

vmax

N m m

m s

. .

...00 kg

and

F mgg = = ( )( ) =4 00 9 80 39 2. . .kg m s N2

13.20 v = −( ) =×

⎛⎝⎜

⎞⎠⎟−

k

mA x2 2 10 0

0 250.

.N m

50.0 10 kg3 mm m m s( ) − ( )⎡⎣ ⎤⎦ =2 20 125 3 06. .



13.21 (a) The motion is simple harmonic because the tire is rotating with constant velocity and you are looking at the uniform circular motion of the “bump” projected on a plane perpendicular to the tire.

(b) Note that the tangential speed of a point on the rim of a rolling tire is the same as the translational speed of the axle. Thus, v vt = =car m s3 00. and the angular velocity of the tire is

ω = = =vt

r

3 00

0 30010 0

.

..

m s

mrad s

Therefore, the period of the motion is

T = = =2 2

10 00 628

πω

π.

.rad s

s

13.22 (a) vt

r

T= = ( )

=2 2 0 2000 628

π π ..

m

2.00 sm s

(b) fT

= = =1 10 500

2.00 sHz.

(c) ω π π= = =2 23 14

T 2.00 srad s.

13.23 The angle of the crank pin is θ ω= t . Its x-coordinate is x A A t= =cos cosθ ω , where A is the distance from the center of the wheel to the crank pin. This is of the correct form to describe simple harmonic motion. Hence, one must conclude that the motion is indeed simple harmonic.

13.24 The period of vibration for an object-spring system is T m k= 2π . Thus, if T = 0 223. s and m = = × −35 4 35 4 10 3. .g kg, the force constant of the spring is

km

T= =

×( )( )

=−

4 4 35 4 10

0 22328 1

2

2

2 3

2

π π .

..

kg

sN mm

13.25 The spring constant is found from

kF

x

mg

xs= = =

( )( )×

=−

0 010 9 802

. ..

kg m s

3.9 10 m

2

2 55 N m

When the object attached to the spring has mass m = 25 g, the period of oscillation is

Tm

k= = =2 2

0 0250 63π π ..

kg

2.5 N ms

x�0

x�0x(t)

x(t)

A

wt

x�0

x�0x(t)

x(t)

A

wt


640 Chapter 13

13.26 The springs compress 0.80 cm when supporting an additional load of m = 320 kg. Thus, the spring constant is

kmg

x= =

( )( )×

= ×−

3 9 80

803 9 105

20 kg m s

0 10 m

2

2

.

.. N m

When the empty car, M = ×2 0 103. kg, oscillates on the springs, the frequency will be

fT

k

M= = = ×

×=1 1

2

1

2

3 9 10

2 0 102 2

5

3π π.

..

N m

kgHz

13.27 (a) The period of oscillation is T m k= 2π where k is the spring constant and m is the mass of the object attached to the end of the spring. Hence,

T = =20 250

9 51 0π .

..

kg

N ms

(b) If the cart is released from rest when it is 4.5 cm from the equilibrium position, the amplitude of oscillation will be A = = × −4 5 4 5 10 2. .cm m. The maximum speed is then given by

vmax ..

..= = = ×( ) =−A A

k

mω 4 5 10

9 5

0 2500 282 m

N m

kgm s

(c) When the cart is 14 cm from the left end of the track, it has a displacement of x = − =14 12 2 0cm cm cm. from the equilibrium position. The speed of the cart at this distance from equilibrium is

v = −( ) = ( ) −k

mA x2 2 29 5

0 2500 045 0 020

.

.. .

N m

kgm mm m s( )⎡⎣ ⎤⎦ =2 0 25.

13.28 The general expression for the position as a function of time for an object undergoing simple harmonic motion with x t= =0 0at is x A t= ( )sin ω . Thus, if x t= ( ) ⋅( )5 2 8 0. sin .cm π , we have that the amplitude is A = 5 2. cm and the angular frequency is ω π= 8 0. rad s.

(a) The period is

T = = =−2 2

8 00 25

πω

ππ.

.s

s1

(b) The frequency of motion is

fT

= = = =−1 1

0 254 0 4 0

.. .

ss Hz1

(c) As discussed above, the amplitude of the motion is A = 5 2. cm .

(d) Note: For this part, your calculator should be set to operate in radians mode.

If x = 2 6. cm, then

ωtx

A= ⎛

⎝⎞⎠ = ⎛

⎝⎞⎠ =− − −sin sin

.sin1 1 12 6

0cm

5.2 cm.. .50 0 52( ) = radians

and

t = = = × =−0 52 0 522 1 10 212. ..

rad rad

8.0 rad ss

ω π×× =−10 213 s ms



13.29 (a) At the equilibrium position, the total energy of the system is in the form of kinetic energy and m Evmax

2 2 = , so the maximum speed is

vmax

.

