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15Oscillatory Motion

CHAPTER OUTLINE

15.1 Motion of an Object Attached to a Spring

15.2 The Particle in Simple Harmonic Motion

15.3 Energy of the Simple Harmonic Oscillator

15.4 Comparing Simple Harmonic Motion with Uniform Circular Motion

15.5 The Pendulum15.6 Damped Oscillations15.7 Forced Oscillations

ANSWERS TO QUESTIONS

Q15.1 Neither are examples of simple harmonic motion, although they are both periodic motion. In neither case is the acceleration proportional to the position. Neither motion is so smooth as SHM. The ball’s acceleration is very large when it is in contact with the fl oor, and the student’s when the dismissal bell rings.

*Q15.2 (i) Answer (c). At 120 cm we have the midpoint between the turning points, so it is the equilibrium position and the point of maximum speed.

(ii) Answer (a). In simple harmonic motion the acceleration is a maximum when the excursion from equilibrium is a maximum.

(iii) Answer (a), by the same logic as in part (ii).

(iv) Answer (c), by the same logic as in part (i).

(v) Answer (c), by the same logic as in part (i).

(vi) Answer (e). The total energy is a constant.

395

Q15.3 You can take φ π= , or equally well, φ π= − . At t = 0, the particle is at its turning point on the negative side of equilibrium, at x A= − .

*Q15.4 The amplitude does not affect the period in simple harmonic motion; neither do constant forces that offset the equilibrium position. Thus a, b, e, and f all have equal periods. The period is pro-portional to the square root of mass divided by spring constant. So c, with larger mass, has larger period than a. And d with greater stiffness has smaller period. In situation g the motion is not quite simple harmonic, but has slightly smaller angular frequency and so slightly longer period. Thus the ranking is c > g > a = b = e = f > d.

*Q15.5 (a) Yes. In simple harmonic motion, one-half of the time, the velocity is in the same direction as the displacement away from equilibrium.

(b) Yes. Velocity and acceleration are in the same direction half the time.

(c) No. Acceleration is always opposite to the position vector, and never in the same direction.

13794_15_ch15_p395-426.indd 39513794_15_ch15_p395-426.indd 395 12/11/06 2:36:42 PM12/11/06 2:36:42 PM

396 Chapter 15

*Q15.6 Answer (e). We assume that the coils of the spring do not hit one another. The frequency will be higher than f by the factor 2 . When the spring with two blocks is set into oscillation in space, the coil in the center of the spring does not move. We can imagine clamping the center coil in place without affecting the motion. We can effectively duplicate the motion of each individual block in space by hanging a single block on a half-spring here on Earth. The half-spring with its center coil clamped—or its other half cut off—has twice the spring constant as the original uncut spring, because an applied force of the same size would produce only one-half the extension distance. Thus

the oscillation frequency in space is 1

2

22

1 2

π⎛⎝

⎞⎠

⎛⎝

⎞⎠ =k

mf . The absence of a force required to

support the vibrating system in orbital free fall has no effect on the frequency of its vibration.

*Q15.7 Answer (c). The equilibrium position is 15 cm below the starting point. The motion is symmet-ric about the equilibrium position, so the two turning points are 30 cm apart.

Q15.8 Since the acceleration is not constant in simple harmonic motion, none of the equations in Table 2.2 are valid.

Equation Information given by equation

x t A t( ) = +( )cos ω φ position as a function of time

v t A t( ) = − +( )ω ω φsin velocity as a function of time

v x A x( ) = ± −( )ω 2 2 1 2 velocity as a function of position

a t A t( ) = − +( )ω ω φ2 cos acceleration as a function of time

a t x t( ) = − ( )ω 2 acceleration as a function of position

The angular frequency ω appears in every equation. It is a good idea to fi gure out the value of angular frequency early in the solution to a problem about vibration, and to store it in calculator memory.

*Q15.9 (i) Answer (e). We have TL

gii= and T

L

g

L

gTf

f ii= = =4

2 . The period gets larger by 2 times, to become 5 s.

(ii) Answer (c). Changing the mass has no effect on the period of a simple pendulum.

*Q15.10 (i) Answer (b). The upward acceleration has the same effect as an increased gravitational fi eld.

(ii) Answer (a). The restoring force is smaller for the same displacement.

(iii) Answer (c).

Q15.11 (a) No force is exerted on the particle. The particle moves with constant velocity.

(b) The particle feels a constant force toward the left. It moves with constant acceleration toward the left. If its initial push is toward the right, it will slow down, turn around, and speed up in motion toward the left. If its initial push is toward the left, it will just speed up.

(c) A constant force towards the right acts on the particle to produce constant acceleration toward the right.

(d) The particle moves in simple harmonic motion about the lowest point of the potential energy curve.


Oscillatory Motion 397

Q15.12 The motion will be periodic—that is, it will repeat. The period is nearly constant as the angu-lar amplitude increases through small values; then the period becomes noticeably larger as θ increases farther.

*Q15.13 The mechanical energy of a damped oscillator changes back and forth between kinetic and poten-tial while it gradually and permanently changes into internal energy.

*Q15.14 The oscillation of an atom in a crystal at constant temperature is not damped but keeps constant amplitude forever.

Q15.15 No. If the resistive force is greater than the restoring force of the spring (in particular, if b mk2 4> ), the system will be overdamped and will not oscillate.

Q15.16 Yes. An oscillator with damping can vibrate at resonance with amplitude that remains constant in time. Without damping, the amplitude would increase without limit at resonance.

Q15.17 Higher frequency. When it supports your weight, the center of the diving board fl exes down less than the end does when it supports your weight. Thus the stiffness constant describing the center

of the board is greater than the stiffness constant describing the end. And then fk

m= ⎛

⎝⎞⎠

1

2π is

greater for you bouncing on the center of the board.

Q15.18 An imperceptibly slight breeze may be blowing past the leaves in tiny puffs. As a leaf twists in the wind, the fi bers in its stem provide a restoring torque. If the frequency of the breeze matches the natural frequency of vibration of one particular leaf as a torsional pendulum, that leaf can be driven into a large-amplitude resonance vibration. Note that it is not the size of the driving force that sets the leaf into resonance, but the frequency of the driving force. If the frequency changes, another leaf will be set into resonant oscillation.

Q15.19 We assume the diameter of the bob is not very small compared to the length of the cord support-ing it. As the water leaks out, the center of mass of the bob moves down, increasing the effective length of the pendulum and slightly lowering its frequency. As the last drops of water dribbleout, the center of mass of the bob hops back up to the center of the sphere, and the pendulumfrequency quickly increases to its original value.

SOLUTIONS TO PROBLEMS

Section 15.1 Motion of an Object Attached to a Spring

P15.1 (a) Since the collision is perfectly elastic, the ball will rebound to the height of 4.00 m and then

repeat the motion over and over again. Thus, the motion is periodic .

(b) To determine the period, we use: x gt= 1

22.

The time for the ball to hit the ground is tx

g= = ( )

=2 2 4 00

9 800 904

.

..

m

m ss2.

This equals one-half the period, so T = ( ) =2 0 904 1 81. .s s .

(c) The motion is not simple harmonic. The net force acting on the ball is a constant given

by F mg= − (except when it is in contact with the ground), which is not in the form of Hooke’s law.


398 Chapter 15

Section 15.2 The Particle in Simple Harmonic Motion

P15.2 (a) x t= ( ) +⎛⎝

⎞⎠5 00 2

6. coscm

π At t = 0, x = ( ) ⎛

⎝⎞⎠ =5 00

64 33. cos .cm cm

π

(b) v = = − ( ) +⎛⎝

⎞⎠

dx

dtt10 0 2

6. sincm s

π At t = 0, v = −5 00. cm s

(c) ad

dtt= = − ( ) +⎛

⎝⎞⎠

v20 0 2

6. coscm s2 π

At t = 0, a = −17 3. cm s2

(d) A = 5 00. cm and T = = =2 2

23 14

πω

π. s

P15.3 x t= ( ) +( )4 00 3 00. cos .m π π Compare this with x A t= +( )cos ω φ to fi nd

(a) ω π π= =2 3 00f .

or f = 1 50. Hz Tf

= =10 667. s

(b) A = 4 00. m

(c) φ π= rad

(d) x t =( ) = ( ) ( ) =0 250 4 00 1 75 2 83. . cos . .s m mπ

P15.4 (a) The spring constant of this spring is

kF

x= = =0 45 9 8

0 3512 6

. .

..

kg m s

mN m

2

we take the x-axis pointing downward, so φ = 0

x A t= =⋅

=cos . cos.

.ω 18 012 6

84 4cmkg

0.45 kg ss2 118 0 446 6 15 8. cos . .cm rad cm=

We choose to solve the parts in a different order.

(d) Now 446 6 71 2 0 497. .rad rad= × +π . In each cycle the object moves 4 18 72( ) = cm, so it

has moved 71 72 18 15 8 51 1cm cm m( ) + −( ) =. . .

(b) By the same steps, k = =0 44 9 8

0 35512 1

. .

..

kg m s

mN m

2

x Ak

mt= = =cos . cos

.

.. . co18 0

12 1

0 4484 4 18 0cm cm ss . .443 5 15 9rad cm= −

(e) 443 5 70 2 3 62. .rad rad= ( ) +π

Distance moved = ( ) + + =70 72 18 15 9 50 7cm cm m. .

(c) The answers to (d) and (e) are not very different given the difference in the data about the two vibrating systems. But when we ask about details of the future, the imprecision in our knowledge about the present makes it impossible to make precise predictions. The two oscillations start out in phase but get completely out of phase.



P15.5 (a) At t = 0, x = 0 and v is positive (to the right). Therefore, this situation corresponds to x A t= sinω

and v v= i cosωt

Since f = 1 50. Hz, ω π π= =2 3 00f .

Also, A = 2 00. cm, so that x t= ( )2 00 3 00. sin .cm π

(b) v vmax . . .= = = ( ) = =i Aω π π2 00 3 00 6 00 18 8cm s . cm s

The particle has this speed at t = 0 and next at tT= =2

1

3s

(c) a Amax . . .= = ( ) = =ω π π2 2 22 00 3 00 18 0 178cm s cm s2 2

This positive value of acceleration fi rst occurs at t T= =3

40 500. s

(d) Since T = 2

3s and A = 2 00. cm, the particle will travel 8.00 cm in this time.

Hence, in 1 003

2. s =⎛

⎝⎞⎠T , the particle will travel 8 00 4 00 12 0. . .cm cm cm+ = .

P15.6 (a) T = =12 02 40

..

s

5s

(b) fT

= = =1 1

2 400 417

.. Hz

(c) ω π π= = ( ) =2 2 0 417 2 62f . . rad s

P15.7 fk

m= =ω

π π2

1

2 or Tf

m

k= =1

2π

Solving for k, km

T= =

( )( ) =4 4 7 00

2 6040 9

2

2

2

2

π π .

..

kg

sN m

*P15.8 (a) For constant acceleration position is given as a function of time by

x x t a ti xi x= + +

= + ( )( ) +

v1

2

0 27 0 14 4 51

2

2

. . .m m s s −−( )( )

= −

0 32 4 5

2 34

2. .

.

m s s

m

2

(b) v vx xi xa t= + = − ( )( ) = −0 14 0 32 4 5 1 30. . . .m s m s s2 mm s

(c) For simple harmonic motion we have instead x A t= +( )cos ω φ and v = − +( )A tω ω φsinwhere a x= −ω 2 , so that − = − ( )0 32 0 272. .m s m2 ω , ω = 1 09. rad s. At t = 0, 0.27 m =

cosA φ and 0 14 1 09. . sinm s s= − ( )A φ. Dividing gives0 14

0 271 09

.

.. tan

m s

ms= − ( ) φ ,

tan .φ = −0 476, φ = −25 5. °. Still at t = 0, 0 27 25 5. cos .m = −( )A ° , A = 0 299. m. Now

at t = 4 5. s,

x = ( ) ( )( ) −⎡⎣ ⎤⎦ =0 299 1 09 4 5 25 5. cos . . .m rad s s ° 00 299 4 90 25 5

0 299 2

. cos . .

. cos

m rad

m

( ) −( )

= ( )

°

555 0 076 3° = − . m

t4.5

x

FIG. P15.8(a)

continued on next page


400 Chapter 15

(d) v = − ( )( ) = +0 299 1 09 255 0 315. . sin .m s m s°

P15.9 x A t= cosω A = 0 05. m v = −A tω ωsin a A t= − ω ω2 cos

If f = =3 600 60rev min Hz, then ω π= −120 1s

vmax . .= ( ) =0 05 120 18 8π m s m s amax . .= ( ) =0 05 120 7 112π m s km s2 2

P15.10 m = 1 00. kg, k = 25 0. N m, and A = 3 00. cm. At t = 0, x = −3 00. cm

(a) ω = = =k

m

25 0

1 005 00

.

.. rad s

so that, T = = =2 2

5 001 26

πω

π.

. s

(b) vmax . . .= = × ( ) =−Aω 3 00 10 5 00 0 1502 m rad s m s

a Amax . . .= = × ( ) =−ω 2 2 23 00 10 5 00 0 750m rad s m s2

(c) Because x = −3 00. cm and v = 0 at t = 0, the required solution is x A t= − cosω

or x t= − ( )3 00 5 00. cos . cm

v = = ( )dx

dtt15 0 5 00. sin . cm s

ad

dtt= = ( )v

75 0 5 00. cos . cm s2

P15.11 (a) ω = = = −k

m

8 00

0 5004 00 1.

..

N m

kgs so position is given by x t= ( )10 0 4 00. sin . cm

From this we fi nd that v = ( )40 0 4 00. cos . t cm s vmax .= 40 0 cm s

a t= − ( )160 4 00sin . cm s2 amax = 160 cm s2

(b) tx= ⎛

⎝⎞⎠

⎛⎝

⎞⎠

−1

4 00 10 01

.sin

. and when x = 6 00. cm, t = 0.161 s.

We fi nd v = ( )[ ] =40 0 4 00 0 161 32 0. cos . . . cm s

a = − ( )[ ] = −160 4 00 0 161 96 0sin . . . cm s2

(c) Using tx= ⎛

⎝⎞⎠

⎛⎝

⎞⎠

−1

4 00 10 01

.sin

.

when x = 0, t = 0 and when x = 8 00. cm, t = 0 232. s

Therefore, ∆t = 0 232. s

t, s4.5

x



*P15.12 We assume that the mass of the spring is negligible and that we are on Earth. Let m represent the mass of the object. Its hanging at rest is described by

ΣFy = 0 −F

g + kx = 0 mg = k(0.183 m) m�k = (0.183 m)�(9.8 N�kg)

The object’s bouncing is described by T = 2π( m�k)1�2 = 2π[(0.183 m)�(9.8 m /s2)]1�2 = 0.859 s

We do have enough information to fi nd the period. Whether the object has small or large mass, the ratio m/k must be equal to 0.183 m/(9.80 m /s2). The period is 0.859 s.

P15.13 The 0.500 s must elapse between one turning point and the other. Thus the period is 1.00 s.

ω π= =26 28

T. s

and vmax . . .= = ( )( ) =ωA 6 28 0 100 0 628s m m s .

Section 15.3 Energy of the Simple Harmonic Oscillator

P15.14 m = 200 g, T = 0 250. s, E = 2 00. J;ω π π= = =2 2

0 25025 1

T .. rad s

(a) k m= = ( ) =ω 2 20 200 126. kg 25.1 rad s N m

(b) EkA

AE

k= ⇒ = = ( )

=2

2

2 2 2 00

1260 178

.. m

P15.15 Choose the car with its shock-absorbing bumper as the system; by conservation of energy,

1

2

1

22 2m kxv = : v = = ×( ) × =−x

k

m3 16 10

5 00 10

102 232

6

3..

.m m s

P15.16 (a) EkA= =

×( )=

−2 2 2

2

250 3 50 10

20 153

N m mJ

..

(b) vmax = Aω where ω = = = −k

m

250

0 50022 4 1

.. s vmax .= 0 784 m s

(c) a Amax . . .= = × ( ) =− −ω 2 2 1 23 50 10 22 4 17 5m s m s2

P15.17 (a) E kA= = ( ) ×( ) =−1

2

1

235 0 4 00 10 28 02 2 2

. . .N m m mJ

(b) v = − = −ω A xk

mA x2 2 2 2

v =×

×( ) − ×( ) =−− −35 0

50 0 104 00 10 1 00 10 13

2 2 2 2.

.. . .002 m s

(c) 1

2

1

2

1

2

1

235 0 4 00 10 3 002 2 2 2 2

m kA kxv = − = ( ) ×( ) −−. . . ××( )⎡⎣

⎤⎦ =−10 12 22 2

. mJ

(d) 1

2

1

215 82 2kx E m= − =v . mJ


402 Chapter 15

P15.18 (a) kF

x= = =20 0

100. N

0.200 mN m

(b) ω = =k

m50 0. rad s so f = =ω

π21 13. Hz

(c) vmax . . .= = ( ) =ωA 50 0 0 200 1 41 m s at x = 0

(d) a Amax . . .= = ( ) =ω 2 50 0 0 200 10 0 m s2 at x A= ±

(e) E kA= = ( )( ) =1

2

1

2100 0 200 2 002 2. . J

(f ) v = − = ( ) =ω A x2 2 250 08

90 200 1 33. . . m s

(g) a x= = ⎛⎝

⎞⎠ =ω 2 50 0

0 200

33 33.

.. m s2

P15.19 Model the oscillator as a block-spring system.

From energy considerations, v2 2 2 2 2+ =ω ωx A

vmax = ωA and v = ωA

2 so

ω ω ωAx A

2

22 2 2 2⎛

⎝⎞⎠ + =

From this we fi nd x A2 23

4= and x A= = ±3

22 60. cm where A = 3 00. cm

P15.20 (a) y y t a tf i yi y= + +v1

22

− = + + −( )

= ⋅ =

11 0 01

29 8

22

9 81 50

2m m s

m s

m

2

2

.

..

t

t s

(b) Take the initial point where she steps off the bridge and the fi nal point at the bottom of her motion.

K U U K U U

mgy kx

g s i g s f+ +( ) = + +( )

+ + = + +0 0 0 01

2

65

2

kg m s 36 m m

N m

29 81

225

73 4

2.

.

= ( )

=

k

k

(c) The spring extension at equilibrium is xF

k= = =65

8 68kg 9.8 m s

73.4 N mm

2

. , so this point is

11 8 68 19 7+ =. .m m below the bridge and the amplitude of her oscillation is

36 − 19.7 = 16.3 m.

(d) ω = = =k

m

73 4

651 06

..

N m

kgrad s




(e) Take the phase as zero at maximum downward extension. We fi nd what the phase was 25 m higher, where x = −8 68. m:

In x A t= cosω 16 3 16 3 0. . cosm m=

− = ⎛⎝

⎞⎠8 68 16 3 1 06. . cos .m m

s

t 1 06 122 2 13. .t

srad= − = −°

t = −2 01. s

Then +2 01. s is the time over which the spring stretches.

(f ) total time = + =1 50 2 01 3 50. . .s s s

P15.21 The potential energy is

U kx kA ts = = ( )1

2

1

22 2 2cos ω

The rate of change of potential energy is

dU

dtkA t t kAs = ( ) − ( )[ ] = −1

22

1

222 2cos sin sinω ω ω ω ωtt

(a) This rate of change is maximal and negative at

22

ω πt = , 2 2

2ω π π

t = + , or in general, 2 22

ω π πt n= + for integer n

Then,

t nn= +( ) = +( )

( )−

πω

π4

4 14 1

4 3 60 1. s

For n = 0, this gives t = 0 218. s while n = 1 gives t = 1 09. s .

All other values of n yield times outside the specifi ed range.

(b) dU

dtkAs

max

. . .= = ( ) ×( )−1

2

1

23 24 5 00 10 32 2 2ω N m m 660 14 61s mW−( ) = .

Section 15.4 Comparing Simple Harmonic Motion with Uniform Circular Motion

P15.22 The angle of the crank pin is θ ω= t. Its x-coordinate is

x A A t= =cos cosθ ωwhere A is the distance from the center of the wheel to the crank pin. This is of the formx A t= +( )cos ω φ , so the yoke and piston rod move with simple harmonic motion.

x = –A x ( )t

Piston

A

ω

FIG. P15.22


404 Chapter 15

P15.23 (a) The motion is simple harmonic because the tire is rotating with constant velocity and you are looking at the motion of the bump projected in a plane perpendicular to the tire.

(b) Since the car is moving with speed v = 3 00. m s, and its radius is 0.300 m, we have

ω = =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

Section 15.5 The Pendulum

P15.24 The period in Tokyo is TL

gTT

T

= 2π

and the period in Cambridge is TL

gCC

C

= 2π

We know T TT C= = 2 00. s

For which, we see L

g

L

gT

T

C

C

=

or g

g

L

LC

T

C

T

= = =0 994 2

0 992 71 001 5

.

..

P15.25 Using the simple harmonic motion model:

A r

g

L

= = =

= = =

θ π

ω

1180

0 262

9 8

13

m 15 m

m s

m

2

°°

.

..113 rad s

(a) vmax . .= = =Aω 0 262 0 820m 3.13 s m s

(b) a Amax . .= = ( ) =ω 2 20 262 2 57m 3.13 s m s2

a rtan = α α = = =a

rtan

22m s

mrad s

2 57

12 57

..

(c) F ma= = =0 25 0 641. .kg 2.57 m s N2

More precisely,

(a) mgh m= 1

22v and h L= −( )1 cosθ

∴ = −( ) =vmax cos .2 1 0 817gL θ m s

g

FIG. P15.25




(b) I mgLα θ= sin

α θ θmax

sinsin .= = =mgL

mL

g

L i2 2 54 rad s2

(c) F mg imax sin . . sin . .= = ( )( ) =θ 0 250 9 80 15 0 0 634° N

The answers agree to two digits. The answers computed from conservation of energy and from Newton’s second law are more precisely correct. With this amplitude the motion of the pendu-lum is approximately simple harmonic.

*P15.26 Note that the angular amplitude 0.032 rad = 1.83 degree is small, as required for the SHM model of a pendulum.

ω π= 2

T: T = = =2 2

4 431 42

πω

π.

. s

ω = g

L: L

g= =( ) =

ω 2 2

9 80

4 430 499

.

.. m

P15.27 Referring to the sketch we have

F mg= − sinθ and tanθ = x

R

For small displacements, tan sinθ θ≈

and Fmg

Rx kx= − = −

Since the restoring force is proportional to the displacement from equilibrium, the motion is simple harmonic motion.

Comparing to F m x= − ω 2 shows ω = =k

m

g

R.

P15.28 (a) The string tension must support the weight of the bob, accelerate it upward, and also provide the restoring force, just as if the elevator were at rest in a gravity fi eld of 9 80 5 00. .+( ) m s2. Thus the period is

TL

g

T

= =

=

2 25 00

3 65

π π .

.

m

14.8 m s

s

2

(b) T =−( ) =2

5 00

5 006 41π .

..

m

9.80 m s m ss2 2

(c) geff = ( ) + ( ) =9 80 5 00 11 02 2

. . .m s m s m s2 2 2

T = = =25 00

4 24π ..

m

11.0 m ss2

P15.29 f = 0 450. Hz, d = 0 350. m, and m = 2 20. kg

T

f

TI

mgdT

I

mgd

I Tmgd

f

=

= =

= =⎛⎝⎜

⎞⎠⎟

1

24

4

1

22

22

;

;π π

π

22

2 2 1 24

2 20 9 80 0 350

4 0 4500

mgd

π π= ( )( )

( ) =−

. . .

. s..944 kg m2⋅

FIG. P15.27

FIG. P15.29


406 Chapter 15

P15.30 (a) From T = total measured time

50

the measured periods are:

Length, m

Period, s

L

T

( )( )

1 000 0 750 0 500

1

. . .

.9996 1 732 1 422. .

(b) TL

g= 2π so g

L

T= 4 2

2

π

The calculated values for g are:

Period, s

m s2

T

g

( )( )

1 996 1 732 1 422

9 91 9 87

. . .

. . 99 76.

Thus, gave2m s This agrees with the accepted v= 9 85. aalue of m s within 0.5%.2g = 9 80.

(c) From Tg

L224=

⎛⎝⎜

⎞⎠⎟

π, the slope of T 2 versus L graph = =4

4 012π

g. s m2 .

Thus, g = =49 85

2πslope

m s2. . This is the same as the value in (b).

P15.31 (a) The parallel axis theorem says directly I I md= +CM2

so TI

mgd

I md

mgd= =

+( )2 2

2

π π CM

(b) When d is very large Td

g→ 2π gets large.

When d is very small TI

mgd→ 2π CM gets large.

So there must be a minimum, found by

dT

dd

d

ddI md mgd

I md

= = +( ) ( )

= +( )

−0 2

2

2 1 2 1 2

2

π

π

CM

CM

11 2 3 2 1 21

22

1

2−⎛

⎝⎞⎠ ( ) + ( ) ⎛

⎝⎞⎠ +− −

mgd mg mgd I mdπ CM22 1 2

2

2 1 2 3 2

2( )

=− +( )

+( ) ( )+

−md

I md mg

I md mgd

π CM

CM

220

2 1 2 3 2

π md mgd

I md mgdCM +( ) ( )=

This requires

− − + =I md mdCM2 22 0

or I mdCM = 2 .

4

3

2

1

00.25 0.5 0.75 1.0

L, m

T2, s2

FIG. P15.30



P15.32 (a) The parallel-axis theorem:

I I Md ML Md

M M

= + = +

= ( ) + (

CM

m m

2 2 2

2

1

121

121 00 1 00. . )) = ⎛

⎝⎞⎠

= =( )

2 13

12

2 213

12 1 00

M

TI

Mgd

M

Mg

m

m

2

2

π π. m

m

12 9.80 m ss2

( )

= ( ) =213

2 09π .