..= = ( )

=2 2 5 83

0 3265 98

E

m

J

kgm s

(b) The period of an object-spring system is T m k= 2π , so the force constant of the spring is

km

T= =

( )( )

=4 4 0 326

0 250206

2

2

2

2

π π .

.

kg

sN m

(c) At the turning points, x A= ± , the total energy of the system is in the form of elastic potential energy, or E kA= 2 2, giving the amplitude as

AE

k= = ( )

=2 2 5 83

2060 238

..

J

N mm

13.30 For a system executing simple harmonic motion, the total energy may be written as E KE PE m kAs= + = =1

22 1

22vmax , where A is the amplitude and vmax is the speed at the

equilibrium position. Observe from this expression, that we may write vmax2 2= kA m.

(a) If v v= 12 max, then E m k x m= + =1

22 1

22 1

22v vmax becomes

1

2 4

1

2

1

2

22 2m kx m

vvmax

max

⎛⎝⎜

⎞⎠⎟

+ =

and gives

xm

k

m

k

k

mA

A2 2 223

4

3

4

3

4= ⎛

⎝⎞⎠ = ⎛

⎝⎞⎠

⎡⎣⎢

⎤⎦⎥

=vmax

or

xA= ± 3

2

(b) If the elastic potential energy is PE Es = 2, we have

1

2

1

2

1

22 2kx kA= ⎛

⎝⎞⎠ or x

A22

2= and x

A= ±2

13.31 (a) At t = 3 50. s,

F kx= − = − ⎛⎝

⎞⎠ ( ) ⎛

⎝⎞⎠5 00 3 00 1 3. . cos

N

mm .58

rad

s.. .50 11 0s N( )⎡

⎣⎢⎤⎦⎥

= − ,

or F = 11 0. N directed to the left .



642 Chapter 13

(b) The angular frequency is

ω = = =k

m

5 00

2 001 58

.

..

N m

kgrad s

and the period of oscillation is

T = = =2 2

1 583 97

πω

π.

.rad s

s

Hence the number of oscillations made in 3.50 s is

Nt

T= = =∆ 3 50

3 970 881

.

..

s

s

13.32 (a) kF

x= =

×=−

7 50250

. N

3.00 10 mN m2

(b) ω = = =k

m

25022 4

N m

0.500 kgrad s.

f = = =ωπ π2

22 4

23 56

..

rad sHz

Tf

= = =1 1

3 560 281

..

Hzs

(c) At t = 0, v = = × −0 5 00 10 2and mx . , so the total energy of the oscillator is

E KE PE m kxs= + = +

= + ( ) × −

1

2

1

2

01

2250 5 00 10

2 2

2

v

N m . mm J( ) =2

0 313.

(d) When x A= , v = = + = +0 0 12

2 so E KE PE kAs .

Thus,

AE

k= = ( )

= × =−2 2 0 3135 00 10 5 002.. .

J

250 N/mm cm

(e) At x = 0, KE m E= =12

2vmax , or

vmax

J

0.500 kgm s= = ( )

=2 2 0 3131 12

E

m

..

aF

m

k A

mmaxmax

N m m

k= = =

( ) ×( )−250 5 00 10

0 500

2.

. ggm s= 25 0 2.




Note: To solve parts (f ) and (g), your calculator should be set in radians mode.

(f ) At t = 0 500. s, Equation 13.14a gives the displacement as

x A t A t k m= ( ) = ( ) = ( ) ( )cos cos . cos .ω 5 00 0 5002

cm s550

0 5000 919

N m

kgcm

..

⎡

⎣⎢

⎤

⎦⎥ =

(g) From Equation 13.14b, the velocity at t = 0 500. s is

v = − ( ) = − ( )= − ×( )−

A t A k m t k mω ωsin sin

.5 00 102502 m

N m

kgs

N m

kg0 5000 500

250

0 500.sin .

.( )⎡

⎣⎢

⎤

⎦⎥⎥ = +1 10. m

and from Equation 13.14c, the acceleration at this time is

a A t A k m t k m= − ( ) = − ( ) ( )= − ×( )−

ω ω2

5 00 10

cos cos

. 2 m2250

0 5000 500

250

0 50

N m

kg s

N m

.cos .

.

⎛⎝⎜

⎞⎠⎟

( )00

4 59 kg

m s2⎡

⎣⎢

⎤

⎦⎥ = − .