(b) For the simple pendulum

T = =21 00

2 01π ..

m

9.80 m ss2 difference = − =2 09 2 01

4 08. .

. %s s

2.01 s

P15.33 T = 0 250. s, I mr= = ×( ) ×( )− −2 3 3 220 0 10 5 00 10. .kg m

(a) I = × ⋅−5 00 10 7. kg m2

(b) Id

dt

2

2

θ κθ= − ; κ ω πI T

= = 2

κ ω π= = ×( )⎛⎝

⎞⎠ = × ⋅− −I 2 7

245 00 10

2

0 2503 16 10.

..

N m

rrad

Section 15.6 Damped Oscillations

P15.34 The total energy is E m kx= +1

2

1

22 2v

Taking the time derivative, dE

dtm

d x

dtkx= +v v

2

2

Use Equation 15.31: md x

dtkx b

2

2 = − − v

dE

dtkx b k x= − −( ) +v v v

Thus, dE

dtb= − <v2 0

We have proved that the mechanical energy of a damped oscillator is always decreasing.

P15.35 θi = 15 0. ° θ t =( ) =1 000 5 50. °

x Ae bt m= − 2 x

x

Ae

Ae

i

bt mb m1 000

21 000 25 50

15 0= = =

−− ( ).

.

ln.

..

.

5 50

15 01 00

1 000

2

21 00 10

⎛⎝

⎞⎠ = − =

− ( )

∴ = ×

b

m

b

m−− −3 1s

θ

FIG. P15.33

FIG. P15.32


408 Chapter 15

P15.36 To show that x Ae tbt m= +( )− 2 cos ω φ

is a solution of − − =kx bdx

dtm

d x

dt

2

2 (1)

where ω = − ⎛⎝

⎞⎠

k

m

b

m2

2

(2)

We take x Ae tbt m= +( )− 2 cos ω φ and compute (3)

dx

dtAe

b

mt Ae tbt m bt m= −⎛

⎝⎞⎠ +( ) −− −2 2

2cos sinω φ ω ω ++( )φ (4)

d x

dt

b

mAe

b

mt Aebt m bt m

2

22 2

2 2= − −⎛

⎝⎞⎠ +( ) −− −cos ω φ ωω ω φsin t +( )⎡

⎣⎢⎤⎦⎥

− −⎛⎝

⎞⎠ +( ) + +− −Ae

b

mt Ae tbt m bt m2 2 2

2ω ω φ ω ω φsin cos (( )⎡

⎣⎢⎤⎦⎥

(5)

We substitute (3), (4) into the left side of (1) and (5) into the right side of (1);

− +( ) + +( ) +− −kAe tb

mAe t b Abt m bt m2

22

2cos cosω φ ω φ ω ee t

bAe

b

mt

bt m

bt m

−

−

+( )

= − −⎛⎝

⎞⎠ +(

2

2

2 2

sin

cos

ω φ

ω φ)) − +( )⎡⎣⎢

⎤⎦⎥

+ +

−

−

Ae t

bAe t

bt m

bt m

2

2

2

ω ω φ

ω ω

sin

sin φφ ω ω φ( ) − +( )−m Ae tbt m2 2 cos

Compare the coeffi cients of Ae tbt m− +( )2 cos ω φ and Ae tbt m− +( )2 sin ω φ :

cosine-term: − + = − −⎛⎝

⎞⎠ − = − −

⎛⎝⎜

⎞⎠

kb

m

b b

mm

b

mm

k

m

b

m

22

2 2

22 2 2 4 4ω ⎟⎟ = − +k

b

m

2

2

sine-term: bb b

bω ω ω ω= + ( ) + ( ) =2 2

Since the coeffi cients are equal, x Ae tbt m= +( )− 2 cos ω φ is a solution of the equation.

P15.37 The frequency if undamped would beω0

42 05 10

10 644 0= = × =k

m

.

..

N m

kgs.

(a) With damping

ω ω= − ⎛⎝

⎞⎠ = ⎛

⎝⎞⎠ −

⎛⎝0

22 2

244

3b

m

1

s

kg

s 2 10.6 kg⎜⎜⎞⎠⎟

= − =

= = =

1 933 96 0 02 44 0

2

44 0

27

. . .

.

1

s

sf

ωπ π

..00 Hz

(b) In x A e tbt m= +( )−0

2 cos ω φ over one cycle, a time T = 2πω

, the amplitude changes from

A0 to A e b m0

2 2− π ω for a fractional decrease of

A A e

Ae e

b m0 0

0

3 10 6 44 0 0 020 21 1−

= − = − =−

− ⋅( ) −π ω

π . . . 11 0 979 98 0 020 0 2 00− = =. . . %




(c) The energy is proportional to the square of the amplitude, so its fractional rate of decrease is twice as fast:

E kA kA e E ebt m bt m= = =− −1

2

1

22

02 2 2

0

We specify

0 05

0 05

20

3

0 03 10 6

3 10 6

3 10 6

.

.

.

.

.

E E e

e

e

t

t

t

t

=

=

=

−

−

+

110 620 3 00

10 6

.ln .

.

= =

=t s

Section 15.7 Forced Oscillations

P15.38 (a) For resonance, her frequency must match

fk

m00

3

2

1

2

1

2

4 30 10

12 52 95= = = × =

ωπ π π

.

..

N m

kgHz

(b) From x A t= cosω , v = = −dx

dtA tω ωsin , and a

d

dtA t= = −v ω ω2 cos , the maximum accel-

eration is Aω 2. When this becomes equal to the acceleration due to gravity, the normal force exerted on her by the mattress will drop to zero at one point in the cycle:

A gω 2 = or Ag g

k m

gm

k= = =

ω 2 � A =

( )( )×

=9 80 12 5

4 30 102 853

. .

..

m s kg

N mcm

2

P15.39 F t= ( )3 00 2. sin π N and k = 20 0. N m

(a) ω π π= =22

Trad s so T = 1 00. s

(b) In this case, ω0

20 0

2 003 16= = =k

m

.

.. rad s

The equation for the amplitude of a driven oscillator,

with b = 0, gives AF

m= ⎛

⎝⎞⎠ −( ) = − ( )⎡⎣ ⎤⎦

− −0 2

02 1 2 2 13

24 3 16ω ω π .

Thus, A = =0 050 9 5 09. .m cm


410 Chapter 15

P15.40 F t kx md x

dt0

2

2sinω − = ω0 = k

m (1)

x A t= +( )cos ω φ (2)

dx

dtA t= − +( )ω ω φsin (3)

d x

dtA t

2

22= − +( )ω ω φcos (4)

Substitute (2) and (4) into (1): F t kA t m A t02sin cos cosω ω φ ω ω φ− +( ) = −( ) +( )

Solve for the amplitude: kA mA t F t F t−( ) +( ) = = −( )ω ω φ ω ω20 0 90cos sin cos °

These will be equal, provided only that φ must be −90° and kA mA F− =ω 20

Thus,

AF m

k m= ( ) −

02

�

� ω

P15.41 From the equation for the amplitude of a driven oscillator with no damping,

AF m F m

f

=−( )

=−

= = ( )−

0

202 2

02

02

102 20 0

ω ω ω ω

ω π π ω. s 22 2

02

02

200

40 0 9 8049 0= =

( ) =

= −( )

−k

m

F mA

F

. / .. s

ω ω

00240 0

9 802 00 10 3 950 49 0 318= ⎛

⎝⎞⎠ ×( ) −( ) =−.

.. . N

P15.42 AF m

b m=

−( ) + ( )ext

ω ω ω202 2 2

With b = 0, AF m F m F m

=−( )

=± −( ) = ±

−ext ext ext

ω ω ω ω ω ω202 2 2

02 2

022

Thus, ω ω202 6 30

0 150

1 7= ± = ± = ±F m

A

k

m

F

mAext ext N m

kg

.

.

. 00

0 440

N

0.150 kg m( )( ).

This yields ω = 8 23. rad s or ω = 4 03. rad s

Then, f = ωπ2

gives either f = 1 31. Hz or f = 0 641. Hz

P15.43 The beeper must resonate at the frequency of a simple pendulum of length 8.21 cm:

fg

L= = =1

2

1

2

9 80

0 082 11 74

π π.

..

m s

mHz

2



Additional Problems

*P15.44 (a) Consider the fi rst process of spring compression. It continues as long as glider 1 is moving faster than glider 2. The spring instantaneously has maximum compression when both glid-ers are moving with the same speed va.

1 2 1 2

FIG. P15.44(a)

Momentum conservation:

m m m mi i f f1 1 2 2 1 1 2 2

0 24 0 74 0

v v v v+ = +

( )( ) +. .kg m s .. . . .36 0 12 0 24 0 36

0

kg m s kg kg( )( ) = ( ) + ( )v va a

..

.. ˆ220 8

0 600 368

kg m s

kgm s

⋅ = =va a

�v i

(b) Energy conservation:

K K U K K U

m m

s i s f

i i

1 2 1 2

1 12

2 221

2

1

20

+ +( ) = + +( )

+ + =v v11

2

1

21

20 24 0 74

1

2

1 22 2

2

m m kxa+( ) +

( )( ) +

v

. .kg m s 00 36 0 12

1

20 60 0 368

2

2

. .

. .

kg m s

kg m s

( )( )

= ( )( ) ++ ( )

= + ( )

1

245

0 068 3 0 040 61

245

2

2

N m

J J N m

x

x

x

. .

==( )⎛

⎝⎜⎞⎠⎟

=2 0 027 7

450 035 1

1 2.

.J

N mm

(c) Conservation of momentum guarantees that the center of mass moves with constant velocity. Imagine viewing the gliders from a reference frame moving with the center of mass. We see the two gliders approach each other with momenta in opposite directions of equal magnitude. Upon colliding they compress the ideal spring and then together bounce, extending and compressing it cyclically.

(d) 1

2

1

20 60 0 368 0 040 6

2mtot CM

2 kg m s Jv = ( )( ) =. . .

(e) 1

2

1

245 0 035 1 0 027 72 2

kA = ( )( ) =N m m J. .


412 Chapter 15

*P15.45 From a = –ω 2x, the maximum acceleration is given by amax

= ω 2A. Then 108 cm /s2 = ω 2(12 cm) ω = 3.00�s.

(a) T = 1�f = 2π�ω = 2π�(3�s) = 2 s.09

(b) f = ω�2π = (3�s)�2ω = 0 477. Hz

(c) vmax

= ωA = (3�s)(12 cm) = 36 0. cm s�

(d) E = (1�2) m vmax

2 = (1�2) m (0.36 m /s)2 = ( . )0 0648m s2 2� m

(e) ω 2 = k�m k = ω 2m = (3�s)2m = ( . )9 00�s2 m

(f ) Period, frequency, and maximum speed are all independent of mass in this situation. The energy and the force constant are directly proportional to mass.

*P15.46 (a) From a = −ω 2x, the maximum acceleration is given by amax

= ω 2A. As A increases, the maximum acceleration increases. When it becomes greater than the acceleration due to gravity, the rock will no longer stay in contact with the vibrating ground, but lag behind as the ground moves down with greater acceleration. We have then

A = g�ω 2 = g�(2πf )2 = g�4π 2f 2 = (9.8 m �s2)�4π 2(2.4�s)2 = 4 31. cm

(b) When the rock is on the point of lifting off, the surrounding water is also barely in free fall. No pressure gradient exists in the water, so no buoyant force acts on the rock.

P15.47 Let F represent the tension in the rod.

(a) At the pivot, F Mg Mg Mg= + = 2

A fraction of the rod’s weight Mgy

L⎛⎝

⎞⎠ as well

as the weight of the ball pulls down on point P. Thus, the tension in the rod at point P is

F Mgy

LMg Mg

y

L= ⎛

⎝⎞⎠ + = +⎛

⎝⎞⎠1

(b) Relative to the pivot, I I I ML ML ML= + = + =rod ball

1

3

4

32 2 2

For the physical pendulum, TI

mgd= 2π where m M= 2 and d is the distance from the

pivot to the center of mass of the rod and ball combination. Therefore,

dM L ML

M M

L= ( ) +

+=

�2 3

4 and T

ML

M g L

L

g=

( ) ( ) =24 3

2 3 4

4

3

22

π π( )�

�

For L = 2 00. m, T = ( )=4

3

2 2 00

9 802 68

π .

..

m

m ss2 .

412 Chapter 1

M

P

pivot

L

y

FIG. P15.47



P15.48 (a) Total energy = = ( )( ) =1

2

1

2100 0 200 2 002 2kA N m m J. .

At equilibrium, the total energy is:

1

2

1

216 0 8 001 2

2 2 2m m+( ) = ( ) = ( )v v v. .kg kg

Therefore,

8 00 2 002. .kg J( ) =v , and v = 0 500. m s

This is the speed of m1 and m

2 at the equilibrium point. Beyond this point, the mass m

2

moves with the constant speed of 0.500 m /s while mass m1 starts to slow down due to the

restoring force of the spring.

(b) The energy of the m1 -spring system at equilibrium is:

1

2

1

29 00 0 500 1 1251

2 2m v = ( )( ) =. . .kg m s J

This is also equal to 1

22k A′( ) , where ′A is the amplitude of the m

1 -spring system.

Therefore,

1

2100 1 1252( ) ′( ) =A . or ′ =A 0 150. m

The period of the m1 -spring system is T

m

k= =2 1 8851π . s

and it takes 1

40 471T = . s after it passes the equilibrium point for the spring to become fully

stretched the fi rst time. The distance separating m1 and m

2 at this time is:

DT

A= ⎛⎝

⎞⎠ − ′ = ( ) − =v

40 500 0 471 0 150 0 08. . . .m s s m 55 6 8 56= . cm

P15.49 The maximum acceleration of the oscillating system is a A Afmax = =ω π2 2 24 . The friction force exerted between the two blocks must be capable of accelerating block B at this rate. Thus, if Block B is about to slip,

f f n mg m Afs s= = = = ( )max µ µ π4 2 2

Ag

fs= = =

µπ π4

0 6 980

4 1 56 622 2 2 2

. ( )

( ..

cm/s

s)c

2

�mm

P15.50 Refer to the diagram in the previous problem. The maximum acceleration of the oscillating sys-tem is a A Afmax = =ω π2 2 24 . The friction force exerted between the two blocks must be capable of accelerating block B at this rate. Thus, if Block B is about to slip,

f f n mg m Afs s= = = = ( )max µ µ π4 2 2 or Ag

fs= µ

π4 2 2

f

n

mg

B

P

B

s

FIG. P15.49


414 Chapter 15

P15.51 Deuterium is the isotope of the element hydrogen with atoms having nuclei consisting of one proton and one neutron. For brevity we refer to the molecule formed by two deuterium atoms as D and to the diatomic molecule of hydrogen-1 as H.

M MD H= 2 ωω

D

H

D

H

H

D

k M

k M

M

M= = =

�

�

1

2 f

fD

H= = ×2

0 919 1014. Hz

*P15.52 (a) A time interval. If the interaction occupied no time, each ball would move with infi nite acceleration. The force exerted by each ball on the other would be infi nite, and that cannot happen.

(b) k = ⏐F⏐�⏐x⏐ = 16 000 N�0.000 2 m = 80 MN m�

(c) We assume that steel has the density of its main constituent, iron, shown in Table 14.1.

Then its mass is ρV = ρ (4�3)πr 3 = (4π�3)(7860 kg�m3)(0.0254 m�2)3 = 0.0674 kg

and K = (1� 2) mv2 = (1� 2) (0.0674 kg)(5 m /s)2 = 0 843. J

(d) Imagine one ball running into an infi nitely hard wall and bouncing off elastically. The original kinetic energy becomes elastic potential energy

0.843 J = (1� 2) (8 × 107 N�m)x2 x = 0 145. mm

(e) The half-cycle is from the equilibrium position of the model spring to maximum compres-sion and back to equilibrium again. The time is one-half the period,

(1� 2)T = (1� 2)2π(m�k)1/2 = π(0.0674 kg�80 × 106 kg�s2)1/2 = 9 12 10 5. × − s

P15.53 (a)

a

a

L

h

Li

FIG. P15.53(a)

(b) TL

g= 2π dT

dt g L

dL

dt= π 1 (1)

We need to fi nd L(t) and dL

dt. From the diagram in (a),

L La h

i= + −2 2

and dL

dt

dh

dt= − ⎛

⎝⎞⎠

1

2

But dM

dt

dV

dtA

dh

dt= = −ρ ρ . Therefore,

dh

dt A

dM

dt

dL

dt A

dM

dt= − =

⎛⎝⎜

⎞⎠⎟

1 1

2ρ ρ (2)

Also,

dLA

dM

dtt L L

L

L

i

i

∫ =⎛⎝⎜

⎞⎠⎟

⎛⎝

⎞⎠ = −1

2ρ (3)

Substituting Equation (2) and Equation (3) into Equation (1):

dT

dt g a

dM

dt L t a dM dti

=⎛⎝⎜

⎞⎠⎟

⎛⎝

⎞⎠ + ( )

πρ ρ1

2

1

22 2� �(( )




(c) Substitute Equation (3) into the equation for the period.

Tg

La

dM

dtti= + ⎛

⎝⎞⎠

2 1

2 2

πρ

Or one can obtain T by integrating (b):

dTg a

dM

dt

dt

L aT

T

i

i

∫ =⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟ + ( )

πρ ρ1

2 1 22 2� ddM dt t

T Tg a

dM

dt

t

i

�

�

( )

− =⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

∫0

2

1

2

2

1

πρ 22

1

22 2ρ ρa dM dtL

a

dM

dtt Li( )( )

⎡

⎣⎢⎢

⎤

⎦⎥⎥

+ ⎛⎝⎜

⎞⎠⎟ −

�ii

⎡

⎣⎢

⎤

⎦⎥

But TL

gT

gL

a

dM

dtti

ii= = + ⎛

⎝⎞⎠2

2 1

2 2π πρ

, so

P15.54 ω π= =k

m T

2

(a) k mm

T= =ω π2

2

2

4 (b) ′ = ′( ) = ′⎛

⎝⎜⎞⎠⎟m

k Tm

T

T

2

2

2

4π

P15.55 We draw a free-body diagram of the pendulum. The force H exerted by the hinge causes no torque about the axis of rotation.

τ α= I and d

dt

2

2

θ α= −

τ θ θ θ= + = −MgL kxh Id

dtsin cos

2

2

For small amplitude vibrations, use the approxi-mations: sin θ ≈ θ, cos θ ≈ 1, and x ≈ s = hθ.

Therefore,

d

dt

MgL kh

I

2

2

22θ θ ω θ= − +⎛

⎝⎜⎞⎠⎟

= − ω π= + =2MgL kh

MLf2 2

fMgL kh

ML= +1

2

2

2π

mg

θ

Hxh

m

L

k

Lsin q

x

kxhcos q

Hy

FIG. P15.55


416 Chapter 15

P15.56 (a) In x A t= +( )cos ω φ , v = − +( )ω ω φA tsin

we have at t = 0 v v= − = −ω φA sin max

This requires φ = 90°, so x A t= +( )cos ω 90°

And this is equivalent to x A t= − sinω

Numerically we have ω = = = −k

m

50

0 510 1N m

kgs

.

and vmax = ωA 20 10 1m s s= ( )− A A = 2 m

So x t= −( ) ( )⎡⎣ ⎤⎦−2 10 1m ssin

(b) In 1

2

1

2

1

22 2 2m kx kAv + = ,

1

23

1

22 2kx m= ⎛

⎝⎞⎠v

implies 1

3

1

2

1

2

1

22 2 2kx kx kA+ =

4

32 2x A=

x A A= ± = ± = ±3

40 866 1 73. . m

(c) ω = g

L L

g= =( ) =

−ω 2 1 2

9 8

100 098 0

..

m s

sm

2

(d) In x t= −( ) ( )⎡⎣ ⎤⎦−2 10 1m ssin

the particle is at x = 0 at t = 0, at 10t = π s, and so on.

The particle is at x = 1 m

when − 1

210 1= ( )⎡⎣ ⎤⎦

−sin s t

with solutions 106

1s−( ) = −tπ

106

1s−( ) = +t π π, and so on.

The minimum time for the motion is ∆ t in 106

∆t = ⎛⎝

⎞⎠

πs

∆t = ⎛⎝

⎞⎠

π60

s = 0.052 4 s

FIG. P15.56(d)



P15.57 (a) At equilibrium, we have

τ∑ = ⎛⎝

⎞⎠ +0

2 0− mgL

kx L

where x0 is the equilibrium compression.

After displacement by a small angle,

τ θ θ∑ = − ⎛⎝

⎞⎠ + = − ⎛

⎝⎞⎠ −( ) =mg

LkxL mg

Lk x L L k L

2 2 02+ −

But,

τ α θ∑ = =I mLd

dt

1

32

2

2

So

d

dt

k

m

2

2

3θ = − θ

The angular acceleration is opposite in direction and proportional to the displacement, so

we have simple harmonic motion with ω 2 3= k

m.

(b) fk

m= = =

( ) =ωπ π π2

1

2

3 1

2

3 100

5 001 23

N m

kgHz

..

P15.58 As it passes through equilibrium, the 4-kg object has speed

vmax .= = =ωAk

mA= 100

42 10 0

N m

kgm m s

In the completely inelastic collision momentum of the two-object system is conserved. So the new 10-kg object starts its oscillation with speed given by

4 6 0 10

4 00

kg 10 m s kg kg

m s

( ) + ( ) = ( )=

v

vmax

max .

(a) The new amplitude is given by 1

2

1

22 2m kAvmax =

10 4 100

1 26

2 2kg m s N m

m

( ) = ( )=

A

A .

Thus the amplitude has decreased by 2 00 1 26 0 735. . .m m m− =

(b) The old period was Tm

k= = =2 2

41 26π π kg

100 N ms.

The new period is T = =210

1001 99π s s2 .

The period has increased by 1 99 1 26 0 730. . .m m s− =

FIG. P15.57



418 Chapter 15

(c) The old energy was 1

2

1

24 10 2002 2

mvmax = ( )( ) =kg m s J

The new mechanical energy is 1

210 4 80

2kg m s J( )( ) =

The energy has decreased by 120 J .

(d) The missing mechanical energy has turned into internal energy in the completely inelastic collision.

P15.59 (a) TL

g= = =2

2 3 00π

ωπ . s

(b) E m= = ( )( ) =1

2

1

26 74 2 06 14 32 2v . . . J

(c) At maximum angular displacement mgh m= 1

22v h

g= v2

20 217= . m

h L L L= − = −( )cos cosθ θ1 cosθ = −1h

L θ = 25 5. °

P15.60 One can write the following equations of motion:

T kx− = 0 (describes the spring)

mg T ma md x

dt− ′ ==

2

2 (for the hanging object)

R T T Id

dt

I

R

d x

dt′ −( ) =

2

2

2

2

θ = (for the pulley)

with I MR= 1

22

Combining these equations gives the equation of motion

m Md x

dtkx mg+ 1

2

2

2⎛⎝

⎞⎠ + =

The solution is x t A tmg

k( ) = +sinω (where

mg

k arises because of the extension of the spring due

to the weight of the hanging object), with frequency

fk

m M M= =

+=

+ωπ π π2

1

2

1

2

100

0 20012

12

N m

kg.

(a) For M = 0 f = 3 56. Hz

(b) For M = 0 250. kg f = 2 79. Hz

(c) For M = 0 750. kg f = 2 10. Hz

FIG. P15.60



P15.61 Suppose a 100-kg biker compresses the suspension 2.00 cm.

Then,

kF

x= = ×−

9804 90 102

4N

2.00 10 mN m

×= .

If total mass of motorcycle and biker is 500 kg, the frequency of free vibration is

fk

m= = × =1

2

1

2

4 90 10

5001 58

4

π π.

.N m

kgHz

If he encounters washboard bumps at the same frequency as the free vibration, resonance will make the motorcycle bounce a lot. It may bounce so much as to interfere with the rider’s control of the machine.

Assuming a speed of 20.0 m /s, we fi nd these ridges are separated by

20 0

1 5812 7 101

1.

.. ~

m s

sm m− =

In addition to this vibration mode of bouncing up and down as one unit, the motorcycle can also vibrate at higher frequencies by rocking back and forth between front and rear wheels, by having just the front wheel bounce inside its fork, or by doing other things. Other spacing of bumps will excite all of these other resonances.

P15.62 (a) For each segment of the spring

dK dm x= ( )1

22v

Also,

v vx

x=�

and dmm

dx=�

Therefore, the total kinetic energy of the block-spring system is

K Mx m

dx Mm= +

⎛⎝⎜

⎞⎠⎟

+⎛⎝

⎞⎠∫

1

2

1

2

1

2 32

2 2

20

2vv

v� �

�

=

(b) ω = k

meff

and 1

2

1

2 32 2m M

meff v v= +⎛

⎝⎞⎠

Therefore,

TM m

k= = +2

23π

ωπ �

P15.63 (a) �F j∑ = −2T sin ˆθ where θ = tan− ⎛

⎝⎞⎠

1 y

L Therefore, for a small displacement

sin tanθ θ� = y

L and

�F∑ = −2Ty

Lj

(b) The total force exerted on the ball is opposite in direction and proportional to its displacement from equilibrium, so the ball moves with simple harmonic motion. For a spring system,

� �F x∑ = −k becomes here

� �F y∑ = − 2T

L

Therefore, the effective spring constant is 2T

L and ω = =k

m

T

mL

2.