13.33 From Equation 13.6,

v = ± −( ) = ± −( )k

mA x A x2 2 2 2 2ω

Hence,

v = ± − ( ) = ± − ( ) = ± ( )ω ω ω ω ω ωA A t A t A t2 2 2 21cos cos sin

From Equation 13.2,

ak

mx A t A t= − = − ( )⎡⎣ ⎤⎦ = − ( )ω ω ω ω2 2cos cos

13.34 (a) The height of the tower is almost the same as the length of the pendulum. From

T L g= 2π , we obtain

LgT= =

( )( )=

2

2

2

24

9 80 15 5

459 6

π π. .

.m s s

m2

(b) On the Moon, where g = 1 67. m s2, the period will be

TL

g= = =2 2

59 637 5π π .

.m

1.67 m ss2

13.35 The period of a pendulum is the time for one complete oscillation and is given by T g= 2π � , where � is the length of the pendulum.

(a) T = ( ) ⎛⎝

⎞⎠ =3 00 60

11

. min

120 oscillations

s

min..50 s

(b) The length of the pendulum is

� =⎛⎝⎜

⎞⎠⎟

= ( ) ( )⎛⎝⎜

⎞⎠⎟

gT 2

2

2

249 80

1 50

4π π.

.m s

s2 == 0 559. m


644 Chapter 13

13.36 The period in Tokyo is T L gT T T= 2π and the period in Cambridge is T L gC C C= 2π .

We know that T TT C= = 2 000. s from which, we see that

L

g

L

gT

T

C

C

= , or g

g

L

LC

T

C

T

= = =0 994 2

0 992 71 001 5

.

..

13.37 (a) The period of the pendulum is T L g= 2π . Thus, on the Moon where the free-fall

acceleration is smaller, the period will be longer and the clock will run slow .

(b) The ratio of the pendulum’s period on the Moon to that on Earth is

T

T

L g

L g

g

gMoon

Earth

Moon

Earth

Earth

Moon

= = =2

2

9ππ

..

..

80

1 632 45=

Hence, the pendulum of the clock on Earth makes 2.45 “ticks” while the clock on the Moon is making 1.00 “tick.” After the Earth clock has ticked off 24.0 h and again reads 12:00

midnight, the Moon clock will have ticked off 24 0 2 45 9 80. . . h h= and will read 9 48: AM .

13.38 (a) The lower temperature will cause the pendulum to contract. The shorter length will produce a smaller period, so the clock will run faster or gain time .

(b) The period of the pendulum is T L g0 02= π at 20°C, and at –5.0°C it is T L g= 2π .

The ratio of these periods is T T L L0 0= .

From Chapter 10, the length at –5.0°C is L L L T= + ( )0 0αAl ∆ , so

L

L T0

6 1

1

1

1

1 24 10 5 0=

+ ( ) =+ × ( )⎡⎣ ⎤⎦ − −− −αAl °C °C∆ . 220

1

0 999 41 000 6

°C[ ]= =

..

This gives

T

T

L

L0 0 1 000 6 1 000 3= = =. .

Thus in one hour (3 600 s), the chilled pendulum will gain 1 000 3 1 3 600 1 1. .−( )( ) =s s .

13.39 (a) From T L g= 2π , the length of a pendulum with period T is L gT= 2 24π .

On Earth, with T = 1 0. s,

L =( )( )

= =9 80 1 0

40 25 25

2

2

. ..

m s sm cm

2

π

If T = 1 0. s on Mars,

L =( )( )

= =3 1 0

40 094 9 4

2

2

.7 m s sm cm

2 .. .

π

(b) The period of an object on a spring is T m k= 2π , which is independent of the local free-fall acceleration. Thus, the same mass will work on Earth and on Mars. This mass is

mk T= =

( )( )=

2

2

2

24

10 1 0

40 25

π πN m s

kg.

.



13.40 The apparent free-fall acceleration is the vector sum of the actual free-fall acceleration and the negative of the elevator’s acceleration. To see this, consider an object that is hanging from a vertical string in the elevator and appears to be at rest to the elevator passengers. These

passengers believe the tension in the string is the negative of the object’s weight, or � �T g= −m app

where �gapp is the apparent free-fall acceleration in the elevator.

An observer located outside the elevator applies Newton’s second law to this object by writing

Σ� � � �F T g a= + =m m e where

�ae is the acceleration of the elevator and all its contents. Thus,

� � � �T a g g= −( ) = −m me app, which gives

� � �g g aapp = − e .

(a) If we choose downward as the positive direction, then �ae = − 5 00. m s2 in this case and �

gapp2 2m s m s= +( ) = +9 80 5 00 14 8. . . (downward). The period of the pendulum is

TL

g= = =2 2

5 003 65π π

app2

m

14.8 m ss

..

(b) Again choosing downward as positive, �ae = 5 00. m s2 and

�gapp

2 2m s m s= − = +( . . ) .9 80 5 00 4 80

(downward) in this case. The period is now given by

TL

g= = =2 2

5 006 41π π

app2

m

4.80 m ss

..

(c) If �ae = 5 00. m s2 horizontally, the vector sum

� � �g g aapp = − e is

as shown in the sketch at the right. The magnitude is

gapp2 2 2m s m s m s= ( ) + ( ) =5 00 9 80 11 0

2 2. . .

and the period of the pendulum is

TL

g= = =2 2

5 004 24π π

app2

m

11.0 m ss

..

13.41 (a) The distance from the bottom of a trough to the top of a crest is twice the amplitude of the wave. Thus, 2 8 26A = . cm and A = 4 13. cm .

(b) The horizontal distance from a crest to a trough is a half wavelength. Hence,

λ2

5 20= . cm and λ = 10 4. cm

(c) The period is

Tf

= = = ×−−1 1

18 05 56 10 2

..

ss1

(d) The wave speed is v = = ( )( ) = =−λ f 10 4 18 0 187 1 871. . .cm s cm s m s .

→g

�→ae

→gapp

→g

�→ae

→gapp

8.26 cm

5.20 cm

Figure P13.41

8.26 cm

5.20 cm

Figure P13.41


646 Chapter 13

13.42 (a) The amplitude is the magnitude of the maximum displacement from equilibrium at x =( )0 .

Thus, A = 2 00. cm .

(b) The period is the time for one full cycle of the motion. Therefore, T = 4 00. s .

(c) The period may be written as T = 2π ω , so the angular frequency is

ω π π π= = =2 2

4 00 2T . srad s

(d) The total energy may be expressed as E m kA= =12

2 12

2vmax . Thus, vmax = A k m , and since ω = k m , this becomes vmax = ωA and yields

vmax .= = ⎛⎝

⎞⎠ ( ) =ω π πA

22 00rad s cm cm s

(e) The spring exerts maximum force, F k x= , when the object is at maximum distance from equilibrium, i.e., at x A= = 2 00. cm. Thus, the maximum acceleration of the object is

aF

m

kA

mAmax

max . .= = = = ⎛⎝

⎞⎠ ( ) =ω π2

2

22 00 4rad s cm 993 cm s2

(f) The general equation for position as a function of time for an object undergoing simple harmonic motion with t x= =0 0when is x A t= ( )sin ω . For this oscillator, this becomes

x t= ( ) ⎛⎝

⎞⎠2 00

2. sincm

π

13.43 (a) The period and the frequency are reciprocals of each other. Therefore,

Tf

= = =×

= ×−−1 1

101 9

1

101 9 109 81 101

9

. ..

MHz ss6 == 9 81. ns

(b) λ = = ××

=−vf

3 00 10

101 9 102 94

8

6 1

.

..

m s

sm

13.44 (a) The frequency of a transverse wave is the number of crests that pass a given point each second. Thus, if 5.00 crests pass in 14.0 seconds, the frequency is

f = = =−5 00

14 00 357 0 3571.

.. .

ss Hz

(b) The wavelength of the wave is the distance between successive maxima or successive minima. Thus, λ = 2 76. m and the wave speed is

v = = ( )( ) =−λ f 2 76 0 357 0 9851. . .m s m s

x (cm)

t (s)

1.00

2.00

0.00

�1.00

�2.00

0 1 2 3 4 5 6

Figure P13.42

x (cm)

t (s)

1.00

2.00

0.00

�1.00

�2.00

0 1 2 3 4 5 6

Figure P13.42



13.45 The speed of the wave is

v = = =∆∆

x

t

425

10 042 5

cm

scm s

..

and the frequency is

f = =40 0

30 01 33

.

..

vib

sHz

Thus,

λ = = =vf

42 5

1 3331 9

.

..

cm s

Hzcm

13.46 From v = λ f , the wavelength (and size of smallest detectable insect) is

λ = =×

= × =−vf

3405 67 5 67

m s

60.0 10 Hz10 m mm3

3. .

13.47 The frequency of the wave (that is, the number of crests passing the cork each second) is f = −2 00. s 1 and the wavelength (distance between successive crests) is λ = 8 50. cm. Thus, the wave speed is

v = = ( )( ) = =−λ f 8 50 2 00 17 0 0 170. . . .cm s cm s m s1

and the time required for the ripples to travel 10.0 m over the surface of the water is

∆ ∆t

x= = =v

10 058 8

..

m

0.170 m ss

13.48 (a) When the boat is at rest in the water, the speed of the wave relative to the boat is the same as the speed of the wave relative to the water, v = 4 0. m s. The frequency detected in this case is

f = = =vλ

4 00 20

..

m s

20 mHz

(b) Taking eastward as positive, � � �v v vwave,boat wave,water boat,water= − (see the discussion of relative

velocity in Chapter 3 of the textbook) gives

�vwave,boat m s m s m s= + − −( ) = +4 0 1 0 5 0. . . and

� �v vboat,wave wave,boat m s= = 5 0.