FIG. P15.62

FIG. P15.63


420 Chapter 15

P15.64 (a) Assuming a Hooke’s Law type spring,

F Mg kx= =

and empirically

Mg x= −1 74 0 113. .

so

k = 1 74 6. %N m ±

M x Mg, kg , m , N

0.020 0

0.040 0

0.050 0

0.060 0

0.070 00

0.080 0

0.17

0.293

0.353

0.413

0.471

0.493

0.196

0..392

0.49

0.588

0.686

0.784

(b) We may write the equation as theoretically

Tk

Mk

ms2

2 24 4

3= +π π

and empirically

T M2 21 7 0 058 9= . .+

so

k = ±4

21 71 82 3

2π.

. %= N m

Time, s , s , kg s

7.03

9.62

10.67

11.67

12.

2T M T 2 ,

552

13.41

0.703

0.962

1.067

1.167

1.252

1.341

0.020 00

0.040 0

0.050 0

0.060 0

0.070 0

0.080 0

0.494

0.925

1..138

1.362

1.568

1.798

The k values 1 74 6. %N m ±

and 1 82 3. %N m ± differ by 4% so they agree .

(c) Utilizing the axis-crossing point, ms = ⎛⎝

⎞⎠ = ±3

0 058 9

21 78

.

.kg grams 12%

in agreement with 7.4 grams.

420 Chapter 15

FIG. P15.64



P15.65 (a) ∆ ∆K U+ = 0

Thus, K U K Utop top bot bot+ = +

where K Utop bot= = 0

Therefore, mgh I= 1

22ω , but

h R R R

R

= − = −( )

=

cos cosθ θ

ω

1

v

and IMR mr

mR= + +2 2

2

2 2

Substituting we fi nd

mgRMR mr

mRR

mgR

11

2 2 2

1

2 22

2

2−( ) = + +⎛⎝⎜

⎞⎠⎟

−

cos

c

θ v

oosθ( ) = + +⎡⎣⎢

⎤⎦⎥

M mr

R

m

4 4 2

2

22v

and v22 24

1

2= −( )

+ +( )gRM m r R

cosθ� �

so v = −( )+ +

21

22 2

Rg

M m r R

cosθ� �

(b) TI

m gdT

= 2πCM

m m MT = + dmR M

m MCM = + ( )+

0

TMR mr mR

mgR= + +

212

2 12

2 2

π

P15.66 (a) We require AeAbt m− =2

2 e bt m+ =2 2

or bt

m22= ln or

0 100

2 0 3750 693

.

..

kg s

kg( ) =t ∴ =t 5 20. s

The spring constant is irrelevant.

(b) We can evaluate the energy at successive turning points, where

cos ω φt +( ) = ±1 and the energy is 1

2

1

22 2kx kA e bt m= − . We require

1

2

1

2

1

22 2kA e kAbt m− = ⎛

⎝⎞⎠

or e bt m+ = 2 ∴ = = ( )=t

m

b

ln .

..

2 0 375

0 1002 60

kg 0.693

kg ss

(c) From E kA= 1

22, the fractional rate of change of energy over time is

dE dt

E

d dt kA

kA

k A dA dt

k

� � �=( ) ( )1

22

12

2

12

12

2( )= ( )

AA

dA dt

A2 2= �

two times faster than the fractional rate of change in amplitude.


mv

M

θ

R

θ

FIG. P15.65


422 Chapter 15

P15.67 (a) When the mass is displaced a distance x from equilibrium, spring 1 is stretched a distance x

1 and

spring 2 is stretched a distance x2.

By Newton’s third law, we expect

k x k x1 1 2 2= .

When this is combined with the requirement that

x x x= +1 2,

we fi nd xk

k kx1

2

1 2

=+

⎡

⎣⎢

⎤

⎦⎥

The force on either spring is given by Fk k

k kx ma1

1 2

1 2

=+

⎡

⎣⎢

⎤

⎦⎥ =

where a is the acceleration of the mass m.

This is in the form F k x maeff= =

and Tm

k

m k k

k keff

= =+( )

2 2 1 2

1 2

π π

(b) In this case each spring is distorted by the distance x which the mass is displaced. Therefore,the restoring force is

F k k x= − +( )1 2 and k k keff = +1 2

so that Tm

k k=

+( )21 2

π

P15.68 Let � represent the length below water at equilibrium and M the tube’s mass:

F Mg r gy∑ = − + =0 02⇒ ρπ �

Now with any excursion x from equilibrium

− + −( ) =Mg r x g Maρπ 2 �

Subtracting the equilibrium equation gives

− =

= −⎛⎝⎜

⎞⎠⎟

= −

ρπ

ρπ ω

r gx Ma

ar g

Mx x

2

22

The opposite direction and direct proportionnality of to imply SHMa x with angular frequency

ω ρ

πω

πρ

=

= = ⎛⎝

⎞⎠

πr g

M

Tr

M

g

2

2 2

The acceleration a = −ρπr 2gx�M is a negative constant times the displacement from equilibrium.

Tr

M

g= 2 π

ρ

FIG. P15.67



P15.69 (a) Newton’s law of universal gravitation is FGMm

r

Gm

rr= − = − ⎛

⎝⎞⎠2 2

34

3π ρ

Thus, F Gm r= − ⎛⎝

⎞⎠

4

3πρ

Which is of Hooke’s law form with k Gm= 4

3πρ

(b) The sack of mail moves without friction according to − ⎛⎝

⎞⎠ =4

3πρGmr ma

a Gr r= − ⎛⎝

⎞⎠ = −4

32πρ ω

Since acceleration is a negative constant times excursion from equilibrium, it executes SHM with

ω πρ= 4

3

G and period T

G= =2 3π

ωπ

ρ

The time for a one-way trip through the earth is T

G2

3

4= π

ρ

We have also gGM

R

G R

RGRe

e

e

ee= = =2

3

2

4

3

4

3

π ρ πρ

so4

3

ρπ

G g

Re

= ( ) and

T R

ge

2

6 37 102 53 10 42 2

63= = × = × =π π .

. .m

9.8 m ss2 mmin

P15.70 (a) The block moves with the board in what we take as the positive x direction, stretching the spring until the spring force −kx is equal in magnitude to the maximum force of static friction

µ µs sn mg= . This occurs at xmg

ks= µ

(b) Since v is small, the block is nearly at the rest at this break point. It starts almost immedi-ately to move back to the left, the forces on it being −kx and +µ

kmg. While it is sliding the

net force exerted on it can be written as

− + = − + = − −⎛⎝

⎞⎠ = −kx mg kx

k mg

kk x

mg

kkxk

k krelµ µ µ

where xrel

is the excursion of the block away from the point µk mg

k.

Conclusion: the block goes into simple harmonic motion centered about the equilibrium

position where the spring is stretched by µk mg

k.

(d) The amplitude of its motion is its original displacement, Amg

k

mg

ks k= −µ µ

. It fi rst comes to

rest at spring extension µ µ µk k smg

kA

mg

k− =

−( )2. Almost immediately at this point it

latches onto the slowly-moving board to move with the board. The board exerts a force of static friction on the block, and the cycle continues.



424 Chapter 15

(c) The graph of the motion looks like this:

FIG. P15.70(c)

(e) The time during each cycle when the block is moving with the board is 2 2A mg

ks k

v v=

−( )µ µ.

The time for which the block is springing back is one half a cycle of simple harmonic motion,

1

22π πm

k

m

k

⎛⎝⎜

⎞⎠⎟

= . We ignore the times at the end points of the motion when the speed of

the block changes from v to 0 and from 0 to v. Since v is small compared to 2A

m kπ �, these

times are negligible. Then the period is

Tmg

k

m

ks k=

−( )2 µ µπ

v+

(f ) T =−( )( )( )

( )2 0 4 0 25 0 3 9 8

0 024 12

. . . .

.

kg m s

m s

2

NN m

kg

12 N ms s s( ) = + =+ π 0 3

3 06 0 497 3 56.

. . .

Then

fT

= =10 281. Hz

(g) Tmg

k

m

ks k=

−( )+

2 µ µπ

v increases as m increases, so the frequency decreases .

(h) As k increases, T decreases and f increases .

(i) As v increases, T decreases and f increases .

( j) As µ µs k−( ) increases, T increases and f decreases .

ANSWERS TO EVEN PROBLEMS

P15.2 (a) 4.33 cm (b) −5 00. cm s (c) −17 3. cm s2 (d) 3.14 s; 5.00 cm

P15.4 (a) 15.8 cm (b) –15.9 cm (c) The patterns of oscillation diverge from each other, starting out in phase but becoming completely out of phase. To calculate the future we would need exact knowledge of the present, an impossibility. (d) 51.1 m (e) 50.7 m

P15.6 (a) 2.40 s (b) 0.417 Hz (c) 2.62 rad �s

P15.8 (a) −2.34 m (b) −1.30 m /s (c) −0.076 3 m (d) 0.315 m �s



P15.10 (a) 1.26 s (b) 0 150. m s; 0 750. m s2 (c) x = −3 cm cos 5t; v = ⎛⎝

⎞⎠

155

cm

ssin t;

a t= ⎛⎝

⎞⎠

755

cm

s2 cos

P15.12 Yes. Whether the object has small or large mass, the ratio m�k must be equal to 0.183 m �(9.80 m �s2). The period is 0.859 s.

P15.14 (a) 126 N m (b) 0.178 m

P15.16 (a) 0.153 J (b) 0 784. m s (c) 17 5. m s2

P15.18 (a) 100 N m (b) 1.13 Hz (c) 1 41. m s at x = 0 (d) 10 0. m s2 at x = ± A (e) 2.00 J(f ) 1 33. m s (g) 3 33. m s2

P15.20 (a) 1.50 s (b) 73.4 N�m (c) 19.7 m below the bridge (d) 1.06 rad �s (e) 2.01 s (f ) 3.50 s

P15.22 The position of the piston is given by x = Acos ωt.

P15.24 g

gC

T

= 1 001 5.

P15.26 1.42 s; 0.499 m

P15.28 (a) 3.65 s (b) 6.41 s (c) 4.24 s

P15.30 (a) see the solution (b), (c) 9 85. m s2, agreeing with the accepted value within 0.5%

P15.32 (a) 2.09 s (b) 4.08%

P15.34 see the solution


P15.38 (a) 2.95 Hz (b) 2.85 cm


P15.42 either 1.31 Hz or 0.641 Hz

P15.44 (a) 0.368 i m /s (b) 3.51 cm (c) Conservation of momentum for the glider-spring-glider system requires that the center of mass move with constant velocity. Conservation of mechanical energy for the system implies that in the center-of-mass reference frame the gliders both oscillate after they couple together. (d) 40.6 mJ (e) 27.7 mJ

P15.46 (a) 4.31 cm (b) When the rock is on the point of lifting off, the surrounding water is also barely in free fall. No pressure gradient exists in the water, so no buoyant force acts on the rock.

P15.48 (a) 0 500. m s (b) 8.56 cm

P15.50 Ag

fs= µ

π4 2 2


426 Chapter 15

P15.52 (a) A time interval. If the interaction occupied no time, the force exerted by each ball on the other would be infi nite, and that cannot happen. (b) 80.0 MN�m (c) 0.843 J. (d) 0.145 mm(e) 9.12 × 10−5 s

P15.54 (a) km

T= 4 2

2

π (b) ′ = ′⎛

⎝⎜⎞⎠⎟m m

T

T

2

P15.56 (a) x = –(2 m)sin(10 t) (b) at x = ±1.73 m (c) 98.0 mm (d) 52.4 ms

P15.58 (a) The amplitude is reduced by 0.735 m (b) The period increases by 0.730 s (c) The energy decreases by 120 J (d) Mechanical energy is converted into internal energy in the perfectly inelastic collision.

P15.60 (a) 3.56 Hz (b) 2.79 Hz (c) 2.10 Hz

P15.62 (a) 1

2 32M

m+⎛⎝

⎞⎠ v (b) T

M m

k= +

23π �

P15.64 see the solution (a) k = 1 74 6. %N m ± (b) 1 82 3. %N m ± ; they agree (c) 8 g 12%± ;it agrees

P15.66 (a) 5.20 s (b) 2.60 s (c) see the solution

P15.68 The acceleration a = −ρπr2gx/M is a negative constant times the displacement from equilibrium.

Tr

M

g= 2 π

ρ

P15.70 see the solution (f ) 0.281 Hz (g) decreases (h) increases (i) increases ( j) decreases


16Wave Motion

CHAPTER OUTLINE

16.1 Propagation of a Disturbance16.2 The Traveling Wave Model16.3 The Speed of Waves on Strings16.5 Rate of Energy Transfer by Sinusoidal

Waves on Strings16.6 The Linear Wave Equation


Q16.1 As the pulse moves down the string, the particles of the string itself move side to side. Since the medium—here, the string—moves perpendicular to the direction of wave propagation, the wave is transverse by defi nition.

Q16.2 To use a slinky to create a longitudinal wave, pull a few coils back and release. For a transverse wave, jostle the end coil side to side.

*Q16.3 (i) Look at the coeffi cients of the sine and cosine functions: 2, 4, 6, 8, 8, 7. The ranking is d = e > f > c > b > a.

(ii) Look at the coeffi cients of x. Each is the wave number, 2π�λ , so the smallest k goes with the largest wavelength. The ranking is d > a = b = c > e > f.

(iii) Look at the coeffi cients of t. The absolute value of each is the angular frequency ω = 2πf. The ranking is f > e > a = b = c = d.

(iv) Period is the reciprocal of frequency, so the ranking is the reverse of that in part iii: d = c =b = a > e > f.

(v) From v = fλ = ω�k, we compute the absolute value of the ratio of the coeffi cient of t to the coeffi cient of x in each case. From a to f respectively the numerical speeds are 5, 5, 5, 7.5, 5, 4. The ranking is d > a = b = c = e > f.

*Q16.4 From v = Tµ

, we must increase the tension by a factor of 4 to make v double. Answer (b).

*Q16.5 Answer (b). Wave speed is inversely proportional to the square root of linear density.

*Q16.6 (i) Answer (a). Higher tension makes wave speed higher.

(ii) Answer (b). Greater linear density makes the wave move more slowly.

Q16.7 It depends on from what the wave refl ects. If refl ecting from a less dense string, the refl ected part of the wave will be right side up.

Q16.8 Yes, among other things it depends on. The particle speed is described by vy,max

v= = =ω π π

λA fA

A2

2.

Here v is the speed of the wave.

427


428 Chapter 16

*Q16.9 (a) through (d): Yes to all. The maximum particle speed and the wave speed are related by

vy,max

v

= = =ω π πλ

A fAA

22

. Thus the amplitude or the wavelength of the wave can be adjusted

to make either vy,max

or v larger.

Q16.10 Since the frequency is 3 cycles per second, the period is 1

3 second = 333 ms.

Q16.11 Each element of the rope must support the weight of the rope below it. The tension increases with

height. (It increases linearly, if the rope does not stretch.) Then the wave speed v = T

µ increases

with height.

*Q16.12 Answer (c). If the frequency does not change, the amplitude is increased by a factor of 2. The wave speed does not change.

*Q16.13 (i) Answer a. As the wave passes from the massive string to the less massive string, the wave

speed will increase according to v = T

µ.

(ii) Answer c. The frequency will remain unchanged. However often crests come up to the boundary they leave the boundary.

(iii) Answer a. Since v = f λ , the wavelength must increase.

Q16.14 Longitudinal waves depend on the compressibility of the fl uid for their propagation. Transverse waves require a restoring force in response to shear strain. Fluids do not have the underlying structure to supply such a force. A fl uid cannot support static shear. A viscous fl uid can tempo-rarily be put under shear, but the higher its viscosity the more quickly it converts input work into internal energy. A local vibration imposed on it is strongly damped, and not a source of wave propagation.

Q16.15 Let ∆t t ts p= − represent the difference in arrival times of the two waves at a station at distance

d t ts s p p= =v v from the hypocenter. Then

d ts p

= −⎛

⎝⎜⎞

⎠⎟

−

∆ 1 11

v v. Knowing the distance from the

fi rst station places the hypocenter on a sphere around it. A measurement from a second sta-tion limits it to another sphere, which intersects with the fi rst in a circle. Data from a third non-collinear station will generally limit the possibilities to a point.

Q16.16 The speed of a wave on a “massless” string would be infi nite!


Section 16.1 Propagation of a Disturbance

P16.1 Replace x by x t x t− = −v 4 5.

to get yx t

=−( ) +⎡⎣ ⎤⎦

6

4 5 32.

428 Chapter


Wave Motion 429

*P16.2

(a)

(b)

The graph (b) has the same amplitude and wavelength as graph (a). it differs just by being shifted toward larger x by 2.40 m. The wave has traveled 2.40 m to the right.

P16.3 (a) The longitudinal P wave travels a shorter distance and is moving faster, so it will arrive

at point B fi rst.

(b) The wave that travels through the Earth must travel

a distance of 2 30 0 2 6 37 10 30 0 6 37 106 6R sin . . sin . .° °= ×( ) = ×m m

at a speed of 7 800 m /s

Therefore, it takes 6 37 10

8176. × =m

7 800 m ss

The wave that travels along the Earth’s surface must travel

a distance of s R R= = ⎛⎝

⎞⎠ = ×θ π

36 67 106rad m.

at a speed of 4 500 m /s

Therefore, it takes 6 67 10

4 5001 482

6. × = s

The time difference is 665 11 1s min= .

P16.4 The distance the waves have traveled is d t t= ( ) = ( ) +( )7 80 4 50 17 3. . .km s km s s where t is the travel time for the faster wave.

Then, 7 80 4 50 4 50 17 3. . . .−( )( ) = ( )( )km s km s st

or t =( )( )

−( ) =4 50 17 3

7 80 4 5023 6

. .

. ..

km s s

km ss

and the distance is d = ( )( ) =7 80 23 6 184. .km s s km .

FIG. P16.2


430 Chapter 16

P16.5 (a) Let u t x= − +10 34

π π π

du

dt

dx

dt= − =10 3 0π π at a point of constant phase

dx

dt= =10

33 33. m s

The velocity is in the positive -directionx .

(b) y 0 100 0 0 350 0 3004

0 05. , . sin . .( ) = ( ) − +⎛⎝

⎞⎠ = −m π π

44 8 5 48m cm= − .

(c) k = =23

πλ

π : λ = 0 667. m ω π π= =2 10f : f = 5 00. Hz

(d) vy

y

tt x=

∂∂

= ( )( ) − +⎛⎝

⎞⎠0 350 10 10 3

4. cosπ π π π

vy, . .max m s= ( )( ) =10 0 350 11 0π

Section 16.2 The Traveling Wave Model

*P16.6 (a) a wave

(b) later by T�4

(c) A is 1.5 times larger

(d) λ is 1.5 times larger

(e) λ is 2�3 as large

P16.7 f = =40 0 4

3

. vibrations

30.0 sHz v = =425

42 5cm

10.0 scm s.

λ = = = =vf

42 531 9 0 319

43

.. .

cm s

Hzcm m

P16.8 Using data from the observations, we have λ = 1 20. m and f = 8 00

12 0

.

. s

Therefore, v = = ( )⎛⎝

⎞⎠ =λ f 1 20

8 00

12 00 800.

.

..m

sm s

P16.9 y x t= ( ) −( )0 020 0 2 11 3 62. sin . .m in SI units A = 2 00. cm

k = 2 11. rad m λ π= =22 98

k. m

ω = 3 62. rad s f = =ωπ2

0 576. Hz

v = = = =fk

λ ωπ

π2

2 3 62

2 111 72

.

.. m s


Wave Motion 431

P16.10 v = = ( )( ) = =f λ 4 00 60 0 240 2 40. . .Hz cm cm s m s

*P16.11 (a) ω π π= = ( ) =−2 2 5 31 41f s rad s.

(b) λ = = =−

vf

204 001

m s

5 sm.

k = = =2 2

41 57

πλ

πm

rad m.

(c) In y A kx t= − +( )sin ω φ we take A = 12 cm. At x = 0 and t = 0 we have y = ( )12 cm sinφ.

To make this fi t y = 0, we take φ = 0. Then

y x t= ( ) ( ) − ( )( )12 0 1 57 31 4. sin . .cm rad m rad s

(d) The transverse velocity is ∂∂

= − −( )y

tA kx tω ωcos

Its maximum magnitude is Aω = ( ) =12 31 4 3 77cm rad s m s. .

(e) at t

A kx t A kx tyy=

∂∂

= ∂∂

− −( )( ) = − −( )vω ω ω ωcos sin2

The maximum value is Aω 2 1 20 12 31 4 118= ( )( ) =−. .m s m s2

P16.12 At time t, the phase of y x t= ( ) −( )15 0 0 157 50 3. cos . .cm at coordinate x is

φ = ( ) − ( )0 157 50 3. .rad cm rad sx t. Since 60 03

. ° = πrad, the requirement for point B is that

φ φ πB A= ±

3rad, or (since xA = 0),

0 157 50 3 0 50 3. . .rad cm rad s rad s( ) − ( ) = − ( )x t tB ±± π3

rad

This reduces to xB = ±( ) = ±π rad

rad cmcm

3 0 1576 67

.. .

P16.13 y x t= −( )0 250 0 300 40 0. sin . . m

Compare this with the general expression y A kx t= −( )sin ω

(a) A = 0 250. m

(b) ω = 40 0. rad s

(c) k = 0 300. rad m

(d) λ π π= = =2 2

0 30020 9

k ..

rad mm

(e) v = = ⎛⎝

⎞⎠ = ⎛

⎝⎞⎠ ( ) =f λ ω

πλ

π2

40 0

220 9 133

..

rad sm m ss

(f) The wave moves to the right, in direction+x .


432 Chapter 16

*P16.14 (a) See fi gure at right.

(b) T = = =2 2

50 30 125

πω

π.

. s is the time from one peak

to the next one.

This agrees with the period found in the examplein the text.

P16.15 (a) A y= = =max . .8 00 0 080 0cm m k = =( ) = −2 2

0 8007 85 1π

λπ

..

mm

ω π π π= = ( ) =2 2 3 00 6 00f . . rad s

Therefore, y A kx t= +( )sin ω

Or (where y t0 0,( ) = at t = 0) y x t= ( ) +( )0 080 0 7 85 6. sin . π m

(b) In general, y x t= + +( )0 080 0 7 85 6. sin . π φ

Assuming y x, 0 0( ) = at x = 0 100. m

then we require that 0 0 080 0 0 785= +( ). sin . φ

or φ = −0 785.

Therefore, y x t= + −( )0 080 0 7 85 6 0 785. sin . .π m

P16.16 (a)

(b) k = = =2 2

0 35018 0

πλ

π.

.m

rad m

Tf

= = =1 1

12 00 083 3

..

ss

ω π π= = =2 2 12 0 75 4f . .s rad s

v = = ( )( ) =f λ 12 0 0 350 4 20. . .s m m s

(c) y A kx t= + +( )sin ω φ specializes to

y x t= + +( )0 200 18 0 75 4. sin . .m m s φ

at x = 0, t = 0 we require

− × = +( )

= − = −

−3 00 10 0 200

8 63 0 151

2. . sin

. .

m m φφ ° rrad

so y x t x t, . sin . . .( ) = ( ) + −(0 200 18 0 75 4 0 151m m s rad))

0.0–0.1–0.2

0.10.2

y (m)

x (m)

t = 0

0.2

0.4

FIG. P16.16(a)

y (cm)

t (s)

10

00.1 0.2

–10

FIG. P16.14


Wave Motion 433

P16.17 y x t= ( ) +⎛⎝

⎞⎠0 120

84. sinm

π π

(a) v = dy

dt: v = ( )( ) +⎛

⎝⎞⎠0 120 4

84. cosπ π πx t

v 0 200 1 51. .s, 1.60 m m s( ) = −

ad

dt= v : a x t= −( )( ) +⎛

⎝⎞⎠0 120 4

842. sinm π π π

a 0 200 0. s, 1.60 m( ) =

(b) k = =π πλ8

2 : λ = 16 0. m

ω π π= =42

T: T = 0 500. s

v = = =λT

16 032 0

..

m

0.500 sm s

P16.18 (a) Let us write the wave function as y x t A kx t, sin( ) = + +( )ω φ

y A0 0 0 020 0, sin .( ) = =φ m

dy

dtA

0 0

2 00,

cos .= = −ω φ m s

Also, ω π π π= = =2 2

0 025 080 0

T ..

ss

A xii2 2

22

0 020 02 00

80 0= + ⎛

⎝⎞⎠ = ( ) +

⎛⎝

vω π

..

.m

m s

s⎜⎜⎞⎠⎟

2

A = 0 021. 5 m

(b) A

A

sin

cos

.

/ .. tan

φφ π

φ=−

= − =0 020 0

2 80 02 51

Your calculator’s answer tan . .− −( ) = −1 2 51 1 19 rad has a negative sine and positive cosine, just the reverse of what is required. You must look beyond your calculator to fi nd

φ π= − =1 19 1 95. .rad rad

(c) vy A, . . .max m s m s= = ( ) =ω π0 021 5 80 0 5 41

(d) λ = = ( )( ) =vxT 30 0 0 025 0 750. . .m s 0 s m

k = = =2 2

0 750

πλ

π. m

8.38 m ω π= 80 0. s

y x t x t, . sin . .( ) = ( ) + +0 021 5 8 38 80 0m rad m rad sπ 11 95. rad( )


434 Chapter 16

P16.19 (a) f = =( ) =v

λ1 00

2 000 500

.

..

m s

mHz

ω π π= = ( ) =2 2 0 500f . s 3.14 rad s

(b) k = = =2 2

2 003 14

πλ

π.