Thus,

f = = =vboat,wave m s

20 mHz

λ5 0

0 25.

.

13.49 The down and back distance is 4.00 m 4.00 m 8.00 m+ = .

The speed is then

v = = ( )= =

d

tFtotal m

sm s

4 8 00

0 80040 0

.

.. µ

Now,

µ = = = × −m

L

0 2005 00 10 2..

kg

4.00 mkg m

so

F = = ×( )( ) =−µv2 2 25 00 10 40 0 80 0. . .kg m m s N


648 Chapter 13

13.50 The speed of the wave is

v = = =∆∆

x

t

20 0

0 80025 0

.

..

m

sm s

and the mass per unit length of the rope is µ = =m L 0 350. kg m. Thus, from v = F µ , we obtain

F = = ( ) ( ) =v2 225 0 0 350 219µ . .m s kg m N

13.51 (a) The speed of transverse waves in the cord is v = F µ , where µ = m L is the mass per unit

length. With the tension in the cord being F = 12 0. N, the wave speed is

v = = = = ( )( )=F F

m L

FL

mµ12 0 6 30

0 15022 4

. .

..

N m

kgmm s

(b) The time to travel the length of the cord is

∆tL= = =v

6 300 281

..

m

22.4 m ss

13.52 (a) In making 6 round trips, the pulse travels the length of the line 12 times for a total distance of 144 m. The speed of the pulse is then

v = = = ( )=∆

∆ ∆x

t

L

t

12 12 12 0

2 9648 6

.

..

m

sm s

(b) The speed of transverse waves in the line is v = F µ , so the tension in the line is

Fm

L= = ⎛

⎝⎞⎠ = ⎛

⎝⎞⎠ ( )µv v2 2 20 375

48 6.

.kg

12.0 mm s == 73 8. N

13.53 (a) The mass per unit length is

µ = = =m

L

0 060 0

5 000 012 0

.

..

kg

m kg m

From v = F µ , the required tension in the string is

F = = ( ) ( ) =v2 250 0 0 012 0 30 0µ . . .m s kg m N

(b) v = = =F

µ8 00

25 8.

. N

0.012 0 kg m m s



13.54 The mass per unit length of the wire is

µ = = × = ×−

−m

L

4

12 50

.00 10 kg

.60 m10 kg m

33.

and the speed of the pulse is

v = = =L

t∆1 60

0 036 144 3

.

..

m

sm s

Thus, the tension in the wire is

F = = ( ) ×( ) =−v2 24 2 50 4 91µ 4.3 m s 10 kg m N3. .

But, the tension in the wire is the weight of a 3.00-kg object on the Moon. Hence, the local free-fall acceleration is

gF

m= = =4 91

3 001 64

.

..

N

kgm s2

13.55 The period of the pendulum is T L g= 2π , so the length of the string is

LgT= =

( )( )=

2

2

2

24

9 80 2 00

40 993

π π. .

.m s s

m2

Then mass per unit length of the string is then

µ = = =m

L

0 060 00 060 4

..

kg

0.993 m

kg

m

When the pendulum is vertical and stationary, the tension in the string is

F M g= = ( )( ) =ball2kg m s N5 00 9 80 49 0. . .

and the speed of transverse waves in it is

v = = =F

µ49 0

28 5.

.N

0.060 4 kg mm s

13.56 If µ1 1= m L is the mass per unit length for the fi rst string, then µ µ2 2 1 12 2= = =m L m L is that of the second string. Thus, with F F F2 1= = , the speed of waves in the second string is

v v22 1 1

1

22 2 2 5 00 7 07= = =

⎛

⎝⎜⎞

⎠⎟= = ( ) =F F F

µ µ µ. .m s mm s

13.57 (a) The tension in the string is F mg= = ( )( ) =3 0 9 80 29. .kg m s N2 . Then, from v = F µ , the mass per unit length is

µ = =( )

=F

v2 2

290 051

N

24 m skg m.

(b) When m = 2.00 kg, the tension is

F mg= = ( )( ) =2 0 9 80 20. .kg m s N2

and the speed of transverse waves in the string is

v = = =F

µ20

20N

0.051 kg mm s


650 Chapter 13

13.58 If the tension in the wire is F, the tensile stress is Stress F A= , so the speed of transverse waves in the wire may be written as

v = = ⋅ =⋅( )

F A Stress

m L

Stress

m A Lµ

But, A L V⋅ = = volume, so m A L⋅( ) = =ρ density. Thus, v = Stress

ρ.