.m

rad m

(c) y A kx t= − +( )sin ω φ becomes

y x t= ( ) − +( )0 100 3 14 3 14 0. sin . .m m s

(d) For x = 0 the wave function requires

y t= ( ) −( )0 100 3 14. sin .m s

(e) y t= ( ) −( )0 100 4 71 3 14. sin . .m rad s

(f) vy

y

tx t= ∂

∂= −( ) −( )0 100 3 14 3 14 3 14. . cos . .m s m s

The cosine varies between +1 and –1, so

vy ≤ 0 314. m s

P16.20 (a) At x = 2 00. m, y t= ( ) −( )0 100 1 00 20 0. sin . .m rad Because this disturbance varies

sinusoidally in time, it describes simple harmonic motion.

(b) y x t A kx t= ( ) −( ) = −( )0 100 0 500 20 0. sin . . sinm ω

so ω = 20 0. rad s and f = =ωπ2

3 18. Hz

Section 16.3 The Speed of Waves on Strings

P16.21 The down and back distance is 4 00 4 00 8 00. . .m m m+ = .

The speed is then v = = ( )= =

d

t

Ttotal m

sm s

4 8 00

0 80040 0

.

..

µ

Now, µ = = × −0 2005 00 10 2..

kg

4.00 mkg m

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

P16.22 (a) ω π π= = ( ) =2 2 500 3140f rad s, k = = =ωv

3140

19616 0. rad m

y x t= ×( ) −( )−2 00 10 16 0 31404. sin .m

(b) v = =× −196

4 10 10 3m skg m

T

.

T = 158 N


Wave Motion 435

P16.23 v = = ⋅×

=−

T

µ1 350

5 00 105203

kg m s

kg mm s

2

.

P16.24 v = T

µ

T A r

T

= = =

= ( )( ) × −

µ ρ ρπ

π

v v v2 2 2 2

48 920 7 50 10kg m3 . mm m s

N

( ) ( )=

2 2200

631T

P16.25 T Mg= is the tension; v = = = =T Mg

m L

MgL

m

L

tµ �is the wave speed.

Then, MgL

m

L

t=

2

2

and gLm

Mt= =

×( )×

−

−2

31 60 4 00 10

3 00 10

. .

.

m kg

kg 3.61 22 2 1 64s

m s2

( ) = .

P16.26 The period of the pendulum is TL

g= 2π

Let F represent the tension in the string (to avoid confusion with the period) when the pendulum is vertical and stationary. The speed of waves in the string is then:

v = = =F Mg

m L

MgL

mµ �

Since it might be diffi cult to measure L precisely, we eliminate LT g

=2π

so

v = =Mg

m

T g Tg M

m2 2π π

P16.27 Since µ

is constant, µ = =T T2

22

1

12v v

and

T T22

1

2

1

230 0

20 06 00=

⎛⎝⎜

⎞⎠⎟

=⎛⎝⎜

⎞⎠⎟

vv

.

..

m s

m sNN N( ) = 13 5.

P16.28 From the free-body diagram mg T= 2 sinθ

Tmg=

2sinθ

The angle θ is found from cosθ = =3 8

2

3

4

L

L

�

�

∴ =θ 41 4. °

(a) v = T

µ v = =

×( )−

mg

2 41 4

9 80

2 8 00 10 43µ sin .

.

. sin°

m s

kg m

2

11 4. °

⎛

⎝⎜

⎞

⎠⎟ m

or v =⎛⎝⎜

⎞⎠⎟

30 4.m s

kgm

(b) v = =60 0 30 4. . m and m = 3 89. kg

FIG. P16.28


436 Chapter 16

P16.29 If the tension in the wire is T, the tensile stress is

Stress = = ( )T

AT Aso stress

The speed of transverse waves in the wire is

v = = ( )= =

T A

m L m AL mµStress Stress Stress

Volume� � �==

Stress

ρ where ρ is the density. The maximum velocity occurs when the stress is a maximum:

vmax

.= × =2 70 101

8 Pa

7 860 kg m85 m s3

*P16.30 (a) f has units Hz s= 1 , so Tf

= 1 has units of seconds, s . For the other T we have T = µv2,

with units kg

m

m

s

kg m

sN

2

2 2= ⋅ = .

(b) The fi rst T is period of time; the second is force of tension.

P16.31 The total time is the sum of the two times.

In each wire tL

LT

= =v

µ

Let A represent the cross-sectional area of one wire. The mass of one wire can be written both as m V AL= =ρ ρ and also as m L= µ .

Then we have µ ρ πρ= =Ad 2

4

Thus, t Ld

T=

⎛⎝⎜

⎞⎠⎟

πρ 2 1 2

4

For copper, t = ( ) ( )( ) ×( )( )( )

⎡

⎣⎢⎢

⎤

⎦

−

20 08 920 1 00 10

4 150

3 2

..π

⎥⎥⎥

=

1 2

0 137. s

For steel, t = ( ) ( )( ) ×( )( )( )

⎡

⎣⎢⎢

⎤

⎦

−

30 07 860 1 00 10

4 150

3 2

..π

⎥⎥⎥

=

1 2

0 192. s

The total time is 0 137 0 192 0 329. . .+ = s

Section 16.5 Rate of Energy Transfer by Sinusoidal Waves on Strings

P16.32 f = = =vλ

30 0

0 50060 0

.

.. Hz ω π π= =2 120f rad s

P = = ⎛⎝

⎞⎠ ( ) ( )1

2

1

2

0 180

3 60120 0 100 302 2 2 2µω πA v

.

.. .. .0 1 07( ) = kW


Wave Motion 437

P16.33 Suppose that no energy is absorbed or carried down into the water. Then a fi xed amount of power is spread thinner farther away from the source. It is spread over the circumference 2πr of an expanding circle. The power-per-width across the wave front

P2πr

is proportional to amplitude squared so amplitude is proportional to

P2πr

P16.34 T = constant; v = T

µ; P = 1

22 2µω A v

(a) If L is doubled, v remains constant and P is constant .

(b) If A is doubled and ω is halved, P � ω 2 2A remains constant .

(c) If λ and A are doubled, the product ωλ

2 22

2AA

� remains constant, so P remains constant .

(d) If L and λ are halved, then ωλ

22

1� is quadrupled, so P is quadrupled .

(Changing L doesn’t affect P .)

P16.35 A = × −5 00 10 2. m µ = × −4 00 10 2. kg m P = 300 W T = 100 N

Therefore, v = =T

µ50 0. m s

P = 1

22 2µω A v : ω 2 = 2P _____ µ A2v

2 2 2

2 300

4 00 10 5 00 10 50 0= ( )

×( ) ×( )− −. . .(( )

ωωπ

=

= =

346

255 1

rad s

Hzf .

P16.36 µ = = × −30 0 30 0 10 3. .g m kg m

λω π

=

= = =

=

−

1 50

50 0 2 314

2 0 150

1

.

. :

. :

m

Hz s

m

f f

A A == × −7 50 10 2. m

(a) y A x t= −⎛⎝

⎞⎠sin

2πλ

ω

y x t= ×( ) −( )−7 50 10 4 19 3142. sin .

(b) P = = ×( )( ) ×( )− −1

2

1

230 0 10 314 7 50 10

32 2 3 2 2 2µω A v . .114

4 19.⎛⎝

⎞⎠ W P = 625 W

FIG. P16.36


438 Chapter 16

P16.37 (a) v = = = = =fk k

λ ωπ

π ω2

2 50 0

0 80062 5

.

..m s m s

(b) λ π π= = =2 2

0 8007 85

k ..m m

(c) f = =50 0

27 96

..

πHz

(d) P = = ×( )( ) ( )−1

2

1

212 0 10 50 0 0 150 62 52 2 3 2 2µω A v . . . .(( ) =W W21 1.

P16.38 Comparing y t x= − +⎛⎝

⎞⎠0 35 10 3

4. sin π π π

with y A kx t A t kx= − +( ) = − − +( )sin sinω φ ω φ π

we have k

m= 3π

, ω π= 10 s, A = 0 35. m. Then v = = = = =f fk

λ π λπ

ω ππ

22

10

33 33

s

mm s. .

(a) The rate of energy transport is

P = = ×( )( ) ( )−1

2

1

275 10 10 0 35 32 2 3 2 2µω πA v kg m s m. .. .33 15 1m s W=

(b) The energy per cycle is

Eλ = P T Aµω λ π= = ×( )( )−1

2

1

275 10 10 0 352 2 3 2

kg m s m.(( ) =2 23 02

ππm

3J.

P16.39 Originally,

P

P

P

02 2

02 2

02 2

1

2

1

2

1

2

=

=

=

µω

µωµ

ω µ

A

AT

A T

v

The doubled string will have doubled mass-per-length. Presuming that we hold tension constant, it can carry power larger by 2 times.

21

22

02 2P = ω µA T

*P16.40 As for a string wave, the rate of energy transfer is proportional to the square of the amplitude and to the speed. The rate of energy transfer stays constant because each wavefront carries con-stant energy and the frequency stays constant. As the speed drops the amplitude must increase.

We write P = F Av 2 where F is some constant. With no absorption of energy,

F A F Av v

v

bedrock bedrock2

mudfill mudfill2

bedro

=

cck

mudfill

mudfill

bedrock

mudfill

mudvv

v= =

A

A

25

ffill

= 5

The amplitude increases by 5.00 times.


Wave Motion 439

Section 16.6 The Linear Wave Equation

P16.41 (a) A = +( )7 00 3 00 4 00. . . yields A = 40 0.

(b) In order for two vectors to be equal, they must have the same magnitude and the same

direction in three-dimensional space. All of their components must be equal.

Thus, 7 00 0 3 00. ˆ ˆ . ˆ ˆ ˆ ˆi j k i j k+ + = + +A B C requires A B C= = =7 00 0 3 00. , , .and .

(c) In order for two functions to be identically equal, they must be equal for every value

of every variable. They must have the same graphs.

In

A B Cx Dt E x t+ + +( ) = + + +cos . cos . . .0 7 00 3 00 4 00 2 0mm 00( )

the equality of average values requires that A = 0 . The equality of maximum values requires

B = 7 00. mm . The equality for the wavelength or periodicity as a function of x requires

C = 3 00. rad m . The equality of period requires D = 4 00. rad s , and the equality of

zero-crossings requires E = 2 00. rad .

P16.42 The linear wave equation is ∂∂

= ∂∂

2

2 2

2

2

1y

x

y

tv

If y eb x t= −( )v

then ∂∂

= − −( )y

tb eb x tv v and

∂∂

= −( )y

xbeb x tv

∂∂

= −( )2

22 2y

tb eb x tv v and

∂∂

= −( )2

22y

xb eb x tv

Therefore, ∂∂

= ∂∂

2

22

2

2

y

t

y

xv , demonstrating that eb x t−( )v is a solution.

P16.43 The linear wave equation is 12

2

2

2

2v∂∂

= ∂∂

y

t

y

x

To show that y b x t= −( )[ ]ln v is a solution, we fi nd its fi rst and second derivatives with respect

to x and t and substitute into the equation.

∂∂

=−( ) −( )y

t b x tb

1

vv ∂

∂=

− −( )−( )

= −−( )

2

2

2

2 2

2

2

1y

t

b

b x t x t

vv

vv

∂∂

= −( )[ ]−y

xb x t bv 1 ∂

∂= − − = −

−−

2

2

2

2

1y

x

b

bx t

x t( )

( )v

v

Then 1 1 12

2

2 2

2

2 2

2

2v v

v

v v∂∂

=−( )−( ) = −

−( ) = ∂∂

y

t x t x t

y

x so the given wave function is a solution.


440 Chapter 16

P16.44 (a) From y x t= +2 2 2v ,

evaluate ∂∂

=y

xx2 ∂

∂=

2

2 2y

x

∂∂

=y

ttv2 2 ∂

∂=

2

222

y

tv

Does ∂∂

= ∂∂

2

2 2

2

2

1y

t

y

tv?

By substitution, we must test 21

222=

vv and this is true, so the wave function does satisfy

the wave equation.

(b) Note

1

2

1

22 2x t x t+( ) + −( )v v = + + + − +1

2

1

2

1

2

1

22 2 2 2 2 2x x t t x x t tv v v v

= +x t2 2 2v as required.

So

f x t x t+( ) = +( )v v1

22 and g x t x t−( ) = −( )v v

1

22

(c) y x t= sin cos v makes

∂∂

=y

xx tcos cos v ∂

∂= −

2

2

y

xx tsin cos v

∂∂

= −y

tx tv vsin sin ∂

∂= −

2

22y

tx tv vsin cos

Then∂∂

= ∂∂

2

2 2

2

2

1y

x

y

tv

becomes − = −sin cos sin cosx t x tv

vv v

12

2 which is true as required.

Note sin sin cos cos sinx t x t x t+( ) = +v v v

sin sin cos cos sinx t x t x t−( ) = −v v v

So sin cosx t f x t g x tv v v= +( ) + −( ) with

f x t x t+( ) = +( )v v1

2sin and g x t x t−( ) = −( )v v

1

2sin

Additional Problems

P16.45 Assume a typical distance between adjacent people ~1 m.

Then the wave speed is v = ∆∆

x

t~ ~

110

m

0.1 sm s

Model the stadium as a circle with a radius of order 100 m. Then, the time for one circuit around the stadium is

Tr

=( )

=2 2 10

1063 1

2π πv

~ ~m s

s min


Wave Motion 441

*P16.46 (a) From y = 0.150 m sin(0.8x – 50t)

we compute dy�dt = 0.150 m (–50) cos(0.8x – 50t)

and a = d2y�dt2 = –0.150 m (–50�s)2 sin(0.8x – 50t)

Then amax

= 375 m/s2

(b) For the 1-cm segment with maximum force acting on it, ΣF = ma = [12 g�(100 cm)] 1 cm 375 m /s2 = 0.045 0 N

We fi nd the tension in the string from v = f λ = ω �k = (50�s)�(0.8�m) = 62.5 m /s = (T�µ)1�2

T = v2µ = (62.5 m /s)2(0.012 kg�m) = 46.9 N.

The maximum transverse force is very small compared to the tension, more than athousand times smaller.

P16.47 The equation v = λ f is a special case of

speed = (cycle length)(repetition rate)

Thus,

v = ×( )( ) =−19 0 10 24 0 0 4563. . .m frame frames s m s

P16.48 (a) 0 175 0 350 99 6. . sin .m m rad s= ( ) ( )⎡⎣ ⎤⎦t

∴ ( )⎡⎣ ⎤⎦ =sin . .99 6 0 5rad s t

The smallest two angles for which the sine function is 0.5 are 30° and 150°, i.e., 0.523 6 rad and 2.618 rad.

99 6 0 523 61. .rad s rad( ) =t , thus t1 5 26= . ms

99 6 2 6182. .rad s rad( ) =t , thus t2 26 3= . ms

∆t t t� 2 1 26 3 5 26 21 0− = − =. . .ms ms ms

(b) Distance traveled by the wave = ⎛⎝

⎞⎠ =

⎛⎝⎜

⎞⎠⎟

× −ωk

t∆ 99 6

1 2521 0 10 3.

..

rad s

rad ms m( ) = 1 68. .

P16.49 Energy is conserved as the block moves down distance x:

K U U E K U U

Mgx

g s g s+ +( ) + = + +( )+ + + = +

top bottom∆

0 0 0 0 001

22

2+

=

kx

xMg

k

(a) T kx Mg= = = ( )( ) =2 2 2 00 9 80 39 2. . .kg m s N2

(b) L L x LMg

k= + = +0 0

2

L = + =0 50039 2

0 892..

.mN

100 N mm

(c) v = =T TL

mµ

v

v

= ××

=

−

39 2 0 892

10

83 6

3

. .

.

N m

5.0 kg

m s


442 Chapter 16

P16.50 Mgx kx= 1

22

(a) T kx Mg= = 2

(b) L L x LMg

k= + = +0 0

2

(c) v = = = +⎛⎝

⎞⎠

T TL

m

Mg

mL

Mg

kµ2 2

0

*P16.51 (a) The energy a wave crest carries is constant in the absence of absorption. Then the rate at which energy passes a stationary point, which is the power of the wave, is constant. The power is proportional to the square of the amplitude and to the wave speed. The speed decreases as the wave moves into shallower water near shore, so the amplitude must increase.

(b) For the wave described, with a single direction of energy transport, the intensity is the same at the deep-water location 1 and at the place 2 with depth 9 m. To express the constant inten-sity we write

A A A gd

A

12

1 22

2 22

2

2221 8 9 8

v v= =

( ) =. .m 200 m s m s22 m

m s

m s

m s

( )=

=⎛⎝⎜

⎞⎠⎟

9

9 39

1 8200

9 39

22

2

A

A

.

..

11 2

8 31= . m

(c) As the water depth goes to zero, our model would predict zero speed and infi nite amplitude. In fact the amplitude must be fi nite as the wave comes ashore. As the speed decreases the wavelength also decreases. When it becomes comparable to the water depth, or smaller, our formula gd for wave speed no longer applies.

P16.52 Assuming the incline to be frictionless and taking the positive x-direction to be up the incline:

F T Mgx∑ = − =sinθ 0 or the tension in the string is T Mg= sinθ

The speed of transverse waves in the string is then v = = =T Mg

m L

MgL

mµθ θsin sin

�

The time interval for a pulse to travel the string’s length is ∆tL

Lm

MgL

mL

Mg= = =

v sin sinθ θ


Wave Motion 443

*P16.53 (a) In P = 12

2 2µω A v where v is the wave speed, the quantity ω A is the maximum particle

speed vymax

. We have µ = 0.5 × 10–3 kg�m and v = (T�µ)1�2 = (20 N�0.5 × 10–3 kg�m)1�2 =

200 m /s

Then P = 12 (0.5 × 10–3 kg�m) v2

ymax (200 m �s) = (0.050 0 kg/s)v2

y, max

(b) The power is proportional to the square of tthe maximum particle speed.

(c) In time t = (3 m)�v = (3 m)�(200 m /s) = 1.5 × 10–2 s, all the energy in a 3-m length of string goes past a point. Therefore the amount of this energy is

E = P t = (0.05 kg�s) v2y, max

(0.015 s) = 7.5 × 10−4 kg vy,max2

The mass of this section is m3 = (0.5 × 10–3 kg�m)3 m = 1.5 × 10–3 kg so (1�2)m

3 = 7.5 × 10−4 kg

and E = (1/2) m3v2

y,max = K

max. The string also contains potential energy. We could write its

energy as Umax

or as Uavg

+ Kavg

(d) E = P t = (0.05 kg�s) v2y, max

(6 s) = 0.300 kg vy,max2

P16.54 v = T

µand in this case T mg= ; therefore, m

g= µv2

Now v = f λ implies v = ωk

so that

mg k

= ⎛⎝

⎞⎠ =

−µ ω ππ

2 10 250

9 80

18

0 750

.

. .

kg m

m s

s2 mm

kg−

⎡⎣⎢

⎤⎦⎥

=1

2

14 7.

P16.55 Let M = mass of block, m = mass of string. For the block, F ma∑ = implies Tm

rm rb= =v2

2ω

The speed of a wave on the string is then

v

v

= = =

= =

= = =

T M r

m rr

M

m

tr m

M

tm

M

µω ω

ω

θ ω

2

1

0 003 2

�

. kg

00.450 kgrad= 0 084 3.

P16.56 (a) µ ρ ρ= = =dm

dLA

dx

dxA

v = = =+( )[ ] =

+( )⎡⎣ ⎤⎦− −

T T

A

T

ax b

T

xµ ρ ρ ρ 10 103 2 cm2

With all SI units,

v =+( )⎡⎣ ⎤⎦

− − −

T

xρ 10 10 103 2 4m s

(b) v x= − −=

( ) +( )( )⎡⎣ ⎤⎦=0 2 4

24 0

2 700 0 10 1094 3

.. m s

v x= − − −=

( ) +( )( )⎡⎣ ⎤⎦=10 0 2 2 4

24 0

2 700 10 10 1066.

..77 m s


444 Chapter 16

P16.57 v = T

µ where T xg= µ , to support the weight of a length x, of rope.

Therefore, v = gx

But v = dx

dt, so that dt

dx

gx=

and tdx

gx g

x L

g

L L

= = =∫0

12 0

12

P16.58 At distance x from the bottom, the tension is Tmxg

LMg= ⎛

⎝⎞⎠ + , so the wave speed is:

v = = = + ⎛⎝

⎞⎠ =T TL

mxg

MgL

m

dx

dtµ

(a) Then t dt xgMgL

mdx

t L

= = + ⎛⎝

⎞⎠

⎡⎣⎢

⎤⎦⎥∫ ∫

−

0

1 2

0

tg

xg MgL m

x

x L

=+ ( )⎡⎣ ⎤⎦

=

=

11 2

12

0

tg

LgMgL

m

MgL

m= +⎛

⎝⎞⎠ − ⎛

⎝⎞⎠

⎡

⎣⎢

⎤

⎦⎥

2 1 2 1 2

tL

g

m M M

m= + −⎛

⎝⎜⎞⎠⎟

2

(b) When M = 0, as in the previous problem, tL

g

m

m

L

g= −⎛

⎝⎜⎞⎠⎟

=20

2

(c) As m → 0 we expand m M Mm

MM

m

M

m

M+ = +⎛

⎝⎞⎠ = + − +

⎛⎝⎜

⎞⎠⎟

1 11

2

1

8

1 2 2

2…

to obtain tL

g

M m M m M M

m=

+ ( ) − ( ) + −⎛

⎝⎜

⎞

⎠⎟2

12

18

2 3 2 …

tL

g

m

M

mL

Mg≈

⎛⎝⎜

⎞⎠⎟

=21

2

P16.59 (a) The speed in the lower half of a rope of length L is the same function of distance (from the

bottom end) as the speed along the entire length of a rope of length L

2⎛⎝

⎞⎠

.

Thus, the time required = ′2

L

g with ′ =L

L

2

and the time required = =⎛⎝⎜

⎞⎠⎟

22

0 707 2L

g

L

g.

It takes the pulse more that 70% of the total time to cover 50% of the distance.

(b) By the same reasoning applied in part (a), the distance climbed in τ is given by dg= τ 2

4

For τ = =t L

g2, we fi nd the distance climbed = L

4.

In half the total trip time, the pulse has climbed 1

4 of the total length.


Wave Motion 445

P16.60 (a) Consider a short section of chain at the top of the loop. A free-body diagram is shown. Its length is s = R(2θ) and its mass is µ R2θ. In the frame of reference of the center of the loop, Newton’s second law is

F may y∑ = 220

202

Tm

R

R

Rsinθ µ θ

down down= =v v

For a very short section, sinθ θ= and T = µ v02

(b) The wave speed is v v= =T

µ 0

(c) In the frame of reference of the center of the loop, each pulse moves with equal speed clockwise and counterclockwise.

In the frame of reference of the ground, once pulse moves backward at speed v v v0 02+ = and the other forward at v v0 0− = .

The one pulse makes two revolutions while the loop makes one revolution and the other pulse does not move around the loop. If it is generated at the six-o’clock position, it will stay at the six-o’clock position.

P16.61 Young’s modulus for the wire may be written as Y = T A

L L

�

�∆, where T is the tension maintained in

the wire and ∆ L is the elongation produced by this tension. Also, the mass density of the wire

may be expressed as ρ µ=A

The speed of transverse waves in the wire is then

v = = =( )T T A

A

Y L L

µ µ ρ�

�

�∆

and the strain in the wire is ∆L

L=

ρv2

Y

If the wire is aluminum and v = 100 m s, the strain is

∆L

L=

×( )( )×

=2 70 10 100

7 00 103

3 2

10

.

.

kg m m s

N m

3

2 ..86 10 4× −

R2θ

θθ

T T

FIG. P16.60(a)

v0 v0 v0

v v

FIG. P16.60(c1)

v0 v0 v0

FIG. P16.60(c2)


446 Chapter 16

P16.62 (a) Assume the spring is originally stationary throughout, extended to have a length L much greater than its equilibrium length. We start moving one end forward with the speed v at which a wave propagates on the spring. In this way we create a single pulse of compression that moves down the length of the spring. For an increment of spring with length dx and mass dm, just as the pulse swallows it up, F ma∑ =

becomes kdx adm= or k

dm dxa

�=

But dm

dx= µ so a

k=µ

Also, ad

dt t= =v v

when vi = 0

But L = vt, so aL

= v2

Equating the two expressions for a, we have k

Lµ= v2

or v = kL

µ

(b) Using the expression from part (a) v = = =( )( )

=kL kL

mµ

2 2100 2 00

0 40031 6

N m m

kgm

.

.. ss

P16.63 (a) P x A A ek k

A ebx( ) = = ⎛⎝

⎞⎠ =−1

2

1

2 22 2 2

02 2

3

02µω µω ω µω

v −−2bx

(b) P 02

3

02( ) =

µωk

A

(c) P (x)

_____ P (0) e bx= −2

P16.64 v = = =4 450468 130

km

9.50 hkm h m s

dg

= =( )( ) =v2 2130

9 801 730

m s

m sm2.