When the stress is at its maximum, the speed of waves in the wire is

vmaxmax .=

( )= ×

×=

Stress

ρ2 70 10

10

9 Pa

7.86 kg m3 3 5586 m s

13.59 (a) The speed of transverse waves in the line is v = F µ , with µ = m L being the mass per unit length. Therefore,

v = = = = ( )( )=F F

m L

FL

mµ12 5 38 0

2 6513 4

. .

..

N m

kgm ss

(b) The worker could throw an object, such as a snowball, at one end of the line to set up a pulse, and use a stopwatch to measure the time it takes a pulse to travel the length of the line. From this measurement, the worker would have an estimate of the wave speed, which in turn can be used to estimate the tension.

13.60 (a) In making n round trips along the length of the line, the total distance traveled by the pulse

is ∆x n L nL= ( ) =2 2 . The wave speed is then

v = =∆x

t

nL

t

2

(b) From v = F µ as the speed of transverse waves in the line, the tension is

FM

L

nL

t

M

L

n L

t= = ⎛

⎝⎞⎠

⎛⎝

⎞⎠ = ⎛

⎝⎞⎠

⎛⎝⎜

⎞⎠⎟

=µv22 2 2

2

2 4 4nn ML

t

2 2

2

13.61 (a) Constructive interference produces the maximum amplitude

′ = + =A A Amax .1 2 0 50 m

(b) Destructive interference produces the minimum amplitude

′ = − =A A Amin .1 2 0 10 m

13.62 We are given that x A t t= ( ) = ( ) ( )cos . cos .ω π0 25 0 4m .

(a) By inspection, the amplitude is seen to be A = 0 25. m

(b) The angular frequency is ω π= 0 4. rad s. But ω = k m , so the spring constant is

k m= = ( )( ) =ω π2 20 30 0 4 0 47. . .kg rad s N m




(c) Note: Your calculator must be in radians mode for part (c).

At t = 0 30. s, x = ( ) ( )( )⎡⎣ ⎤⎦ =0 25 0 4 0 30 0 23. cos . . .m rad s s mπ

(d) From conservation of mechanical energy, the speed at displacement x is given by

v = −ω A x2 2 . Thus, at t = 0 30. s, when x = 0 23. m, the speed is

v = ( ) ( ) − ( ) =0 4 0 25 0 23 0 122 2. . . .π rad s m m m s

13.63 (a) The period of a vibrating object-spring system is T m k= =2 2π ω π , so the spring constant is

km

T= =

( )( )

=4 4 2 00

0 600219

2

2

2

2

π π .

.

kg

sN m

(b) If T = 1 05. s for mass m2, this mass is

mkT

2

2

2

2

24

219 1 05

46 12= =

( )( )=

π πN m s

kg.

.

13.64 (a) The period is the reciprocal of the frequency, or

Tf

= = = × =−−1 1

1965 10 10 5 101

3

ss ms. .

(b) λ = = =−vsound m s

sm

f

343

1961 751 .

13.65 (a) The period of a simple pendulum is T g= 2π � , so the period of the fi rst system is

Tg1 2 2

0 7001 68= = =π π� .

.m

9.80 m ss2

(b) The period of mass-spring system is T m k= 2π , so if the period of the second system is

T T2 1= , then 2 2π πm k g= � and the spring constant is

kmg= =

( )( )=

�

1 20 9 80

0 70016 8

. .

..

kg m s

mN m

2

13.66 Since the spring is “light,” we neglect any small amount of energy lost in the collision with the spring, and apply conservation of mechanical energy from when the block fi rst starts until it comes to rest again. This gives

KE PE PE KE PE PEg s f g s i+ +( ) = + +( ) , or 0 0

1

20 02+ + = + +kx mghimax

Thus,

xmgh

ki

max

2kg m s m= =

( )( )( )2 2 0 500 9 80 2 00

20

. . .

...

00 990

N mm=


652 Chapter 13

13.67 Choosing PEg = 0 at the initial height of the block, conservation of mechanical energy gives

KE PE PE KE PE PEg s f g s i+ +( ) = + +( ) , or

1

2

1

202 2m mg x kxv + −( ) + = ,

where v is the speed of the block after falling distance x.

(a) When v = 0, the non zero solution to the energy equation from above gives

1

22kx mgxmax max= or k

mg

x= =

( )( )=2 2 3 00 9 80

0 100588

max

2kg m s

mN

. .