Wave Motion 447

P16.65 (a) µ x( ) is a linear function, so it is of the form µ x mx b( ) = +

To have µ µ0 0( ) = we require b = µ0. Then µ µ µL mLL( ) = = + 0

so mL

L=−µ µ0

Then µµ µ

µxx

LL( ) =

−( )+0

0

(b) From v = dx

dt, the time required to move from x to x + dx is dx

v. The time required to move

from 0 to L is

∆

∆

tdx dx

T Tx dx

tT

x

L L L

L

= = = ( )

=−( )

∫ ∫ ∫v0 0 0

0

1

1

�µµ

µ µLL L

dxLL

L

L

+⎛

⎝⎜⎞⎠⎟

−⎛⎝⎜

⎞⎠⎟ −

⎛⎝⎜

⎞⎠⎟∫ µ µ µ

µ µ0

1 2

0

00

∆

∆

tT

L x

LL

L

L

=−

⎛⎝⎜

⎞⎠⎟

−( )+

⎛

⎝⎜⎞⎠⎟

1 1

0

00

3 2

32 0

µ µµ µ

µ

ttL

T

tL

LL

L L L

=−( ) −( )

=−( ) +

2

3

2

0

3 203 2

0

µ µµ µ

µ µ µ µ µ∆ 00 0

0 0

0 0

3

2

3

+( )−( ) +( )

=+ +

+

µ

µ µ µ µ

µ µ µ µµ

T

tL

T

L L

L L

L

∆µµ0

⎛

⎝⎜⎞

⎠⎟


P16.2 See the solution. The graph (b) has the same amplitude and wavelength as graph (a). It differs just by being shifted toward larger x by 2.40 m. The wave has traveled 2.40 m to the right.

P16.4 184 km

P16.6 See the solution

P16.8 0 800. m s

P16.10 2 40. m s

P16.12 ±6 67. cm

P16.14 (a) see the solution (b) 0.125 s, in agreement with the example

P16.16 (a) see the solution (b) 18.0 m; 83 3. ms; 75 4. rad s; 4 20. m s(c) 0 2 18 75 4 0 151. sin . .m( ) + −( )x t

P16.18 (a) 0.021 5 m (b) 1.95 rad (c) 5 41. m s (d) y x t x t, . sin . . .( ) = ( ) + +( )0 021 5 8 38 80 0 1 95m π


448 Chapter 16

P16.20 (a) see the solution (b) 3.18 Hz

P16.22 (a) y x t= ( ) −( )0 2 16 3140. sinmm (b) 158 N

P16.24 631 N

P16.26 v = Tg M

m2π

P16.28 (a) v =⋅

⎛⎝⎜

⎞⎠⎟

30 4.m

s kgm (b) 3 89. kg

P16.30 (a) s and N (b) The fi rst T is period of time; the second is force of tension.

P16.32 1.07 kW

P16.34 (a), (b), (c) P is a constant (d) P is quadrupled

P16.36 (a) y x t= ( ) −( )0 075 0 4 19 314. sin . (b) 625 W

P16.38 (a) 15.1 W (b) 3.02 J

P16.40 As for a string wave, the rate of energy transfer is proportional to the square of the amplitude and to the speed. The rate of energy transfer stays constant because each wavefront carries constant energy and the frequency stays constant. As the speed drops the amplitude must increase. It increases by 5.00 times.


P16.44 (a) see the solution (b) 1

2

1

22 2x t x t+( ) + −( )v v (c)

1

2

1

2sin sinx t x t+( ) + −( )v v

P16.46 (a) 375 m /s2 (b) 0.0450 N. This force is very small compared to the 46.9-N tension, more than a thousand times smaller.

P16.48 (a) 21.0 ms (b) 1.68 m

P16.50 (a) 2Mg (b) LMg

k0

2+ (c) 2 2

0

Mg

mL

Mg

k+⎛

⎝⎞⎠

P16.52 ∆tmL

Mg=

sinθ

P16.54 14.7 Kg

P16.56 (a) v =+( )− −

T

xρ 10 107 6 in SI units (b) 94 3. m s; 66 7. m s

P16.58 See the solution.

P16.60 (a) µ v02 (b) v0 (c) One travels 2 rev and the other does not move around the loop.

P16.62 (a) see the solution (b) 31.6 m /s

P16.64 130 m s; 1.73 km


17Sound Waves

CHAPTER OUTLINE

17.1 Speed of Sound Waves17.2 Periodic Sound Waves17.3 Intensity of Periodic Sound Waves17.4 The Doppler Effect17.5 Digital Sound Recording17.6 Motion Picture Sound


*Q17.1 Answer (b). The typically higher density would by itself make the speed of sound lower in a solid compared to a gas.

Q17.2 We assume that a perfect vacuum surrounds the clock. The sound waves require a medium for them to travel to your ear. The hammer on the alarm will strike the bell, and the vibration will spread as sound waves through the body of the clock. If a bone of your skull were in contact with the clock, you would hear the bell. However, in the absence of a surrounding medium like air or water, no sound can be radiated away. A larger-scale example of the same effect: Colossal storms raging on the Sun are deathly still for us.

What happens to the sound energy within the clock? Here is the answer: As the sound wave travels through the steel and plastic, traversing joints and going around corners, its energy is con-verted into additional internal energy, raising the temperature of the materials. After the sound has died away, the clock will glow very slightly brighter in the infrared portion of the electromagnetic spectrum.

Q17.3 If an object is 1

2 meter from the sonic ranger, then the sensor would have to measure how long it

would take for a sound pulse to travel one meter. Since sound of any frequency moves at about343 m s, then the sonic ranger would have to be able to measure a time difference of under 0.003 seconds. This small time measurement is possible with modern electronics. But it would be more expensive to outfi t sonic rangers with the more sensitive equipment than it is to print “do not

use to measure distances less than 1

2 meter” in the users’ manual.

Q17.4 The speed of sound to two signifi cant fi gures is 340 m s. Let’s assume that you can measure time

to 1

10 second by using a stopwatch. To get a speed to two signifi cant fi gures, you need to measure

a time of at least 1.0 seconds. Since d t= v , the minimum distance is 340 meters.

*Q17.5 (i) Answer (b). The frequency increases by a factor of 2 because the wave speed, which is depen-dent only on the medium through which the wave travels, remains constant.

(ii) Answer (c).

*Q17.6 (i) Answer (c). Every crest in air produces one crest in water immediately as it reaches the interface, so there must be 500 in every second.

(ii) Answer (a). The speed increases greatly so the wavelength must increase.

449


450 Chapter 17

Q17.7 When listening, you are approximately the same distance from all of the members of the group. If different frequencies traveled at different speeds, then you might hear the higher pitched frequen-cies before you heard the lower ones produced at the same time. Although it might be interesting to think that each listener heard his or her own personal performance depending on where they were seated, a time lag like this could make a Beethoven sonata sound as if it were written by Charles Ives.

*Q17.8 Answer (a). We suppose that a point source has no structure, and radiates sound equally in all directions (isotropically). The sound wavefronts are expanding spheres, so the area over which the sound energy spreads increases according to A r= 4 2π . Thus, if the distance is tripled, the area increases by a factor of nine, and the new intensity will be one-ninth of the old intensity. This answer according to the inverse-square law applies if the medium is uniform and unbounded.

For contrast, suppose that the sound is confi ned to move in a horizontal layer. (Thermal strati-fi cation in an ocean can have this effect on sonar “pings.”) Then the area over which the sound energy is dispersed will only increase according to the circumference of an expanding circle: A rh= 2π , and so three times the distance will result in one third the intensity.

In the case of an entirely enclosed speaking tube (such as a ship’s telephone), the area perpen-dicular to the energy fl ow stays the same, and increasing the distance will not change the intensity appreciably.

*Q17.9 Answer (d). The drop in intensity is what we should expect according to the inverse-square law:

4π r 1 2 P

1 and 4π r 2

2 P2 should agree. (300 m)2(2 μW�m2) and (950 m)2(0.2 μW�m2) are 0.18 W and

0.18 W, agreeing with each other.

*Q17.10 Answer (c). Normal conversation has an intensity level of about 60 dB.

*Q17.11 Answer (c). The intensity is about 10−13 W�m2.

Q17.12 Our brave Siberian saw the fi rst wave he encountered, light traveling at 3 00 108. × m s. At the same moment, infrared as well as visible light began warming his skin, but some time was required to raise the temperature of the outer skin layers before he noticed it. The meteor produced compres-sional waves in the air and in the ground. The wave in the ground, which can be called either sound or a seismic wave, traveled much faster than the wave in air, since the ground is much stiffer against compression. Our witness received it next and noticed it as a little earthquake. He was no doubt unable to distinguish the P and S waves from each other. The fi rst air-compression wave he received was a shock wave with an amplitude on the order of meters. It transported him off his doorstep. Then he could hear some additional direct sound, refl ected sound, and perhaps the sound of the falling trees.

Q17.13 As you move towards the canyon wall, the echo of your car horn would be shifted up in frequency; as you move away, the echo would be shifted down in frequency.

*Q17.14 In f ′ = (v + vo)f�(v − v

s) we can consider the size of the fraction (v + v

o)�(v − v

s) in each case.

The positive direction is defi ned to run from the observer toward the source.

In (a), 340�340 = 1 In (b), 340�(340 − 25) = 1.08 In (c), 340�(340 + 25) = 0.932 In (d),(340 + 25)�340 = 1.07 In (e), (340 − 25)�340 = 0.926 In (f ), (340 + 25)�(340 + 25) = 1 In (g), (340 − 25)�(340 − 25) = 1. In order of decreasing size we have b > d > a = f = g > c > e.

450 Chapter 17


Sound Waves 451

*Q17.15 (i) Answer (c). Both observer and source have equal speeds in opposite directions relative to the medium, so in f ′ = (v + v

o)f�(v − v

s) we would have something like (340 − 25)f�(340 − 25) = f.

(ii) Answer (a). The speed of the medium adds to the speed of sound as far as the observer is concerned, to cause an increase in λ = v�f.

(iii) Answer (a).

Q17.16 For the sound from a source not to shift in frequency, the radial velocity of the source relative to the observer must be zero; that is, the source must not be moving toward or away from the observer. The source can be moving in a plane perpendicular to the line between it and the observer. Other possibilities: The source and observer might both have zero velocity. They might have equal velocities relative to the medium. The source might be moving around the observer on a sphere of constant radius. Even if the source speeds up on the sphere, slows down, or stops, the frequency heard will be equal to the frequency emitted by the source.


Section 17.1 Speed of Sound Waves

*P17.1 Since v vlight sound>> we have d ≈ ( )( ) =343 16 2 5 56m s s km. .

We do not need to know the value of the speed of light. As long as it is very large, the travel time for the light is negligible compared to that for the sound.

P17.2 v = = ××

=B

ρ2 80 10

13 6 101 43

10

3

.

.. km s

*P17.3 The sound pulse must travel 150 m before refl ection and 150 m after refl ection. We have d t= v

td= = =v

3000 196

m

1 533 m ss.

P17.4 (a) At 9 000 m, ΔT = ⎛⎝

⎞⎠ −( ) = −9 000

1501 00 60 0. .° °C C

so T = −30 0. °C

Using the chain rule:

d

dt

d

dT

dT

dx

dx

dt

d

dT

dT

dx

v vv

vv= = = ( )⎛

⎝⎞0 607

1

150. ⎠⎠ = v

247, so dt

d= ( )247 sv

v

dt

d

t

t

f

i

i

f

0

247

247

∫ ∫= ( )

= ( ) ⎛⎝⎜

⎞⎠⎟

=

s

s

vv

v

v

v

v

ln 2247331 5 0 607 30 0

331 5 0 607 30 0s( ) + ( )

+ −ln

. . .

. . .(( )⎡⎣⎢

⎤⎦⎥

t = 27 2. s for sound to reach ground.

(b) th= =

+ ( )[ ] =v

9 000

331 5 0 607 30 025 7

. . .. s

It takes longer when the air cools off than if it were at a uniform temperature.

Sound Waves 451


452 Chapter 17

P17.5 Sound takes this time to reach the man: 20 0 1 75

3435 32 10 2. ..

m m

m ss

−( )= × −

so the warning should be shouted no later than 0 300 5 32 10 0 3532. . .s s s+ × =−

before the pot strikes.

Since the whole time of fall is given by y gt= 1

22: 18 25

1

29 80 2. .m m s2= ( ) t

t = 1 93. s

the warning needs to come 1 93 0 353 1 58. . .s s s− =

into the fall, when the pot has fallen 1

29 80 1 58 12 22. . .m s s m2( )( ) =

to be above the ground by 20 0 12 2 7 82. . .m m m− =

P17.6 It is easiest to solve part (b) fi rst:

(b) The distance the sound travels to the plane is d hh h

s = + ⎛⎝

⎞⎠ =2

2

2

5

2 The sound travels this distance in 2.00 s, so

dh

s = = ( )( ) =5

2343 2 00 686m s s m.

giving the altitude of the plane as h = ( )=2 686

5614

mm

(a) The distance the plane has traveled in 2.00 s is v 2 002

307. s m( ) = =h

Thus, the speed of the plane is: v = =307153

m

2.00 sm s

P17.7 Let x1 represent the cowboy’s distance from the nearer canyon wall and x2 his distance from the

farther cliff. The sound for the fi rst echo travels distance 2 1x . For the second, 2 2x . For the third,

2 21 2x x+ . For the fourth echo, 2 2 21 2 1x x x+ + .

Then

2 2

3401 922 1x x− =

m ss. and 2 2 2

3401 471 2 2x x x+ − =

m ss.

Thus

x1

1

2340 250= =m s 1.47 s m and

2

3401 92 1 472x

m ss s= +. . ; x2 576= m

(a) So x x1 2 826+ = m

(b) 2 2 2 2 2

3401 471 2 1 1 2x x x x x+ + − +( )

=m s

s.

452 Chapter 17


Sound Waves 453

Section 17.2 Periodic Sound Waves

*P17.8 (a) The speed gradually changes from v = (331 m �s)( 1 + 27°C�273°C)1�2 = 347 m �s to (331 m �s) (1 + 0�273°C)1�2 = 331 m �s, a 4.6% decrease. The cooler air at the samepressure is more dense.

(b) The frequency is unchanged, because every wave crest in the hot air becomes one crest without delay in the cold air.

(c) The wavelength decreases by 4.6%, from v�f = (347 m �s)�(4 000�s) = 86.7 mm to (331 m �s)�(4 000�s) = 82.8 mm. The crests are more crowded together when they move slower.

*P17.9 (a) If f = 2 4. MHz, λ = =×

=vf

1 500

2 4 100 6256

m s

smm

..

(b) If f = 1 MHz, λ = = =vf

1 500

101 506

m s

smm.

If f = 20 MHz, λ μ=×

=1 500

2 1075 07

m s

sm.

P17.10 ΔP smax max= ρ ωv

sP

maxmax

.

.= =

×( )( )

−Δρ ωv

4 00 10

1 20 343

3 N m

kg m

2

3 m s sm( )( ) ×( ) = ×−

−

2 10 0 101 55 10

3 110

π ..

P17.11 (a) A = 2 00. mμ

λ π= = =2

15 70 400 40 0

.. .m cm

v = = =ωk

858

15 754 6

.. m s

(b) s = ( )( ) − ( ) ×( )⎡⎣ ⎤−2 00 15 7 0 050 0 858 3 00 10 3. cos . . . ⎦⎦ = −0 433. mμ

(c) vmax . .= = ( )( ) =−Aω μ2 00 858 1 721m s mm s

P17.12 (a) ΔPx t

= ( ) −⎛⎝

⎞⎠1 27

340. sinPa

m s

π π (SI units)

The pressure amplitude is: ΔPmax .= 1 27 Pa

(b) ω π π= =2 340f s, so f = 170 Hz

(c) k = =2πλ

π m, giving λ = 2 00. m

(d) v = = ( )( ) =λ f 2 00 170. m Hz 340 m s

Sound Waves 453


454 Chapter 17

P17.13 k = =( ) = −2 2

0 10062 8 1π

λπ

..

mm

ω πλ

π= =

( )( ) = × −2 2 343

0 1002 16 104 1v m s

ms

..

Therefore,

ΔP x t= ( ) − ×⎡⎣ ⎤⎦0 200 62 8 2 16 104. sin . .Pa m s

P17.14 (a) The sound “pressure” is extra tensile stress for one-half of each cycle. When it becomes

0 500 13 0 10 6 50 1010 8. % . .( ) ×( ) = ×Pa Pa, the rod will break. Then, ΔP smax max= ρ ωv

sP

maxmax .

.= =

××( )

Δρ ωv

6 50 10

8 92 10 5 0

8

3

N m

kg m

2

3 110 2 5004 63

m s smm( )( ) =

π.

(b) From s s kx t= −( )max cos ω

v

v

= ∂∂

= − −( )

= = ( )

s

ts kx t

s

ω ω

ω π

max

max max

sin

2 500 4s .. .63 14 5mm m s( ) =

(c) I s= ( ) = = ×( )1

2

1

2

1

28 92 10 5

2 2 3ρ ω ρv vvmax max . kg m3 0010 14 52

m s m s( )( ).

= ×4 73 109. W m2

P17.15 ΔP s smax max max= = ⎛⎝

⎞⎠ρ ω ρ π

λv v

v2

λ πρ π= =

( )( ) ×( )−2 2 1 20 343 5 50 10

0

2 2 6v s

Pmax

max

. .

Δ ...

8405 81= m

Section 17.3 Intensity of Periodic Sound Waves

P17.16 The sound power incident on the eardrum is P = IA where I is the intensity of the sound and A = × −5 0 10 5. m2 is the area of the eardrum.

(a) At the threshold of hearing, I = × −1 0 10 12. W m2, and

P = ×( ) ×( ) = ×− − −1 0 10 5 0 10 5 00 1012 5 17. . .W m m W2 2

(b) At the threshold of pain, I = 1 0. W m2, and

P = × = ×− −( . / )( . ) .1 0 5 0 10 5 00 102 5 2 5W m m W

P17.17 β =⎛⎝⎜

⎞⎠⎟

=××

⎛⎝

−

−10 104 00 10

1 00 100

6

12log log.

.

I

I ⎜⎜⎞⎠⎟

= 66 0. dB

454 Chapter 17


Sound Waves 455

P17.18 The power necessarily supplied to the speaker is the power carried away by the sound wave:

P = ( ) =

= (

1

22

2 1 20

2 2 2 2

2

ρ ω π ρ

π

A s A f sv vmax max

. kg m3 )) ⎛⎝

⎞⎠ ( )( ) × −π 0 08

343 600 0 12 102

2 2..

m

2m s 1 s m(( ) =

221 2. W

P17.19 I s= 1

22 2ρω maxv

(a) At f = 2 500 Hz, the frequency is increased by a factor of 2.50, so the intensity (at constant

smax) increases by 2 50 6 252. .( ) = .

Therefore, 6 25 0 600 3 75. . .( ) = W m2

(b) 0 600. W m2

P17.20 The original intensity is I s f s12 2 2 2 21

22= =ρω π ρmax maxv v

(a) If the frequency is increased to ′f while a constant displacement amplitude is maintained, the new intensity is

I f s22 2 22= ′( )π ρv max

so I

I

f s

f s

f

f2

1

2 2

2 2 2

22

2=

′( ) = ′⎛⎝⎜

⎞⎠⎟

π ρπ ρ

vv

max

max

or

If

fI2

2

1= ′⎛⎝⎜

⎞⎠⎟

(b) If the frequency is reduced to ′ =ff

2 while the displacement amplitude is doubled, the new

intensity is

If

s f s I22

22 2 2 2

122

2 2= ⎛⎝

⎞⎠ ( ) = =π ρ π ρv vmax max

or the intensity is unchanged .

P17.21 (a) For the low note the wavelength is λ = = =vf

343

146 82 34

m s

sm

..

For the high note λ = =343

8800 390

m s

sm.

We observe that the ratio of the frequencies of these two notes is 880

5 99Hz

146.8 Hz= . nearly

equal to a small integer. This fact is associated with the consonance of the notes D and A.

(b) β =⎛⎝⎜

⎞⎠⎟

=−1010

7512dBW m

dB2logI

gives I = × −3 16 10 5. W m2

I

P

P

=

= × ( )−

Δ

Δ

max

max . .

2

5

2

3 16 10 2 1 20 3

ρv

W m kg m2 3 443 0 161m s Pa( ) = .

for both low and high notes.


Sound Waves 455


456 Chapter 17 456 Chapter 17

(c) I s f s= ( ) =1

2

1

24

2 2 2 2ρ ω ρ πv vmax max

sI

fmax =2 2 2π ρv

for the low note, smax

.

. .= × −3 16 10

2 1 20 343

1

146 8

5

2

W m

kg m m s s

2

3π

== × = ×−

−6 24 10

146 84 25 10

57.

..m m

for the high note, smax

..= × = ×

−−6 24 10

7 09 105

8

880m m

(d) With both frequencies lower (numerically smaller) by the factor 146 8

134 3

880

804 91 093

.

. ..= = ,

the wavelengths and displacement amplitudes are made 1.093 times larger, and the pressure amplitudes are unchanged.

P17.22 We begin with β22

0

10=⎛⎝⎜

⎞⎠⎟

logI

I and β1

1

0

10=⎛⎝⎜

⎞⎠⎟

logI

I so β β2 1

2

1

10− =⎛⎝⎜

⎞⎠⎟

logI

I

Also, Ir2

224

=Pπ

and Ir1124

=Pπ

giving I

I

r

r2

1

1

2

2

=⎛⎝⎜

⎞⎠⎟

Then, β β2 11

2

2

1

2

10 20− =⎛⎝⎜

⎞⎠⎟

=⎛⎝⎜

⎞⎠⎟

log logr

r

r

r

P17.23 (a) I112 10 121 00 10 10 1 00 101= ×( ) = ×(− ( ) −. .W m W m2 2β ))1080 0 10.

or I141 00 10= × −. W m2

I212 10 121 00 10 10 1 00 102= ×( ) = ×(− ( ) −. .W m W m2 2β ))1075 0 10.

or I24 5 51 00 10 3 16 10= × = ×− −. .. W m W m2 2

When both sounds are present, the total intensity is

I I I= + = × + × = ×− −1 2

4 51 00 10 3 16 10 1 32 10. . .W m W m2 2 −−4 W m2

(b) The decibel level for the combined sounds is

β = ××

⎛⎝⎜

⎞⎠⎟

=−

−101 32 10

1 00 1010

4

12log.

.

W m

W m

2

2 llog . .1 32 10 81 28×( ) = dB


Sound Waves 457 Sound Waves 457

P17.24 In Ir

=P

4 2π, intensity I is proportional to

12r

, so between locations 1 and 2: I

I

r

r2

1

12

22=

In I s= ( )1

22ρ ωv max , intensity is proportional to smax

2 , so I

I

s

s2

1

22

12=

Then, s

s

r

r2

1

2

1

2

2⎛⎝⎜

⎞⎠⎟

=⎛⎝⎜

⎞⎠⎟

or 1

2

21

2

2

⎛⎝

⎞⎠ =

⎛⎝⎜

⎞⎠⎟

r

r giving r r2 12 2 50 0 100= = ( ) =. m m

But, r d22 250 0= ( ) +. m

yields d = 86 6. m

P17.25 (a) 120 1010 12 2dB dB

W m= ⎡

⎣⎢

⎤⎦⎥−log

I

Ir

rI

= =

= = ( ) =

1 004

4

6 000

2.

.

W m

W

4 1.00 W m

2

2

P

Pπ

π π..691 m

We have assumed the speaker is an isotropic point source.

(b) 0 1010 12dB dB

W m2=⎛⎝⎜

⎞⎠⎟−log

I

I

rI

= ×

= =×

−1 00 10

4

6 00

12.

.

W m

W

4 1.00 10 W

2

12

Pπ π − mm

km2( ) = 691

We have assumed a uniform medium that absorbs no energy.

P17.26 We presume the speakers broadcast equally in all directions.

(a) rAC = + =3 00 4 00 5 002 2. . .m m

Ir

= =×

( )= ×

−−P

4

1 00 10

4 5 003 18 102

3

26

π π.

..

W

mW m22

2

2dBW m

W mβ

β

=×⎛

⎝⎜⎞⎠⎟

=

−

−103 18 10

10

1

6

12log.

00 6 50 65 0dB dB. .=

(b) rBC = 4 47. m

I = ×( ) = ×

=

−−1 50 10

4 4 475 97 10

10

3

26.

..

W

mW m2

π

β ddB

dB

log.

.

5 97 10

10

67 8

6

12

×⎛⎝⎜

⎞⎠⎟

=

−

−

β

(c) I = +3 18 5 97. .W m W m2 2μ μ

β = ×⎛⎝⎜

⎞⎠⎟

=−

−109 15 10

1069 6

6

12dB dBlog.

.



P17.27 Since intensity is inversely proportional to the square of the distance,

I I4 0 4

1

100= .

and I

P0 4

2 2

2

10 0

2 1 20 3430 121.

max .

..= = ( )

( )( ) =Δ

ρvW m22

The difference in sound intensity level is

Δβ =⎛⎝⎜

⎞⎠⎟

= −( ) = −10 10 2 00 20 04log . .I

Ikm

0.4 km

ddB

At 0.400 km,

β0 4 12100 121

10110 8. log

..=

⎛⎝⎜

⎞⎠⎟

=−

W m

W mdB

2

2

At 4.00 km,

β β β4 0 4 110 8 20 0 90 8= + = −( ) =. . . .Δ dB dB

Allowing for absorption of the wave over the distance traveled,

′ = − ( )( ) =β β4 4 7 00 3 60 65 6. . .dB km km dB

This is equivalent to the sound intensity level of heavy traffi c.

P17.28 (a) E t r It= P = = ( ) ×( )−4 4 100 7 00 10 0 2002 2 2π π m W m s2. .(( ) = 1 76. kJ

(b) β = ××

⎛⎝⎜

⎞⎠⎟

=−

−107 00 10

1 00 10108

2

12log.

.dB

P17.29 β = ⎛⎝

⎞⎠−10

10 12logI I = ⎡⎣ ⎤⎦( )( ) −10 1010 12β W m2

I 120 1 00dB2W m( ) = . ; I 100

21 00 10dB2W m( )

−= ×. ; I 10111 00 10dB

2W m( )−= ×.

(a) P = 4 2πr I so that r I r I12

1 22

2=

r rI

I2 11

2

1 2

23 001 00

1 00 1030=

⎛⎝⎜

⎞⎠⎟

= ( )×

=−..