.mm

(b) When x = =5 00 0 050 0. .cm m, the energy equation gives

v = −22

gxkx

m

or

v = ( )( ) −( )( )

2 9 80 0 050 0588 0 050 0

2

. ..

m s mN m m2

33 000 700

..

kgm s=

13.68 (a) We apply conservation of mechanical energy from just after the collision until the block comes to rest.

KE PE KE PEs f s i+( ) = +( ) gives 0

1

2

1

202 2+ = +k x MVf

or the speed of the block just after the collision is

Vk x

Mf= =

( )( ) =2 2

900 0 050 0

1 001 50

N m m

kgm s

.

..

Now, we apply conservation of momentum from just before impact to immediately after the collision. This gives

m m MVi f

v vbullet bullet( ) + = ( ) +0

or

v vbullet bullet

m sk

( ) = ( ) − ⎛⎝

⎞⎠

= −

f i

M

mV

4001 00. gg

5.00 10 kgm s m s3×

⎛⎝⎜

⎞⎠⎟

( ) =− 1 5 100.




(b) The mechanical energy converted into internal energy during the collision is

∆ ΣE KE KE m mi f i f= − = ( ) − ( ) −1

2

1

2

12 2v vbullet bullet 22

2MV

or

∆E = ×( ) ( ) − ( )⎡⎣ ⎤⎦ −−1

25 00 10 400 1003 2 2. kg m s m s

11

21 00 1 50

2. .kg m s( )( )

∆E = 374 J

13.69 Choose PEg = 0 when the blocks start from rest. Then, using conservation of mechanical energy from when the blocks are released until the spring returns to its unstretched length

gives KE PE PE KE PE PEg s f g s i+ +( ) = + +( ) , or

1

240 0 0 0

1

21 22

1 22m m m g x m g x k xf+( ) + −( ) + = + +v sin °

1

225 30 25 9 80 0 2002+( )[ ] + ( )( ) (kg kg m s m2v f . . ))[ ]

− ( )(

sin

.

40

30 9 80

°

kg m s2 ))( ) = ( )( )0 2001

2200 0 200 2. .m N m m

yielding v f = 1 1. m s .

13.70 (a) When the gun is fi red, the energy initially stored as elastic potential energy in the spring is transformed into kinetic energy of the bullet. Assuming no loss of energy, we have 12

2 12

2m kxiv = , or

v = = ( )×

=−xk

mi 0 2009 80

1 00 1019 83.

.

..m

N m

kgm s

(b) From ∆y t a ty y= +v012

2 , the time required for the pellet to drop 1.00 m to the fl oor, starting

with v0 0y = , is

ty

ay

=( ) = −( )

−=

2 2 1 00

9 800 452

∆ .

..

m

m ss2

The range (horizontal distance traveled during the fl ight) is then

∆x tx= = ( )( ) =v0 19 8 0 452 8 94. . .m s s m


654 Chapter 13

13.71 The free-body diagram at the right shows the forces acting on the balloon when it is displaced distance s L= θ along the circular arc it follows. The net force tangential to this path is

F F B mg B mgxnet = = − + = − −( )Σ sin sin sinθ θ θ

For small angles, sinθ θ≈ = s L

Also, mg V g= ( )ρHe .

and the buoyant force is B V g= ( )ρair . Thus, the net restoring force acting on the balloon is

FVg

Lsnet

air He≈ −−( )⎡

⎣⎢

⎤

⎦⎥

ρ ρ

Observe that this is in the form of Hooke’s law, F k s= − , with k Vg L= −( )ρ ρair He .

Thus, the motion will be simple harmonic and the period is given by

Tf

m

k

V

Vg L= = = =

−( ) =1 22 2 2

πω

π π ρρ ρ

π ρρ

He

air He

He

air −−⎛⎝⎜

⎞⎠⎟ρHe

L

g

This yields

T =−

⎛⎝

⎞⎠

( )( ) =2

0 180

1 29 0 180

3 00

9 80π .

. .

.

.

m

m s211 40. s

13.72 (a) When the block is given some small upward displacement, the net restoring force exerted on it by the rubber bands is

F F Fynet = = −Σ 2 sinθ where tanθ = y

L

For small displacements, the angle θ will be very small. Then sin tanθ θ≈ = y

L, and the net

restoring force is

F Fy

L

F

Lynet = − ⎛

⎝⎞⎠ = − ⎛

⎝⎞⎠2

2

(b) The net restoring force found in part (a) is in the form of Hooke’s law F ky= − , with k F L= 2 . Thus, the motion will be simple harmonic, and the angular frequency is

ω = =k

m

F

m L

2

L

→B

→T

�

�

�

s � L�

Equilibriumposition

m→g

�y

�x

L

→B

→T

�

�

�

s � L�

Equilibriumposition

m→g

�y

�x



13.73 Newton’s law of gravitation is

FGMm

rM r= − = ⎛

⎝⎞⎠2

34

3, where ρ π

Thus,

F Gm r= − ⎛⎝

⎞⎠

4

3πρ

which is of Hooke’s law form, F k r= − , with

k Gm=4

3πρ

13.74 The inner tip of the wing is attached to the end of the spring and always moves with the same speed as the end of the vibrating spring. Thus, its maximum speed is

v vinner,max spring cm= = = ( ) × −

,max ..