..m 00 m

(b) r rI

I2 11

2

1 2

113 001 00

1 00 109=

⎛⎝⎜

⎞⎠⎟

= ( )×

=−..

..m 449 105× m



P17.30 Assume you are 1 m away from your lawnmower and receiving 100 dB sound from it. The intensity

of this sound is given by 100 1010 12dB dB

W m2= −logI

; I = −10 2 W m2. If the lawnmower

radiates as a point source, its sound power is given by I

r=

P4 2π

P = ( ) =−4 1 10 0 1262 2π m W m W2 .

Now let your neighbor have an identical lawnmower 20 m away. You receive from it sound with

intensity I =( ) = × −0 126

2 5 1025.

.W

4 20 mW m2

π. The total sound intensity impinging on you is

10 2 5 10 1 002 5 102 5 2− − −+ × = ×W m W m W m2 2 2. . . So its level is

101 002 5 10

10100 01

2

12dB dBlog.

.× =

−

−

If the smallest noticeable difference is between 100 dB and 101 dB, this cannot be heard as a change from 100 dB.

P17.31 (a) The sound intensity inside the church is given by

β =⎛⎝⎜

⎞⎠⎟

= ( ) ⎛⎝ −

10

101 1010

0

12

ln

ln

I

I

IdB dB

W m2⎜⎜⎞⎠⎟

= ( ) = =− −I 10 10 10 0 012 610 1 12 1 90. . .W m W m2 2 W m2

We suppose that sound comes perpendicularly out through the windows and doors. Then, the radiated power is

P = = ( )( ) =IA 0 012 6 22 0 0 277. . .W m m W2 2

Are you surprised by how small this is? The energy radiated in 20.0 minutes is

E t= = ( )( )⎛⎝

⎞⎠P 0 277 20 0

60 0. .

.J s min

s

1.00 min== 332 J

(b) If the ground refl ects all sound energy headed downward, the sound power, P = 0 277. W, covers the area of a hemisphere. One kilometer away, this area is

A r= = ( ) = ×2 2 1 000 2 102 2 6π π πm m2

The intensity at this distance is

IA

= =×

= × −P 0 2774 41 10 8.

.W

2 10 mW m6 2

2

π and the sound intensity level is

β = ( ) ××

⎛⎝⎜

⎞⎠

−

−104 41 10

1 00 10

8

12dBW m

W m

2

2ln.

. ⎟⎟ = 46 4. dB



Section 17.4 The Doppler Effect

P17.32 (a) ω π π= =⎛⎝⎜

⎞⎠⎟

=2 2115

60 012 0f

min

s minrad s

..

vmax . . .= = ( ) ×( ) =−ωA 12 0 1 80 10 0 021 73rad s m m s

(b) The heart wall is a moving observer.

′ = +⎛⎝

⎞⎠ = ( ) +⎛

⎝f f Ov v

v2 000 000

1 500 0 021 7

1 500Hz

.⎜⎜

⎞⎠⎟

= 2 000 028 9. Hz

(c) Now the heart wall is a moving source.

′′ = ′−

⎛⎝⎜

⎞⎠⎟

= ( )−

f fs

vv v

2 000 0291 500

1 500 0 02Hz

. 11 72 000 057 8

⎛⎝⎜

⎞⎠⎟

= . Hz

*P17.33 (a) ′ =+( )

−( )ff o

s

v vv v

′ = +( )−( ) =f 2 500

343 25 0

343 40 03 04

.

.. kHz

(b) ′ = + −( )− −

⎛⎝⎜

⎞⎠⎟

=f 2 500343 25 0

343 40 02

.

( . ).08 kHzz

(c) ′ = + −( )−

⎛⎝⎜

⎞⎠⎟

=f 2 500343 25 0

343 40 02

.

..62 kHz while police car overtakes

′ = +− −( )

⎛⎝⎜

⎞⎠⎟

=f 2 500343 25 0

343 40 02

.

..40 kHz after police car passes

P17.34 (a) The maximum speed of the speaker is described by

1

2

1

2

20 0

5 000 500

2 2m kA

k

mA

v

v

max

max

.

..

=

= = N m

kgmm m s( ) = 1 00.

The frequencies heard by the stationary observer range from

′ =+

⎛⎝⎜

⎞⎠⎟

f fminmax

vv v

to ′ =−

⎛⎝⎜

⎞⎠⎟

f fmaxmax

vv v

where v is the speed of sound.

′ =+

⎛⎝⎜

⎞⎠⎟

=fmin .440

343 1 00439Hz

343 m s

m s m sHHz

Hz343 m s

m s m s′ =

−⎛⎝⎜

⎞⎠⎟

=fmax .440

343 1 004441 Hz




(b) β π=

⎛⎝⎜

⎞⎠⎟

=⎛⎝⎜

⎞⎠⎟

10 104

0

2

0

dB dBlog logI

I

r

I

P

The maximum intensity level (of 60.0 dB) occurs at r r= =min .1 00 m. The minimum intensity level occurs when the speaker is farthest from the listener (i.e., when r r r A= = + =max min .2 2 00 m).

Thus,

β βπmax min

min

log log− =⎛⎝⎜

⎞⎠⎟

−104

100

2dB dBPI r

PP4 0

2π I rmax

⎛⎝⎜

⎞⎠⎟

or

β βπ

πmax min

min

maxlog− =⎛⎝⎜

⎞⎠

104

4

02

02

dBP

PI r

I r⎟⎟ =

⎛⎝⎜

⎞⎠⎟

102

2dBlog max

min

r

r

This gives:

60 0 10 4 00 6 02. log . .mindB dB dB− = ( ) =β and βmin .= 54 0 dB

P17.35 Approaching ambulance: ′ =−( )f

f

S1 v v

Departing ambulance: ′′ =− −( )( )f

f

S1 v v

Since ′ =f 560 Hz and ′′ =f 480 Hz 560 1 480 1−⎛⎝

⎞⎠ = +⎛

⎝⎞⎠

vv

vv

S S

1 040 80 0

80 0 343

1 04026 4

vv

v

S

S

=

= ( )=

.

..m s m s

P17.36 (a) v = ( ) +⋅

−( ) =331 0 6 10 325m sm

s CC m s.

°°

(b) Approaching the bell, the athlete hears a frequency of ′ = +⎛⎝

⎞⎠f f Ov v

v

After passing the bell, she hears a lower frequency of ′′ =+ −( )⎛

⎝⎜⎞⎠⎟

f f Ov vv

The ratio is ′′′

= −+

=f

fO

O

v vv v

5

6

which gives 6 6 5 5v v v v− = +o o

or vv

O = = =11

325

1129 5

m sm s.

P17.37 ′ =−

⎛⎝⎜

⎞⎠⎟

f fs

vv v

485 512340

340 9 80=

− −( )⎛

⎝⎜⎞

⎠⎟. tfall

485 340 485 9 80 512 340

512 485

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

= −

. t

t

f

f 4485

340

9 801 93⎛

⎝⎞⎠ =

.. s

d gt f121

218 3= = . m: treturn s= =18 3

3400 053 8

..

The fork continues to fall while the sound returns.

t t tftotal fall return s s= + = + =1 93 0 053 8 1 98. . . 55

1

219 3

s

mtotal total fall2d gt= = .



P17.38 (a) Sound moves upwind with speed 343 15−( ) m s. Crests pass a stationary upwind point at frequency 900 Hz.

Then λ = = =vf

328

9000 364

m s

sm.

(b) By similar logic, λ = = +( )=v

f

343 15

9000 398

m s

sm.

(c) The source is moving through the air at 15 m �s toward the observer. The observer is station-ary relative to the air.

′ = +−

⎛⎝⎜

⎞⎠⎟

= +−

⎛⎝

⎞⎠ =f f o

s

v vv v

900343 0

343 1594Hz 11 Hz

(d) The source is moving through the air at 15 m �s away from the downwind fi refi ghter. Her speed relative to the air is 30 m �s toward the source.

′ = +−

⎛⎝⎜

⎞⎠⎟

= +− −( )

⎛⎝⎜

f f o

s

v vv v

900343 30

343 15Hz

⎞⎞⎠⎟

= ⎛⎝

⎞⎠ =900

373

358938Hz Hz

P17.39 (b) sin.

θ = =vvS

1

3 00; θ = 19 5. °

tanθ = h

x; x

h=tanθ

x = = × =20 000

19 55 66 10 56 64m

m kmtan .

. .°

(a) It takes the plane tx

S

= = ×( ) =

v5 66 10

56 34.

.m

3.00 335 m ss to travel this distance.

P17.40 θ = = =− −sin sin.

.1 1 1

1 3846 4

vvS

°

P17.41 The half angle of the shock wave cone is given by sinθ =v

vlight

S

vv

S = =×( ) = ×light m s

sin

.

sin ..

θ2 25 10

53 02 82 10

8

°88 m s

t = 0

a.

h

Observer

b.

h

Observer hears the boom

x

FIG. P17.39(a)



Section 17.5 Digital Sound Recording

Section 17.6 Motion Picture Sound

P17.42 For a 40-dB sound,

40 1010

10

12

8

dB dBW m

W m

2

2

= ⎡⎣⎢

⎤⎦⎥

= =

−

−

logI

IPΔ mmax

max .

2

8

2

2 2 1 20 343 10

ρ

ρ

v

vΔP I= = ( )( ) −kg m m s2 W m N m2 2= 2 87 10 3. × −

(a) code =× =

−2 87 10

28 765 536 7

3.

.

N m

N m

2

2

(b) For sounds of 40 dB or softer, too few digital words are available to represent the wave form with good fi delity.

(c) In a sound wave Δ P is negative half of the time but this coding scheme has no words available for negative pressure variations.

Additional Problems

*P17.43 The gliders stick together and move with fi nal speed given by momentum conservation for the two-glider system:

0.15 kg 2.3 m �s + 0 = (0.15 + 0.2) kg v v = 0.986 m �s

The missing mechanical energy is

(1/ 2)(0.15 kg)(2.3 m /s)2 – (1/ 2)(0.35 kg)(0.986 m /s)2 = 0.397 J – 0.170 J = 0.227 J

We imagine one-half of 227 mJ going into internal energy and half into sound radiated isotropically in 7 ms. Its intensity 0.8 m away is

I = E�At = 0.5(0.227 J)�[4π(0.8 m)2 0.007 s] = 2.01 W�m2

Its intensity level is β = 10 log(2.01�10−12) = 123 dB

The sound of air track gliders latching together is many orders of magnitude less intense. The idea is unreasonable. Nearly all of the missing mechanical energy becomes internal energy in the latch.



*P17.44 The wave moves outward equally in all directions. (We can tell it is outward because of the negative sign in 1.36 r – 2030 t.) Its amplitude is inversely proportional to its distance from the center. Its intensity is proportional to the square of the amplitude, so the intensity follows the inverse-square law, with no absorption of energy by the medium. Its speed is constant at v = f λ = ω �k = (2030�s)�(1.36�m) = 1.49 km �s. By comparison to the table in the chapter, it can be moving through water at 25°C, and we assume that it is. Its frequency is constant at (2030�s)�2π = 323 Hz. Its wavelength is constant at 2π �k = 2π �(1.36 �m) = 4.62 m. Itspressure amplitude is 25.0 Pa at radius 1 m. Its intensity at this distance is

IP

= =Δ max

2

2ρv

( )( )(

225

2 1000 1490

N/m

kg/m m/

2

3 ss)W/m2= 209 μ

so the power of the source and the net power of the wave at all distances is P = I 4π r 2 =×( ) =−2 09 10 4 1 2 634 π. ( .W/m m) mW2 2 .

*P17.45 Model your loud, sharp sound impulse as a single narrow peak in a graph of air pressure versus time. It is a noise with no pitch, no frequency, wavelength, or period. It radiates away from you in all directions and some of it is incident on each one of the solid vertical risers of the bleachers. Suppose that, at the ambient temperature, sound moves at 340 m �s; and suppose that the horizon-tal width of each row of seats is 60 cm. Then there is a time delay of

0 6

0 002.

.m

340 m ss( ) =

between your sound impulse reaching each riser and the next. Whatever its material, each will refl ect much of the sound that reaches it. The refl ected wave sounds very different from the sharp pop you made. If there are twenty rows of seats, you hear from the bleachers a tone with twenty crests, each separated from the next in time by

2 0 6

3400 004

..

m

m ss

( )( ) =

This is the extra time for it to cross the width of one seat twice, once as an incident pulse and once again after its refl ection. Thus, you hear a sound of defi nite pitch, with period about 0.004 s, frequency

1

0 003 5300

.~

sHz a few hundred Hz=

wavelength

λ = =( )

( ) =vf

340

3001 2 100m s

sm m. ~

and duration

20 0 004 10 1. ~s s( ) −

(b) Yes. With the steps narrower, the frequency can be close to 1000 Hz. If the person clap-ping his hands is at the base of the pyramid, the echo can drop somewhat in frequency and in loudness as sound returns, with the later cycles coming from the smaller and more distant upper risers. The sound could imitate some particular bird, and could in fact con-stitute a recording of the call.



*P17.46 (a) The distance is larger by 240�60 = 4 times. The intensity is 16 times smaller at the larger distance, because the sound power is spread over a 42 times larger area.

(b) The amplitude is 4 times smaller at the larger distance, because intensity is proportional to the square of amplitude.

(c) The extra distance is (240 – 60)�45 = 4 wavelengths. The phase is the same at both points, because they are separated by an integer number of wavelengths.

P17.47 Since cos sin2 2 1θ θ+ = , sin cosθ θ= ± −1 2 (each sign applying half the time)

Δ ΔP P kx t s kx t= −( ) = ± − −( )max maxsin cosω ρ ω ωv 1 2

Therefore ΔP s s kx t s s= ± − −( ) = ± −ρ ω ω ρ ωv vmax max maxcos2 2 2 2 2

P17.48 (a) λ = = =−

vf

343

1 4800 2321

m s

sm.

(b) β = = ⎡⎣⎢

⎤⎦⎥−81 0 10

10 12. logdB dBW m2

I

I = ( ) = = ×− − −10 10 10 1 26 1012 8 10 3 90 4W m W m W2 2. . . mm

W m

2

2

=

= =×( )−

1

2

2 2 1 26 10

2 2

2

4

ρ ω

ρ ω

v

v

s

sI

max

max

.

11 20 343 4 1 4808 41 10

2 1 2.

.kg m m s s3( )( ) ( ) = ×

−−

π88 m

(c) ′ =′

= =−λ vf

343

1 3970 2461

m s

sm. Δλ λ λ= ′ − = 13 8. mm

P17.49 The trucks form a train analogous to a wave train of crests with speed v = 19 7. m s and unshifted

frequency f = = −2

3 000 667 1

..

minmin

(a) The cyclist as observer measures a lower Doppler-shifted frequency:

′ = +⎛⎝

⎞⎠ = ( ) + −( )−f f ov v

v0 667

19 7 4 47

19 71.

. .

.min

⎛⎛⎝⎜

⎞⎠⎟

= 0 515. min

(b) ′′ =+⎛

⎝⎜⎞⎠⎟ = ( ) + −( )−f f ov v

v′

0 66719 7 1 561.

. .min

119 70 614

..

⎛⎝⎜

⎞⎠⎟

= min

The cyclist’s speed has decreased very signifi cantly, but there is only a modest increase in the frequency of trucks passing him.

P17.50 (a) The speed of a compression wave in a bar is

v = = × = ×Y

ρ20 0 10

7 8605 04 10

103.

.N m

kg mm s

2

3

(b) The signal to stop passes between layers of atoms as a sound wave, reaching the back end of the bar in time

tL= =

×= × −

v0 800

1 59 10 4..

m

5.04 10 m ss3




(c) As described by Newton’s fi rst law, the rearmost layer of steel has continued to move forward with its original speed vi for this time, compressing the bar by

ΔL ti= = ( ) ×( ) = × =− −v 12 0 1 59 10 1 90 10 14 3. . .m s s m ..90 mm

(d) The strain in the rod is: ΔL

L=

×= ×

−−1 90 10

2 38 103

3..

m

0.800 m

(e) The stress in the rod is: σ = ⎛⎝

⎞⎠ = ×( ) ×( ) =−Y

L

L

Δ20 0 10 2 38 10 47610 3. .N m M2 PPa

Since σ > 400 MPa, the rod will be permanently distorted.

(f ) We go through the same steps as in parts (a) through (e), but use algebraic expressions rather than numbers:

The speed of sound in the rod is v = Y

ρ The back end of the rod continues to move forward at speed vi for a time of t

LL

Y= =

vρ

, traveling distance ΔL ti= v after the front end hits the wall.

The strain in the rod is: ΔL

L

t

L Yi

i= =v

vρ

The stress is then: σ ρ ρ= ⎛⎝

⎞⎠ = =Y

L

LY

YYi i

Δv v

For this to be less than the yield stress, σ y, it is necessary that

vi yYρ σ< or viy

Y<

σρ

With the given numbers, this speed is 10.1 m �s. The fact that the length of the rod divides out means that the steel will start to bend right away at the front end of the rod. There it will yield enough so that eventually the remainder of the rod will experience only stress within the elastic range. You can see this effect when sledgehammer blows give a mushroom top to a rod used as a tent stake.

P17.51 (a) ′ =−( )f f

vv vdiver

so 1− =′

vvdiver f

f ⇒ = −

′⎛⎝⎜

⎞⎠⎟

v vdiver 1f

f

with v = 343 m s, f = 1 800 Hz and ′ =f 2 150 Hz

we fi nd vdiver m s= −⎛⎝⎜

⎞⎠⎟

=343 11 800

2 15055 8.

(b) If the waves are refl ected, and the skydiver is moving into them, we have

′′ = ′+( )

⇒ ′′ =−( )

⎡

⎣⎢

⎤

⎦⎥

+f f f f

v vv

vv v

v vdiver

diver

ddiver( )v

so ′′ = +( )−( ) =f 1 800

343 55 8

343 55 82 500

.

.Hz



P17.52 Let P(x) represent absolute pressure as a function of x. The net force to the right on the chunkof air is + ( ) − +( )P x A P x x AΔ . Atmospheric pressure subtracts out, leaving

− +( ) + ( )⎡⎣ ⎤⎦Δ Δ ΔP x x P x = −∂∂Δ

ΔAP

xxA. The mass of the air is Δ Δm V A x= =ρ ρ Δ and its

acceleration is ∂∂

2

2

s

t. So Newton’s second law becomes

−∂∂

=∂∂

−∂∂

−∂∂

⎛⎝⎜

⎞⎠⎟

=∂∂

ΔΔ Δ

P

xxA A x

s

t

xB

s

x

s

t

ρ

ρ

2

2

2

22

2

2

2

2

B s

x

s

tρ∂∂

=∂∂

Into this wave equation as a trial solution we substitute the wave function s x t s kx t, cosmax( ) = −( )ω We fi nd

∂∂

= − −( )

∂∂

= − −

s

xks kx t

s

xk s kx t

max

max

sin

cos

ω

ω2

22 (( )

∂∂

= + −( )

∂∂

= −

s

ts kx t

s

ts kx

ω ω

ω

max

max

sin

cos2

22 −−( )ωt

B s

x

s

tρ∂∂

= ∂∂

2

2

2

2 becomes − −( ) = − −( )Bk s kx t s kx t

ρω ω ω2 2

max maxcos cos

This is true provided B

fρ

πλ

π44

2

22 2=

The sound wave can propagate provided it has λρ

2 2 2fB= =v ; that is, provided it propagates with

speed v = B

ρP17.53 When observer is moving in front of and in the same direction as the source, ′ = −

−f f O

S

v vv v

where

vO and vS are measured relative to the medium in which the sound is propagated. In this case the ocean current is opposite the direction of travel of the ships and

vO = − −( ) = =45 0 10 0 55 0 15 3. . . .km h km h km h m s , annd

km h km h km h mvS = − −( ) = =64 0 10 0 74 0 20 55. . . . ss

Therefore,

′ = ( ) −−

f 1 200 01 520 15 3

1 520 20 55.

.

.Hz

m s m s

m s mm sHz= 1 204 2.

P17.54 Use the Doppler formula, and remember that the bat is a moving source. If the velocity of the insect is vx,

40 4 40 0340 5 00 340

340 5 00 340. .

.

.=

+( ) −( )−( ) +

vv

x

x(( ) Solving,

vx = 3 31. m s

Therefore,

the bat is gaining on its prey at 1.69 m s .

P(x)A P(x + Δx)A

FIG. P17.52



P17.55 103 1010 12dB dB

W m2= ⎡⎣⎢

⎤⎦⎥−log

I

(a) Ir

= × = =( )

−2 00 104 4 1 6

22 2.

.W m

m2 P P

π π

P = 0 642. W

(b) effi ciency = =sound output power

total input power

0 642. WW

150 W= 0 004 28.

P17.56 (a)

(b) λ = = =−

vf

343

1 0000 3431

m s

sm.

(c) ′ =′

= −⎛⎝

⎞⎠ = −( )

=−λ v v v vvf f

S 343 40 0

1 00001

..

m s

s3303 m

(d) ′′ =′′

= +⎛⎝

⎞⎠ = +( )

=−λ v v v vvf f

S 343 40 0

1 000 1

. m s

s00 383. m

(e) ′ = −−

⎛⎝⎜

⎞⎠⎟

= ( ) −( )f f O

S

v vv v

1 000343 30 0

34Hz

m s.

33 40 01 03

−( ) =.

.m s

kHz

*P17.57 (a) The sound through the metal arrives first, bbecause it moves faster than sound in air.

(b) Each travel time is individually given by Δ t = L �v. Then the delay between the pulses’ arrivals

is Δt L L= −⎛⎝⎜

⎞⎠⎟

=−1 1

v vv v

v vair cu

cu air

air cu

and the length of the bar is L t=−

=( ) ×( )v v

v vair cu

cu air

m s m sΔ

331 3 56 10

3 5

3.

660 331365

−( ) =m s

m/s)Δ Δt t(

(c) L = (365 m �s)(0.127 s) = 46.3 m

(d) The answer becomes Lt

r

=−

Δ1

3311

m s� v

where vr is the speed of sound in the rod. As v

r

goes to infi nity, the travel time in the rod becomes negligible. The answer approaches(331 m �s)Δ t, which is just the distance that the sound travels in air during the delay time.

P17.58 P P2 1

1

20 0=

. β β1 2

1

2

10− = logPP

80 0 10 20 0 13 0

67 0

2

2

. log . .

.

− = = +

=

β

β dB

468 Chapter 17

FIG. P17.56(a)



P17.59 (a) θ =⎛

⎝⎜⎞

⎠⎟=

×⎛⎝

⎞− −sin sin.

1 13

331

20 0 10

vvsound

obj⎠⎠ = 0 948. °

(b) ′ =×

⎛⎝

⎞⎠ =−θ sin

..1

3

1 533

20 0 104 40°

P17.60 Let T represent the period of the source vibration, and E be the energy put into each wavefront.

Then Pav = E

T. When the observer is at distance r in front of the source, he is receiving a spherical

wavefront of radius vt, where t is the time since this energy was radiated, given by v vt t rS− = .

Then, t

r

S

=−v v

The area of the sphere is 442

2 2

2π πv

vv v

tr

S

( ) =−( )

. The energy per unit area over the spherical wavefront

is uniform with the value E

A

T

rS=

−( )Pav v v

v

2

2 24π. The observer receives parcels of energy with the

Doppler shifted frequency ′ =−

⎛⎝⎜

⎞⎠⎟

=−( )f f

TS S

vv v

vv v

, so the observer receives a wave with intensity

IE

Af

T

r TS=

⎛⎝⎜

⎞⎠⎟

′ =−( )⎛

⎝⎜⎜

⎞

⎠⎟⎟ −

Pav

v v

vv

v v

2

2 24πSS

S

r( )⎛

⎝⎜

⎞

⎠⎟ =

−⎛

⎝⎜⎞

⎠⎟P

av

4 2πv v

v

P17.61 For the longitudinal wave vL

Y=⎛⎝⎜

⎞⎠⎟ρ

1 2

For the transverse wave vT

T=⎛⎝⎜

⎞⎠⎟μ

1 2

If we require vv

L

T

= 8 00. , we have TY= μ

ρ64 0. where μ = m

L and

ρπ

= =mass

volume

m

r L2

This gives

Tr Y

= =×( ) ×( )−π π2 3 2 10

64 0

2 00 10 6 80 10

64.

. .m N m2

...

01 34 104= × N

Sound Waves 469



P17.62 (a) If the source and the observer are moving away from each other, we have: θ θS = =0 180°, and since cos180 1° = − , we get Equation (17.13) with negative values for both v

O and v

S.

(b) If vO

= 0 m /s then ′ =−

f fS S

vv v cosθ

Also, when the train is 40.0 m from the intersection, and the car is 30.0 m from the intersection,

cosθS = 4

5

so ′ =− ( ) ( )f343

343 0 800 25 0500

m s

m s m sHz

. .

or ′ =f 531 Hz

Note that as the train approaches, passes, and departs from the intersection, θS varies from

0° to 180° and the frequency heard by the observer varies between the limits

′ =−

=−

f fS

max cos .

vv v 0

343

343 25 0500

°

m s

m s m sHzz Hz

m s

m s

( ) =

′ =−

=

539

180

343

343f f

Smin cos

vv v ° ++

( ) =25 0

500 466. m s

Hz Hz

P17.63 (a) The time required for a sound pulse to travel

distance L at speed v is given by tL L

Y= =

v ρ Using this expression we fi nd

tL

Y

L1

1

1 1

1

107 00 10 2 7001 96= =

×( ) ( )=

ρ ..

N m kg m2 3××( )

= − = −

×

−10

1 50 1 50

1 60

41

21

2 2

1

L

tL

Y

L

s

m m. .

.ρ 110 11 3 1010 3N m kg m2 3( ) ×( ).

or t L23 4

11 26 10 8 40 10= × − ×( )− −. . s

t

t

3

3

1 50

8 800

4 24

=×( ) ( )

= ×

.

.

m

11.0 10 N m kg m10 3 3

110 4− s

We require t t t1 2 3+ = , or

1 96 10 1 26 10 8 40 10 4 24 1041

3 41

4. . . .× + × − × = ×− − − −L L

This gives L1 1 30= . m and L2 1 50 1 30 0 201= − =. . . m

The ratio of lengths is then L

L1

2

6 45= .

(b) The ratio of lengths L

L1

2

is adjusted in part (a) so that t t t1 2 3+ = . Sound travels the two paths

in equal time and the phase difference Δφ = 0 .

470 Chapter 17

FIG. P17.63

L1 L2

L3


Sound Waves 471


P17.2 1 43. km s

P17.4 (a) 27.2 s (b) longer than 25.7 s, because the air is cooler

P17.6 (a) 153 m s (b) 614 m

P14.8 (a) The speed decreases by 4.6%, from 347 m �s to 331 m �s. (b) The frequency is unchanged, because every wave crest in the hot air becomes one crest without delay in the cold air. (c) The wavelength decreases by 4.6%, from 86.7 mm to 82.8 mm. The crests are more crowded together when they move slower.

P17.10 1 55 10 10. × − m

P17.12 (a) 1 27. Pa (b) 170 Hz (c) 2.00 m (d) 340 m s

P17.14 (a) 4.63 mm (b) 14 5. m s (c) 4 73 109. × W m2

P17.16 (a) 5 00 10 17. × − W (b) 5 00 10 5. × − W

P17.18 21.2 W

P17.20 (a) If

fI2

2

1= ′⎛⎝⎜

⎞⎠⎟

(b) I I2 1=


P17.24 86.6 m

P17.26 (a) 65.0 dB (b) 67.8 dB (c) 69.6 dB

P17.28 (a) 1 76. kJ (b) 108 dB

P17.30 no

P17.32 (a) 2 17. cm s (b) 2 000 028 9. Hz (c) 2 000 057 8. Hz

P17.34 (a) 441 Hz; 439 Hz (b) 54.0 dB

P17.36 (a) 325 m s (b) 29 5. m s

P17.38 (a) 0.364 m (b) 0.398 m (c) 941 Hz (d) 938 Hz

P17.40 46 4. °

P17.42 (a) 7 (b) For sounds of 40 dB or softer, too few digital words are available to represent the wave form with good fi delity. (c) In a sound wave ΔP is negative half of the time but this coding scheme has no words available for negative pressure variations.

P17.44 The wave moves outward equally in all directions. Its amplitude is inversely proportional to its distance from the center so that its intensity follows the inverse-square law, with no absorption of energy by the medium. Its speed is constant at 1.49 km �s, so it can be moving through water at 25°C, and we assume that it is. Its frequency is constant at 323 Hz. Its wavelength is constant at 4.62 m. Its pressure amplitude is 25.0 Pa at radius 1 m. Its intensity at this distance is 209 μW�m2, so the power of the source and the net power of the wave at all distances is 2.63 mW.


472 Chapter 17

P17.46 (a) The intensity is 16 times smaller at the larger distance, because the sound power is spread over a 42 times larger area. (b) The amplitude is 4 times smaller at the larger distance, because intensity is proportional to the square of amplitude. (c) The phase is the same at both points, because they are separated by an integer number of wavelengths.

P17.48 (a) 0 232. m (b) 84 1. nm (c) 13.8 mm

P17.50 (a) 5 04. km s (b) 159 sμ (c) 1.90 mm (d) 0.002 38 (e) 476 MPa (f ) see the solution


P17.54 The gap between bat and insect is closing at 1.69 m s.

P17.56 (a) see the solution (b) 0.343 m (c) 0.303 m (d) 0.383 m (e) 1 03. kHz

P17.58 67 0. dB


P17.62 (a) see the solution (b) 531 Hz


18Superposition and Standing Waves

CHAPTER OUTLINE

18.1 Superposition and Interference18.2 Standing Waves18.3 Standing Waves in a String Fixed

at Both Ends18.4 Resonance18.5 Standing Waves in Air Columns18.6 Standing Waves in Rod and

Membranes18.7 Beats: Interference in Time18.8 Nonsinusoidal Wave Patterns


Q18.1 No. Waves with all waveforms interfere. Waves with other wave shapes are also trains of disturbance that add together when waves from different sources move through the same medium at the same time.

*Q18.2 (i) If the end is fi xed, there is inversion of the pulse upon refl ection. Thus, when they meet, they cancel and the amplitude is zero. Answer (d).

(ii) If the end is free, there is no inversion on refl ection. When they meet, the amplitude is 2 2 0 1 0 2A = ( ) =. . mm . Answer (b).

*Q18.3 In the starting situation, the waves interfere constructively. When the sliding section is moved out by 0.1 m, the wave going through it has an extra path length of 0.2 m = λ �4, to show partial interference. When the slide has come out 0.2 m from the starting confi guration, the extra path length is 0.4 m = λ �2, for destructive interference. Another 0.1 m and we are at r

2 − r

1 = 3λ �4

for partial interference as before. At last, another equal step of sliding and one wave travels one wavelength farther to interfere constructively. The ranking is then d > a = c > b.

Q18.4 No. The total energy of the pair of waves remains the same. Energy missing from zones of destructive interference appears in zones of constructive interference.

*Q18.5 Answer (c). The two waves must have slightly different amplitudes at P because of their different distances, so they cannot cancel each other exactly.

Q18.6 Damping, and non–linear effects in the vibration turn the energy of vibration into internal energy.

*Q18.7 The strings have different linear densities and are stretched to different tensions, so they carry string waves with different speeds and vibrate with different fundamental frequencies. They are all equally long, so the string waves have equal wavelengths. They all radiate sound into air, where the sound moves with the same speed for different sound wavelengths. The answer is (b) and (e).

*Q18.8 The fundamental frequency is described by fL1 2

= v, where v =

⎛⎝⎜

⎞⎠⎟

T

μ

1 2

(i) If L is doubled, then f L11� − will be reduced by a factor

1

2. Answer (f ).

(ii) If μ is doubled, then f11 2� μ− will be reduced by a factor

1

2. Answer (e).

(iii) If T is doubled, then f T1 � will increase by a factor of 2. Answer (c).

473


*Q18.9 Answer (d). The energy has not disappeared, but is still carried by the wave pulses. Each par-ticle of the string still has kinetic energy. This is similar to the motion of a simple pendulum. The pendulum does not stop at its equilibrium position during oscillation—likewise the particles of the string do not stop at the equilibrium position of the string when these two waves superimpose.

*Q18.10 The resultant amplitude is greater than either individual amplitude, wherever the two waves are nearly enough in phase that 2Acos(φ �2) is greater than A. This condition is satisfi ed whenever the absolute value of the phase difference φ between the two waves is less than 120°. Answer (d).

Q18.11 What is needed is a tuning fork—or other pure-tone generator—of the desired frequency. Strike the tuning fork and pluck the corresponding string on the piano at the same time. If they are pre-cisely in tune, you will hear a single pitch with no amplitude modulation. If the two pitches are a bit off, you will hear beats. As they vibrate, retune the piano string until the beat frequency goes to zero.

*Q18.12 The bow string is pulled away from equilibrium and released, similar to the way that a guitar string is pulled and released when it is plucked. Thus, standing waves will be excited in the bow string. If the arrow leaves from the exact center of the string, then a series of odd harmonics will be excited. Even harmonies will not be excited because they have a node at the point where the string exhibits its maximum displacement. Answer (c).

*Q18.13 (a) The tuning fork hits the paper repetitively to make a sound like a buzzer, and the paper effi ciently moves the surrounding air. The tuning fork will vibrate audibly for a shorter time.

(b) Instead of just radiating sound very softly into the surrounding air, the tuning fork makes the chalkboard vibrate. With its large area this stiff sounding board radiates sound into the air with higher power. So it drains away the fork’s energy of vibration faster and the fork stops vibrating sooner.

(c) The tuning fork in resonance makes the column of air vibrate, especially at the antinode of displacement at the top of the tube. Its area is larger than that of the fork tines, so it radiates louder sound into the environment. The tuning fork will not vibrate for so long.

(d) The tuning fork ordinarily pushes air to the right on one side and simultaneously pushes air to the left a couple of centimeters away, on the far side of its other time. Its net disturbance for sound radiation is small. The slot in the cardboard admits the ‘back wave’ from the far side of the fork and keeps much of it from interfering destructively with the sound radiated by the tine in front. Thus the sound radiated in front of the screen can become noticeably louder. The fork will vibrate for a shorter time.

All four of these processes exemplify conservation of energy, as the energy of vibration of the fork is transferred faster into energy of vibration of the air. The reduction in the time of audible fork vibration is easy to observe in case (a), but may be challenging to observe in the other cases.

Q18.14 Walking makes the person’s hand vibrate a little. If the frequency of this motion is equal to the natural frequency of coffee sloshing from side to side in the cup, then a large–amplitude vibra-tion of the coffee will build up in resonance. To get off resonance and back to the normal case of a small-amplitude disturbance producing a small–amplitude result, the person can walk faster, walk slower, or get a larger or smaller cup. Alternatively, even at resonance he can reduce the amplitude by adding damping, as by stirring high–fi ber quick–cooking oatmeal into the hot coffee. You do not need a cover on your cup.

474 Chapter 18


*Q18.15 The tape will reduce the frequency of the fork, leaving the string frequency unchanged. If the bit of tape is small, the fork must have started with a frequency 4 Hz below that of the string, to end up with a frequency 5 Hz below that of the string. The string frequency is 262 + 4 = 266 Hz, answer (d).

Q18.16 Beats. The propellers are rotating at slightly different frequencies.


Section 18.1 Superposition and Interference

P18.1 y y y x t x= + = −( ) + −1 2 3 00 4 00 1 60 4 00 5 0 2. cos . . . sin . .000t( ) evaluated at the given x values.

(a) x = 1.00, t = 1.00 y = ( ) + +( ) = −3 00 2 40 4 00 3 00 1 6. cos . . sin . .rad rad 55 cm

(b) x = 1.00, t = 0.500 y = +( ) + +( ) = −3 00 3 20 4 00 4 00 6. cos . . sin . .rad rad 002 cm

(c) x = 0.500, t = 0 y = +( ) + +( ) =3 00 2 00 4 00 2 50 1 1. cos . . sin . .rad rad 55 cm

P18.2

P18.3 (a) y f x t1 = −( )v , so wave 1 travels in the +x direction

y f x t2 = +( )v , so wave 2 travels in the −x direction

(b) To cancel, y y1 2 0+ = : 5

3 4 2

5

3 4 6 22 2x t x t−( ) += +

+ −( ) +

3 4 3 4 6

3 4 3 4 6

2 2x t x t

x t x t

−( ) = + −( )− = ± + −( )

from the positive root, 8 6t = t = 0 750. s

(at t = 0 750. s, the waves cancel everywhere)

(c) from the negative root, 6 6x = x = 1 00. m

(at x = 1.00 m, the waves cancel always)

Superposition and Standing Waves 475

FIG. P18.2


P18.4 Suppose the waves are sinusoidal.

The sum is 4 00 4 00 90 0. sin . sin .cm cm( ) −( ) + ( ) − +kx t kx tω ω °(( )

2 4 00 45 0 45 0. sin . cos .cm( ) − +( )kx tω ° °

So the amplitude is 8 00 45 0 5 66. cos . .cm cm( ) =° .

P18.5 The resultant wave function has the form

y A kx t= ⎛⎝

⎞⎠ − +⎛

⎝⎞⎠2

2 20 cos sinφ ω φ

(a) A A= ⎛⎝

⎞⎠ = ( ) −⎡

⎣⎢⎤⎦⎥=2

22 5 00

4

29 240 cos . cos .

φ πm

(b) f = = =ωπ

ππ2

1 200

2600 Hz

P18.6 (a) Δx = + − = − =9 00 4 00 3 00 13 3 00 0 606. . . . . m

The wavelength is λ = = =vf

343

3001 14

m s

Hzm.

Thus, Δx

λ= =

0 606

1 140 530

.

..

of a wave,

or Δφ π= ( ) =2 0 530 3 33. . rad

(b) For destructive interference, we want Δ Δx

fx

λ= =0 500.

v

where Δx is a constant in this set up. fx

= =( )

=v

2

343

2 0 606283

Δ .Hz

P18.7 We suppose the man’s ears are at the same level as the lower speaker. Sound from the upper

speaker is delayed by traveling the extra distance L d L2 2+ − .

He hears a minimum when this is 2 1

2

n −( )λ with n = 1 2 3, , , …

Then,

L d Ln

f2 2 1 2+ − =

−( )v

L dn

fL

L dn

fL

n

2 2

2 22 2

22

1 2

1 2 2 1 2

+ =−( ) +

+ =−( ) + +

−(

v

v ))

=− −( )

−( ) =

v

vv

L

f

Ld n f

n fn

2 2 2 21 2

2 1 21 2 3, , ,…

This will give us the answer to (b). The path difference starts from nearly zero when the man is very far away and increases to d when L = 0. The number of minima he hears is the greatest

integer solution to dn

f≥

−( )1 2 v

n = greatest integer ≤ +df

v1

2


476 Chapter 18


(a) df

v+ =

( )( ) + =1

2

4 00 200

330

1

22 92

..

m s

m s

He hears two minima.

(b) With n = 1,

L

d f

f=

− ( )( ) =

( ) − ( )2 2 2 2 2 21 2

2 1 2

4 00 330 4 2vv

. m m s 000

330 200

9 28

2s

m s s

m

( )( )

=L .

With n = 2,

Ld f

f=

− ( )( ) =

2 2 2 23 2

2 3 21 99

vv

. m

P18.8 Suppose the man’s ears are at the same level as the lower speaker. Sound from the upper speaker

is delayed by traveling the extra distance Δr L d L= + −2 2 .

He hears a minimum when Δr n= −( )⎛⎝⎜

⎞⎠⎟2 1

2

λ with n = 1 2 3, , ,…

Then,

L d L nf

2 2 1

2+ − = −⎛

⎝⎞⎠⎛⎝⎜

⎞⎠⎟

v

L d nf

L

L d nf

2 2

2 22

1

2

1

2

+ = −⎛⎝

⎞⎠⎛⎝⎜

⎞⎠⎟+

+ = −⎛⎝

⎞⎠

⎛⎝

v

v⎜⎜

⎞⎠⎟

+ −⎛⎝

⎞⎠⎛⎝⎜

⎞⎠⎟

+2

221

2n

fL L

v

d nf

nf

L22 2

1

22

1

2− −⎛⎝

⎞⎠

⎛⎝⎜

⎞⎠⎟

= −⎛⎝

⎞⎠⎛⎝⎜

⎞⎠⎟

v v (1)

Equation 1 gives the distances from the lower speaker at which the man will hear a minimum. The path difference Δr starts from nearly zero when the man is very far away and increases to d when L = 0.

(a) The number of minima he hears is the greatest integer value for which L ≥ 0. This is the

same as the greatest integer solution to d nf

≥ −⎛⎝

⎞⎠⎛⎝⎜

⎞⎠⎟

1

2

v, or

number of minima heard greatest intege= nmax = rr ≤ df

v⎛⎝⎜

⎞⎠⎟ +

1

2

(b) From equation 1, the distances at which minima occur are given by

Ld n f

n fnn =

− −( ) ( )−( )( ) =

2 2 21 2

2 1 21 2

vv

where , ,…,, maxn



P18.9 (a) φ1 20 0 5 00 32 0 2 00= ( )( ) − ( )(. . . .rad cm cm rad s s)) = 36 0. rad

φ1 25 0 5 00 40 0 2 00= ( )( ) − ( )(. . . .rad cm cm rad s s)) =

= = =

45 0

9 00 516 156

.

.

rad

radiansΔφ ° °

(b) Δφ = − − −[ ] = − +20 0 32 0 25 0 40 0 5 00 8 00. . . . . .x t x t x t

At t = 2.00 s, the requirement is

Δφ π= − + ( ) = +( )5 00 8 00 2 00 2 1. . .x n for any integer n.

For x < 3.20, − 5.00x + 16.0 is positive, so we have

− + = +( )

= − +( )5 00 16 0 2 1

3 202 1

5 00

. .

..

x n

xn

ππ

, or

The smallest positive value of x occurs for n = 2 and is

x = − +( )= − =3 20

4 1

5 003 20 0 058 4.

.. .

π π cm

*P18.10 (a) First we calculate the wavelength: λ = = =vf

344

21 516 0

m s

Hzm

..

Then we note that the path difference equals 9 00 1 001

2. .m m− = λ

Point A is one-half wavelength farther from one speaker than from the other. The waves from the two sources interfere destructively, so the receiver records a minimum in sound intensity.

(b) We choose the origin at the midpoint between the speakers. If the receiver is located at point (x, y), then we must solve:

x y x y+( ) + − −( ) + =5 00 5 001

22 2 2 2. . λ

Then, x y x y+( ) + = −( ) + +5 00 5 001

22 2 2 2. . λ

Square both sides and simplify to get: 20 04

5 002

2 2. .x x y− = −( ) +λ λ

Upon squaring again, this reduces to: 400 10 016 0

5 002 24

2 2 2 2x x x y− + = −( )..

.λ λ λ λ+

Substituting λ = 16 0. m, and reducing, 9 00 16 0 1442 2. .x y− =

or x y2 2

16 0 9 001

. .− =

The point should move along the hyperbola 9x2 − 16y2 = 144.

(c) Yes. Far from the origin the equation might as well be 9x2 − 16y2 = 0, so the point can move along the straight line through the origin with slope 0.75 or the straight line through the origin with slope −0.75.

478 Chapter 18


Section 18.2 Standing Waves

P18.11 y x t A kx= ( ) ( ) ( ) =1 50 0 400 200 2 0. sin . cos sin cosm ωtt

Therefore, k = =20 400

πλ

. rad m λ π= =2

0 40015 7

..

rad mm

and ω π= 2 f so f = = =ωπ π2

200

231 8

rad s

radHz.

The speed of waves in the medium is v = = = = =λ λπ

π ωf f

k22

200

0 400500

rad s

rad mm s

.

P18.12 From y A kx t= 2 0 sin cosω we fi nd

∂∂

=y

xA k kx t2 0 cos cosω

∂∂

= −y

tA kx t2 0ω ωsin sin

∂∂

= −2

2 022

y

xA k kx tsin cosω

∂∂

= −2

2 022

y

tA kx tω ωsin cos

Substitution into the wave equation gives − = ⎛⎝

⎞⎠ −( )2

120

22 0

2A k kx t A kx tsin cos sin cosω ω ωv

This is satisfi ed, provided that v = ωk

. But this is true, because v = = =λ λπ

π ωf f

k22

P18.13 The facing speakers produce a standing wave in the space between them, with the spacing between nodes being

dfNN

m s

sm= = = ( ) =−

λ2 2

343

2 8000 2141

v.

If the speakers vibrate in phase, the point halfway between them is an antinode of pressure at a distance from either speaker of

1 25

0 625.

.m

2=

Then there is a node at 0 6250 214

20 518.

..− = m

a node at 0 518 0 214 0 303. . .m m m− =

a node at 0 303 0 214 0 089 1. . .m m m− =

a node at 0 518 0 214 0 732. . .m m m+ =

a node at 0 732 0 214 0 947. . .m m m+ =

and a node at 0 947 0 214 1 16. . .m m m+ = from either speaker.



*P18.14 (a)

−4

4

02 4 6

x

t = 0y

−4

4

02 4 6

x

t = 20 msy

−4

4

02 4 6

x

t = 5 msy

−4

4

02 4 6

x

t = 10 msy

−4

4

02 4 6

x

t = 15 msy

(b) In any one picture, the distance from one positive-going zero crossing to the next is λ = 4 m.

(c) f = 50 Hz. The oscillation at any point starts to repeat after a period of 20 ms, and f = 1�T.

(d) 4 m. By comparison with the wave function y = (2A sin kx)cos ω t, we identify k = π �2, and

then compute λ = 2π �k.

(e) 50 Hz. By comparison with the wave function y = (2A sin kx)cos ω t, we identify ω = 2πf =100π.

P18.15 y x t1 3 00 0 600= +( )[ ]. sin .π cm; y x t2 3 00 0 600= −( )[ ]. sin .π cm

y y y x t x= + = ( ) ( ) + ( )1 2 3 00 0 600 3 00. sin cos . . sinπ π π ccos .

. sin cos .

0 600

6 00 0 60

ππ

t

y x

( )[ ]= ( ) ( )

cm

cm 00πt( )

(a) We can take cos .0 600 1πt( ) = to get the maximum y.

At x = 0 250. cm, ymax . sin . .= ( ) ( ) =6 00 0 250 4 24cm cmπ

(b) At x = 0 500. cm, ymax . sin . .= ( ) ( ) =6 00 0 500 6 00cm cmπ

(c) Now take cos .0 600 1πt( ) = − to get ymax:

At x = 1 50. cm, ymax . sin . .= ( ) ( ) −( ) =6 00 1 50 1 6 00cm cmπ

480 Chapter 18



(d) The antinodes occur when xn= λ4

n =( )1 3 5, , ,…

But k = =2πλ

π so λ = 2 00. cm

and x1 40 500= =λ. cm as in (b)

x2

3

41 50= =λ. cm as in (c)

x3

5

42 50= =λ. cm

*P18.16 (a) The resultant wave is y A kx t= +⎛⎝

⎞⎠ −⎛

⎝⎞⎠2

2 2sin cos

φ ω φ

The oscillation of the sin(kx + φ �2) factor means that this wave shows alternating nodes and antinodes. It is a standing wave.

The nodes are located at kx n+ =φ π2

so xn

k k= −π φ

2

which means that each node is shifted φ2k

to the left by the phase difference between the traveling waves.

(b) The separation of nodes is Δx nk k

n

k k= +( ) −⎡⎣⎢

⎤⎦⎥− −⎡⎣⎢

⎤⎦⎥

12 2

π φ π φΔx

k= =π λ

2

The nodes are still separated by half a wavelength.

(c) As noted in part (a), the nodes are all shifted by the distance φ�2k to the left.

Section 18.3 Standing Waves in a String Fixed at Both Ends

P18.17 L = 30 0. m; μ = × −9 00 10 3. kg m; T = 20 0. N; fL1 2

= v

where v =⎛⎝⎜

⎞⎠⎟

=T

μ

1 2

47 1. m s

so f147 1

60 00 786= =.

.. Hz f f2 12 1 57= = . Hz

f f3 13 2 36= = . Hz f f4 14 3 14= = . Hz

P18.18 The tension in the string is T = ( )( ) =4 9 8 39 2kg m s N2. .

Its linear density is μ = = × = ×−

−m

L

8 101 6 10

33kg

5 mkg m.

and the wave speed on the string is v = =×

=−

T

μ39 2

10156 53

..

N

1.6 kg mm s

In its fundamental mode of vibration, we have λ = = ( ) =2 2 5 10L m m

Thus, f = = =vλ

156 5

1015 7

..

m s

mHz



P18.19 (a) Let n be the number of nodes in the standing wave resulting from the 25.0-kg mass. Then

n + 1 is the number of nodes for the standing wave resulting from the 16.0-kg mass. For

standing waves, λ = 2L

n, and the frequency is f = v

λ

Thus, fn

L

Tn=2 μ

and also fn

L

Tn= + +1

21

μ

Thus, n

n

T

T

g

gn

n

+ = =( )( ) =

+

1 25 0

16 0

5

41

.

.

kg

kg

Therefore, 4n + 4 = 5n, or n = 4

Then, f =( )

( )( )=4

2 2 00

25 0 9 80

0 002 00.

. .

.m

kg m s

kg m

2

3350 Hz

(b) The largest mass will correspond to a standing wave of 1 loop

(n = 1) so 3501

2 2 00

9 80

0 002 00Hz

m

m s

kg m

2

=( )

( ).

.

.

m

yielding m = 400 kg

P18.20 For the whole string vibrating, dNN = =0 642

. mλ

; λ = 1 28. m

The speed of a pulse on the string is

v = = =fs

λ 3301

1 28 422. m m s

(a) When the string is stopped at the fret,

dNN = =2

30 64

2. m

λ ; λ = 0 853. m

f = = =v

λ422

0 853495

m s

mHz

.

(b) The light touch at a point one third of the way along the string damps out vibration in the two lowest vibration states of the string as a whole. The whole string vibrates

in its third resonance possibility: 3 0 64 32

dNN = =. mλ

λ = 0 427. m

f = = =v

λ422

0 4279

m s

m90 Hz

.

P18.21 dNN = 0 700. m = λ �2

λ

λ

=

= = =×( ) ( )−

1 40

3081 20 10 0 7003

.

. .

m

m sfT

v

(a) T = 163 N

(b) With one-third the distance between nodes, the frequency

is f3 3 220 660= ⋅ Hz = Hz

482 Chapter 18

FIG. P18.20(a)

FIG. P18.20(b)

FIG. P18.21


P18.22 λGGf

= ( ) =2 0 350. mv

; λA AA

Lf

= =2v

L L Lf

fL L

f

fG A GG

AG G

G

A

− = −⎛⎝⎜

⎞⎠⎟

= −⎛⎝⎜

⎞⎠⎟=1 0 350. mm m( ) −⎛

⎝⎞⎠ =1

392

4400 038 2.