Ak

m0 20

4 7 10 44

30 30 100 25

N m

kgcm s

..

×=−

Treating the wing as a rigid bar, all points in the wing have the same angular velocity at any instant in time. As the wing rocks on the fulcrum, the inner tip and outer tips follow circular paths of different radii. Since the angular velocities of the tips are always equal, we may write

ω = =v vouter

outer

inner

innerr r

The maximum speed of the outer tip is then

v vouter,maxouter

innerinner max=

⎛⎝⎜

⎞⎠⎟

=r

r ,

.15 000 25 1 3

mm

3.00 mmcm s cm s⎛

⎝⎞⎠ ( ) =. .

13.75 (a) ω = = =k

m

500

2 0015 8

N m

kgrad s

..

(b) Apply Newton’s second law to the block while the elevator is accelerating:

ΣF F mg may s y= − =

With F kx a gs y= =and 3, this gives kx m g g= +( )3 , or

xmg

k= =

( )( )( ) = ×4

3

4 2 00 9 805 23

. ..

kg m s

3 500 N m

2

110 5 232− =m cm.

m

M

RE

r

F

m

M

RE

r

F


656 Chapter 13

13.76 (a) Note that as the spring passes through the vertical position, the object is moving in a circular arc of radius L yf− . Also, observe that the y-coordinate of the object at this

point must be negative yf <( )0 so the spring is stretched and exerting an upward tension force of magnitude greater than the object’s weight. This is necessary so the object experiences a net force toward the pivot to supply the needed centripetal acceleration in this position. This is summarized by Newton’s second law applied to the object at this

point, stating

ΣFm

L yky mgy

ff=

−= − −v2

(b) Conservation of energy requires that E KE PE PE KE PE PEi g i s i f g f s f= + + = + +, , , , , or

E mgL m mgy kyf f= + + = + +0 01

2

1

22 2v , reducing to 2 2 2mg L y m kyf f−( ) = +v

(c) From the result of part (a), observe that m L y ky mgf fv2 = − − +( )( ). Substituting this into the result from part (b) gives 2 2mg L y L y ky mg kyf f f f( ) ( )( )− = − − + + . After expanding and regrouping terms, this becomes ( ) ( ) ( )2 3 3 02k y mg kL y mgLf f+ − + − = , which is a quadratic equation ay by cf f

2 0+ + = , with

a k= = ( ) = ×2 2 1 250 2 50 103N m N m.

b mg kL= − = ( )( ) − ( )3 3 5 00 9 80 1 250 1 50. . .kg m s N m2 m N( ) = − ×1 73 103.

and

c mgL= − = − ( )( )( ) = −3 3 5 00 9 80 1 50 221. . .kg m s m2 NN m⋅

Applying the quadratic formula, keeping only the negative solution [see the discussion in part (a)] gives

yb b ac

af = − − − =− − ×( ) − − ×( ) −2 3 3 2

4

2

1 73 10 1 73 10 4 2. . ..

.

50 10 221

2 2 50 10

3

3

×( ) −( )×( )

or yf = −0 110. m .

(d) Because the length of this pendulum varies and is longer throughout its motion than a simple pendulum of length L, its period will be longer than that of a simple pendulum.

13.77 The maximum acceleration of the oscillating system is

a A f Amax = = ( )ω π2 22

The friction force exerted between the two blocks must be capable of accelerating block B at this rate. When block B is on the verge of slipping, f f n mg mas s s s= ( ) = = =

max maxµ µ and we must have

a f A gsmax = ( ) =22π µ

Thus,

Ag

fs=

( )=

( )( )( )[ ]

µπ π2

0 600 9 802 2

. . m s

2 1.50 Hz

2

== × =−6 62 10 6 622. .m cm

→Fg � m

→g

→fs

→n

B

→Fg � m

→g

→fs

→n

B


Vibrations and Waves€¦ · Vibrations and Waves CLICKER QUESTIONS Question G1.01 Description:...

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Transcript of Vibrations and Waves€¦ · Vibrations and Waves CLICKER QUESTIONS Question G1.01 Description:...