Thus, L LA G= − = − =0 038 2 0 350 0 038 2 0 312. . . .m m m m, or the fi nger should be placed

31 2. cm from the bridge .

Lf f

TA

A A

= =v2

1

2 μ; dL

dT

f TAA

=4 μ

; dL

L

dT

TA

A

= 1

2

dT

T

dL

LA

A

= =−( ) =2 2

0 600

3 823 84

.

.. %

cm

35.0 cm

P18.23 In the fundamental mode, the string above the rod has only two nodes, at A and B, with an anti-node halfway between A and B. Thus,

λ

θ2= =AB

L

cos or λ

θ= 2L

cos

Since the fundamental frequency is f, the wave speed inthis segment of string is

v = =λθ

fLf2

cos

Also,

v = = =T T

m AB

TL

mμ θcos

where T is the tension in this part of the string. Thus,

2Lf TL

mcos cosθ θ= or

4 2 2

2

L f TL

mcos cosθ θ=

and the mass of string above the rod is:

mT

Lf= cosθ

4 2 [1]

Now, consider the tension in the string. The light rod would rotate about point P if the string exerted any vertical force on it. Therefore, recalling Newton’s third law, the rod must exert only a horizontal force on the string. Consider a free-body diagram of the string segment in contact with the end of the rod.

F T Mg TMg

y∑ = − = ⇒ =sinsin

θθ

0

Then, from Equation [1], the mass of string above the rod is

mMg

Lf

Mg

Lf= ⎛⎝

⎞⎠ =

sin

cos

tanθθ

θ4 42 2


L

M

A

Bq

T

F

Mg

q

FIG. P18.23


P18.24 Let m = ρV represent the mass of the copper cylinder. The original tension in the wire is T1 = mg =

ρVg. The water exerts a buoyant force ρwater

Vg

2⎛⎝

⎞⎠ on the cylinder, to reduce the tension to

T VgV

g Vg2 2 2= − ⎛

⎝⎞⎠ = −⎛

⎝⎞⎠ρ ρ ρ ρ

waterwater

The speed of a wave on the string changes from T1

μ to T2

μ. The frequency changes from

fT

11 1 1= =vλ μ λ

to fT

22 1=μ λ

where we assume λ = 2L is constant.

Then

f

f

T

T2

1

2

1

2 8 92 1 00 2

8 92= =

−= −ρ ρ

ρwater . .

.

f2 3008 42

8 92291= =Hz Hz

.

.

P18.25 Comparing y = (0.002 m) sin ((π rad�m)x) cos ((100 π rad�s)t)

with y A kx t= 2 sin cosω

we fi nd k = = −2 1πλ

π m , λ = 2 00. m, and ω π π= = −2 100 1f s : f = 50 0. Hz

(a) Then the distance between adjacent nodes is dNN m= =λ2

1 00.

and on the string are L

dNN

m

mloops= =3 00

1 003

.

.

For the speed we have v = = ( ) =−f λ 50 2 1001s m m s

(b) In the simplest standing wave vibration, d bNN m= =3 00

2.

λ, λb = 6 00. m

and

fba

b

= = =vλ

100

6 0016 7

m s

mHz

..

(c) In v00=

T

μ, if the tension increases to T Tc = 9 0 and the string does not stretch, the speed

increases to

v vc

T T= = = = ( ) =9

3 3 3 100 3000 00μ μ

m s m s

Then

λcc

af= = =−

v 300

506 001

m s

sm. d c

NN m= =λ2

3 00.

and one loop fi ts onto the string.

484 Chapter 18


Section 18.4 Resonance

P18.26 The wave speed is v = = ( )( ) =gd 9 80 36 1 18 8. . .m s m m s2

The bay has one end open and one closed. Its simplest resonance is with a node of horizontal veloc-ity, which is also an antinode of vertical displacement, at the head of the bay and an antinode of velocity, which is a node of displacement, at the mouth. The vibration of the water in the bay is like that in one half of the pond shown in Figure P18.27.

Then, dNA m= × =210 104

3 λ

and λ = ×840 103 m

Therefore, the period is Tf

= = = × = × =1 840 104 47 10 12

34λ

vm

18.8 m ss h 24. min

The natural frequency of the water sloshing in the bay agrees precisely with that of lunar excita-tion, so we identify the extra-high tides as amplifi ed by resonance.

P18.27 (a) The wave speed is v = =9 153 66

..

m

2.50 sm s

(b) From the fi gure, there are antinodes at both ends of the pond, so the distance between adjacent antinodes

is dAA m= =λ2

9 15.

and the wavelength is λ = 18 3. m

The frequency is then f = = =vλ

3 66

18 30 200

.

..

m s

mHz

We have assumed the wave speed is the same for all wavelengths.

P18.28 The distance between adjacent nodes is one-quarter of the circumference.

d dNN AA

cmcm= = = =λ

2

20 0

45 00

..

so

λ = 10 0. cm

and

f = = = =vλ

900

0 1009 000 9 00

m s

mHz kHz

..

The singer must match this frequency quite precisely for some interval of time to feed enough energy into the glass to crack it.



Section 18.5 Standing Waves in Air Columns

P18.29 (a) For the fundamental mode in a closed pipe, λ = 4L, as inthe diagram.

But v = f λ, therefore Lf

= v4

So,

L = ( ) =−

343

4 2400 3571

m s

sm.

(b) For an open pipe, λ = 2L, as in the diagram.

So,

Lf

= = ( ) =−

v2

343

2 2400 7151

m s

sm.

P18.30 dAA m= 0 320. ; λ = 0 640. m

(a) f = =vλ

531 Hz

(b) λ = =v� f 0 085 0. m; dAA mm= 42 5.

P18.31 The wavelength is λ = = =vf

3431 31

m s

261.6 sm.

so the length of the open pipe vibrating in its simplest (A-N-A) mode is

dA to A m= =1

20 656λ .

A closed pipe has (N-A) for its simplest resonance,

(N-A-N-A) for the second,

and (N-A-N-A-N-A) for the third.

Here, the pipe length is 55

4

5

41 31 1 64dN to A m m= = ( ) =λ. .

P18.32 The air in the auditory canal, about 3 cm long, can vibrate with a node at the closed end and antinode at the open end,

with dN to A cm= =34

λ

so λ = 0 12. m

and f = = ≈vλ

343

0 123

m s

mkHz

.

A small-amplitude external excitation at this frequency can, over time, feed energy into a larger-amplitude resonance vibration of the air in the canal, making it audible.

486 Chapter 18

L

NA

AA N

l/4

l/2

FIG. P18.29

FIG. P18.31


P18.33 For a closed box, the resonant frequencies will have nodes at both sides, so the permitted

wavelengths will be L n n= =( )1

21 2 3λ , , , , … .

i.e.,

Ln n

f= =λ

2 2

v and f

L=

nv2

Therefore, with L = 0.860 m and ′ =L 2 10. m, the resonant frequencies are f nn = ( )206 Hz for

L = 0.860 m for each n from 1 to 9 and ′ = ( )f nn 84 5. Hz for ′ =L 2 10. m for each n from 2 to 23.

P18.34 The wavelength of sound is λ = vf

The distance between water levels at resonance is df

= v2

∴ = =Rt r dr

fπ π2

2

2

v

and tr

Rf=

π 2

2

v

P18.35 For both open and closed pipes, resonant frequencies are equally spaced as numbers. The set of resonant frequencies then must be 650 Hz, 550 Hz, 450 Hz, 350 Hz, 250 Hz, 150 Hz, 50 Hz.

These are odd-integer multipliers of the fundamental frequency of 50 0. Hz . Then the pipe

length is dfNA = = = ( ) =λ

4 4

340

4 501

v m s

s.70 m .

*P18.36 (a) The open ends of the tunnel are antinodes, so dAA

= 2000 m �n with n = 1, 2, 3, … . Then λ = 2d

AA = 4000 m �n. And f = v�λ = (343 m �s)�(4000 m �n) =

0.0858 Hz, with 1, 2, 3,n n= …

(b) It is a good rule. Any car horn would produce several or many of the closely-spaced reso-nance frequencies of the air in the tunnel, so it would be greatly amplifi ed. Other drivers might be frightened directly into dangerous behavior, or might blow their horns also.

P18.37 For resonance in a narrow tube open at one end,

f nL

=v

4 n =( )1 3 5, , ,…

(a) Assuming n = 1 and n = 3,

3844 0 228

=( )

v.

and 3843

4 0 683=

( )v

.

In either case, v = 350 m s .

(b) For the next resonance n = 5, and

Lf

= =( )( ) =−

5

4

5 350

4 3841 141

v m s

sm.


22.8 cm

68.3 cm

f = 384 Hz

warmair

FIG. P18.37


P18.38 The length corresponding to the fundamental satisfi es fL

= v4

: Lf1 4

34

4 5120 167= =

( ) =v

. m.

Since L > 20.0 cm, the next two modes will be observed, corresponding to

fL

= 3

4 2

v and f

L= 5

4 3

v

or

Lf2

3

40 502= =v. m and L

f3

5

40 837= =v. m

*P18.39 Call L the depth of the well and v the speed of sound.

Then for some integer n L n nf

n= −( ) = −( ) =

−( )( )2 1

42 1

4

2 1 343

4 51 51

1

λ v m s

. ss−( )1

and for the next resonance L n nf

n= +( ) −[ ] = +( ) =

+( )( )2 1 1

42 1

4

2 1 343

4 62

2

λ v m s

00 0 1. s−( ) Thus,

2 1 343

4 51 5

2 1 343

4 601

n n−( )( )( ) =

+( )( )−

m s

s

m s

. ..0 1s−( ) and we require an integer solution to

2 1

60 0

2 1

51 5

n n+ = −. .

The equation gives n = =111 5

176 56

.. , so the best fi tting integer is n = 7.

Then the results L =( ) −[ ]( )

( ) =−

2 7 1 343

4 51 521 61

m s

sm

..

and L =( ) +[ ]( )

( ) =−

2 7 1 343

4 60 021 41

m s

sm

..

suggest that we can say

the depth of the well is (21.5 ± 0.1) m. The data suggest 0.6-Hz uncertainty in the frequency measurements, which is only a little more than 1%.

P18.40 (a) For the fundamental mode of an open tube,

Lf

= = = ( ) =−

λ2 2

343

2 8800 1951

v m s

sm.

(b) v = + −( )=331 1

5 00

273328m s m s

.

We ignore the thermal expansion of the metal.

fL

= = =( ) =

v vλ 2

328

2 0 195841

m s

mHz

.

The fl ute is fl at by a semitone.

488 Chapter 18


Section 18.6 Standing Waves in Rod and Membranes

P18.41 (a) fL

= =( )( ) =

v2

5 100

2 1 601 59

.. kHz

(b) Since it is held in the center, there must be a node in the center as well as antinodes at the

ends. The even harmonics have an antinode at the center so only the odd harmonics are present.

(c) fL

= ′ =( )( ) =

v2

3 560

2 1 601 11

.. kHz

P18.42 When the rod is clamped at one-quarter of its length, the vibration pattern reads ANANA and the rod length is L d= 2 AA = λ .

Therefore,

Lf

= = =v 5 100

4 4001 16

m s

Hzm.

Section 18.7 Beats: Interference in Time

P18.43 f T� �v fnew Hz= =110540

600104 4.

Δ f = −110 104 4 5 64� �s s = beats s. .

P18.44 (a) The string could be tuned to either 521 Hz or 525 Hz from this evidence.

(b) Tightening the string raises the wave speed and frequency. If the frequency were originally 521 Hz, the beats would slow down.

Instead, the frequency must have started at 525 Hz to become

526 Hz .

(c) From fT

L L

T= = =vλ

μμ2

1

2

f

f

T

T2

1

2

1

= and Tf

fT T T2

2

1

2

1

2

1

5230 989=

⎛⎝⎜

⎞⎠⎟

= ⎛⎝

⎞⎠ =Hz

526 Hz. 11

The fractional change that should be made in the tension is then

fractional change = − = − = =T T

T1 2

1

1 0 989 0 011 4 1 14. . . % lower

The tension should be reduced by 1.14% .

P18.45 For an echo ′ =+( )−( )f f s

s

v vv v

the beat frequency is f f fb = ′ − .

Solving for fb , gives f fbs

s

=( )−( )2v

v v when approaching wall.

(a) fb = ( ) ( )−( ) =256

2 1 33

343 1 331 99

.

.. Hz beat frequency

(b) When he is moving away from the wall, vs changes sign. Solving for sv gives

vv

sb

b

f

f f=

−= ( )( )( )( ) − =

2

5 343

2 256 53 38. m s



P18.46 Using the 4 and 22

3-foot pipes produces actual frequencies of 131 Hz and 196 Hz and a com-

bination tone at 196 131 65 4−( ) =Hz Hz. , so this pair supplies the so-called missing fundamental. The 4 and 2-foot pipes produce a combination tone 262 131 131−( ) =Hz Hz, so this does not work.

The 22

32and -foot pipes produce a combination tone at 262 196 65 4−( ) =Hz Hz. , so this works.

Also, 4 22

32, , and -foot pipes all playing together produce the 65.4-Hz combination tone.

Section 18.8 Nonsinusoidal Wave Patterns

P18.47 We list the frequencies of the harmonics of each note in Hz:

HarmonicNote 1 2 3 4 5

A 440.00 880.00 1 320.0 1 760.0 2 200.0C# 554.37 1 108.7 1 663.1 2 217.5 2 771.9E 659.26 1 318.5 1 977.8 2 637.0 3 296.3

The second harmonic of E is close the the thhird harmonic of A, and the fourth harmonicc of C# is

close to the fifth harmonic of A..

P18.48 We evaluate

s = + + +100 157 2 62 9 3 105 4sin sin . sin sinθ θ θ θ

+ 51 9. siin . sin . sin5 29 5 6 25 3 7θ θ θ+ +

where s represents particle displacement in nanometers and θ represents the phase of the wave in radians. As θ advances by 2π , time advances by (1�523) s. Here is the result:

Additional Problems

*P18.49 (a) The yo-yo’s downward speed is dL �dt = 0 + (0.8 m �s2)(1.2 s) = 0.960 m �s. The instanta-neous wavelength of the fundamental string wave is given by d

NN = λ �2 = L so λ = 2L and

dλ �dt = 2 dL �dt = 2(0.96 m �s) = 1.92 m �s.

(b) For the second harmonic, the wavelength is equal to the length of the string. Then the rate of

change of wavelength is equal to dL �dt = 0.960 m/s, half as much as for the f irst harmonic .

(c) A yo-yo of different mass will hold the string under different tension to make each string wave vibrate with a different frequency, but the geometrical argument given in parts (a) and

(b) still applies to the wavelength. The answers are unchanged : dλ1�dt = 1.92 m �s and

dλ2 �dt = 0.960 m �s.

490 Chapter 18

FIG. P18.48


*P18.50 (a) Use the Doppler formula

′ =±( )

( )f fs

v vv v

0

∓ With ′=f1 frequency of the speaker in front of student and

′ =f2 frequency of the speaker behind the student.

′= ( ) +( )−( ) =f1 456

343 1 50

343 0458Hz

m s m s

m s

.HHz

Hzm s m s

m s′ = ( ) −( )

+( ) =f2 456343 1 50

343 04

.554 Hz

Therefore, f f fb = ′− ′ =1 2 3 99. Hz

(b) The waves broadcast by both speakers have λ = = =vf

343

4560 752

m s

sm. . The standing

wave between them has dAA = =λ2

0 376. m. The student walks from one maximum to the

next in time Δt = =0 376

1 500 251

.

..

m

m ss, so the frequency at which she hears maxima is

fT

= =13 99. Hz

(c) The answers are identical. The models are equally valid. We may think of the interference of the two waves as interference in space or in time, linked to space by the steady motion of the student.

P18.51 f = 87 0. Hz

speed of sound in air: va = 340 m s

(a) λb = � v = = ( )( )−f bλ 87 0 0 4001. .s m

v = 34 8. m s

(b) λλ

a

a a

L

f

==

⎫⎬⎭

4

v L

fa= = ( ) =−

v4

340

4 87 00 9771

m s

sm

..


FIG. P18.51


P18.52

dNA

dNA

dNA

dNA

dNA

IIIII I

A

A

AA

A

A

N

N

N

N

NN

FIG. P18.52

(a) μ = × = ×−

−5 5 106 40 10

33.

.kg

0.86 mkg m

v = = ⋅×

=−

T

μ1 30

6 40 1014 33

.

..

kg m s

kg mm s

2

(b) In state I, dNA m= =0 8604

.λ

(c) λ = 3 44. m f = = =vλ

14 3

3 444 14

.

..

m s

mHz

In state II, dNA m m= ( ) =1

30 86 0 287. .

λ = ( ) =4 0 287 1 15. .m m f = = =vλ

14 3

1 1512 4

.

..

m s

mHz

In state III, dNA m m= ( ) =1

50 86 0 172. .

f = =( ) =

vλ

14 3

4 0 17220 7

.

..

m s

mHz

P18.53 If the train is moving away from station, its frequency is depressed:

′ = − =f 180 2 00 178. Hz: 178 180343

343=

− −( )v

Solving for v gives v = ( )( )2 00 343

178

.

Therefore, v = 3 85. m s away from station

If it is moving toward the station, the frequency is enhanced:

′ = + =f 180 2 00 182. Hz: 182 180343

343=

− v

Solving for v gives 42 00 343

182= ( )( ).

Therefore, v = 3 77. m s toward the station

P18.54 v = ( )( )×

=−

48 0 2 00

4 80 101413

. .

.m s

dNN m= 1 00. ; λ = 2 00. m; f = =vλ

70 7. Hz

λaa

f= = =v 343

70 74 85

m s

Hzm

..

492 Chapter 18


P18.55 (a) Since the fi rst node is at the weld, the wavelength in the thin wire is 2L or 80.0 cm. The frequency and tension are the same in both sections, so

fL

T= =( ) ×

=−

1

2

1

2 0 400

4 60

2 00 1059 93μ .

.

.. Hz

(b) As the thick wire is twice the diameter, the linear density is 4 times that of the thin wire. ′ =μ 8 00. g m

so ′ =′

Lf

T1

2 μ

′ =

( )( )⎡⎣⎢

⎤⎦⎥ × −L

1

2 59 9

4 60

8 00 10 3.

.

. = 20 0. cm half the length of the thin wire.

*P18.56 The wavelength stays constant at 0.96 m while the wavespeed rises according tov = (T�μ)1/2 = [(15 + 2.86t)�0.0016]1/2 = [9375 + 1786t]1/2 so the frequency rises asf = v�λ = [9375 + 1786t]1/2�0.96 = [10 173 + 1938t]1/2 The number of cycles is f dt ineach incremental bit of time, or altogether

10173 1938

1

193810173 19381 2

0

3 5 1+( ) = +( )∫ t dt t/. //. 2

0

3 51938 dt∫

/ .3 2

0

31

1938

10173 1938

3 2

t=

+[ ]�

55 3 2 3 216954 10173

2906407=

( ) − ( )=

/ /

cycles

P18.57 (a) fn

L

T=2 μ

so ′ =′

= =f

f

L

L

L

L2

1

2

The frequency should be halved to get the same number of antinodes for twice the

length.

(b) ′ =

′n

n

T

T so ′ =

′⎛⎝

⎞⎠ =

+⎡⎣⎢

⎤⎦⎥

T

T

n

n

n

n

2 2

1

The tension must be ′ =+

⎡⎣⎢

⎤⎦⎥

Tn

nT

1

2

(c) ′ = ′′

′f

f

n L

nL

T

T so ′ = ′ ′

′⎛⎝⎜

⎞⎠⎟

T

T

nf L

n fL

2

′ =⋅

⎛⎝

⎞⎠

T

T

3

2 2

2

′ =T

T

9

16 to get twice as many antinodes.



P18.58 (a) For the block:

F T Mgx∑ = − =sin .30 0 0°

so T Mg Mg= =sin .30 01

2°

(b) The length of the section of string parallel to the incline ish

hsin .30 0

2°= . The total length of the string is then 3h .

(c) The mass per unit length of the string is μ = m

h3

(d) The speed of waves in the string is v = = ⎛⎝

⎞⎠⎛⎝

⎞⎠ =T Mg h

m

Mgh

mμ 2

3 3

2

(e) In the fundamental mode, the segment of length h vibrates as one loop. The distance

between adjacent nodes is then d hNN = =λ2

, so the wavelength is λ = 2h.

The frequency is fh

Mgh

m

Mg

mh= = =vλ

1

2

3

2

3

8

(g) When the vertical segment of string vibrates with 2 loops (i.e., 3 nodes), then h = ⎛⎝⎜

⎞⎠⎟2

2

λ

and the wavelength is λ = h

(f ) The period of the standing wave of 3 nodes (or two loops) is

Tf

hm

Mgh

mh

Mg= = = =1 2

3

2

3

λv

(h) f f f fMg

mhb = − = ×( ) = ×( )− −1 02 2 00 10 2 00 103

82 2. . .

P18.59 We look for a solution of the form

5 00 2 00 10 0 10 0 2 00 10 0. sin . . . cos . .x t x t−( ) + −( )

A= ssin . .

sin . . cos

2 00 10 0

2 00 10 0

x t

A x t

− +( )= −( )

φφ A+ ccos . . sin2 00 10 0x t−( ) φ

This will be true if both 5 00. cos= A φ and10 0. sin= A φ,

requiring 5 00 10 02 2 2. .( ) + ( ) = A

A = 11 2. and φ = 63 4. °

The resultant wave 11 2 2 00 10 0 63 4. sin . . .x t− +( )° is sinusoidal.

494 Chapter 18

FIG. P18.58


P18.60 dAA m= = × −λ2

7 05 10 3. is the distance between antinodes.

Then λ = × −14 1 10 3. m

and f = = ××

= ×−

vλ

3 70 102 62 10

35.

.m s

14.1 10 mHz3

The crystal can be tuned to vibrate at 218 Hz, so that binary counters canderive from it a signal at precisely 1 Hz.

P18.61 (a) Let θ represent the angle each slanted rope makes with the vertical.

In the diagram, observe that:

sin.θ = =1 00 2

3

m

1.50 m or θ = 41 8. °

Considering the mass,

Fy∑ = 0: 2T mgcosθ =

or T =( )( )

=12 0 9 80

2 41 878 9

. .

cos ..

kg m sN

2

°

(b) The speed of transverse waves in the string is v = = =T

μ78 9

281. N

0.001 00 kg mm s

For the standing wave pattern shown (3 loops), d = 3

2λ

or λ = ( )=2 2 00

31 33

..

mm

Thus, the required frequency is f = = =vλ

281

1 33211

m s

mHz

.



P18.4 5.66 cm

P18.6 (a) 3.33 rad (b) 283 Hz

P18.8 (a) The number is the greatest integer ≤ ⎛⎝

⎞⎠ +d

f

v1

2 (b) L

d n f

n fn =− −( ) ( )

−( )( )2 2 2

1 2

2 1 2

vv

where

n n= 1 2, , , max…

P18.10 (a) Point A is one half wavelength farther from one speaker than from the other. The waves it receives interfere destructively. (b) Along the hyperbola 9x2 − 16y2 = 144. (c) Yes; along the straight line through the origin with slope 0.75 or the straight line through the origin with slope −0.75.


P18.14 (a) see the solution (b) 4 m is the distance between crests. (c) 50 Hz. The oscillation at any point starts to repeat after a period of 20 ms, and f = 1�T. (d) 4 m. By comparison with equation 18.3,k = π �2, and λ = 2π �k. (e) 50 Hz. By comparison with equation 18.3, ω = 2πf = 100π.


FIG. P18.61

FIG. P18.60


P18.16 (a) Yes. The resultant wave contains points of no motion. (b) and (c) The nodes are still separated by λ �2. They are all shifted by the distance φ � 2k to the left.

P18.18 15.7 Hz

P18.20 (a) 495 Hz (b) 990 Hz

P18.22 31.2 cm from the bridge; 3.84%

P18.24 291 Hz

P18.26 The natural frequency of the water sloshing in the bay agrees precisely with that of lunar excita-tion, so we identify the extra-high tides as amplifi ed by resonance.

P18.28 9.00 kHz

P18.30 (a) 531 Hz (b) 42.5 mm

P18.32 3 kHz; a small-amplitude external excitation at this frequency can, over times, feed energy into a larger-amplitude resonance vibration of the air in the canal, making it audible.

P18.34 Δtr

Rf=π 2

2

v

P18.36 (a) 0.0858 n Hz, with n = 1, 2, 3, … (b) It is a good rule. Any car horn would produce several or many of the closely-spaced resonance frequencies of the air in the tunnel, so it would be greatly amplifi ed. Other drivers might be frightened directly into dangerous behavior, or might blow their horns also.

P18.38 0.502 m; 0.837 m

P18.40 (a) 0.195 m (b) 841 m

P18.42 1.16 m

P18.44 (a) 521 Hz or 525 Hz (b) 526 Hz (c) reduce by 1.14%

P18.46 4-foot and 22

3-foot; 2

2

32and -foot; and all three together


P18.50 (a) 3.99 beats�s (b) 3.99 beats�s (c) The answers are identical. The models are equally valid. We may think of the interference of the two waves as interference in space or in time, linked to space by the steady motion of the student.

P18.52 (a) 14.3 m �s (b) 86.0 cm, 28.7 cm, 17.2 cm (c) 4.14 Hz, 12.4 Hz, 20.7 Hz

P18.54 4.85 m

P18.56 407 cycles

P18.58 (a)1

2Mg (b) 3h (c) m

h3 (d)

3

2

Mgh

m (e)

3

8

Mg

mh (f )

2

3

mh

Mg (g) h (h) 2 00 10

3

82. ×( )− Mg

mh

P18.60 262 kHz

496 Chapter 18


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