Pop5e ISM Chapter13 vp rev ed vp

542 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

13 Mechanical Waves

CHAPTER OUTLINE

13.1 Propagation of a Disturbance

13.2 Analysis Model: Traveling Wave

13.3 The Speed of Transverse Waves on Strings

13.4 Reflection and Transmission

13.5 Rate of Energy Transfer by Sinusoidal Waves on Strings

13.6 Sound Waves

13.7 The Doppler Effect

13.8 Context Connection: Seismic Waves

* An asterisk indicates an item new to this edition.

ANSWERS TO OBJECTIVE QUESTIONS

*OQ13.1 Answer (b). Not all waves are sinusoidal. A sinusoidal wave is a wave of a single frequency. In general, a wave can be a superposition of many sinusoidal waves.

*OQ13.2 Answer (c). The distance between two successive peaks is the wavelength: λ = 2 m, and the frequency is 4 Hz. The frequency, wavelength, and speed of a wave are related by the equation fλ = v.

*OQ13.3 (i) The ranking is (c) = (d) > (e) > (b) > (a). Look at the coefficients of the sine and cosine functions: (a) 4, (b) 6, (c) 8, (d) 8, (e) 7.

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

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

Chapter 13 543

© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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

(v) The ranking is (c) > (a) = (b) = (d) > (e). From v = fλ = ω/k, we compute the absolute value of the ratio of the coefficient of t to the coefficient of x in each case: (a) 5, (b) 5, (c) 7.5, (d) 5, (e) 4.

*OQ13.4 (i) Answer (a). Higher tension makes wave speed higher.

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

*OQ13.5 Answer (b). Wave speed is inversely proportional to the square root of linear density.

*OQ13.6 Answer (b). From v =

Tµ

, we must increase the tension by a factor

of 4 to make v double.

*OQ13.7 Answer (b). A sound wave is a longitudinal vibration that is propagated through a material medium.

*OQ13.8 Answer (e). The speed of sound in air, at atmospheric pressure, is determined by the temperature of the air and does not depend on the frequency of the sound. Sound from siren A will have a wavelength that is half the wavelength of the sound from B, but the speed of the sound (the product of frequency times wavelength) will be the same for the two sirens.

*OQ13.9 Answer (b). The typically higher density would by itself make the speed of sound lower in a solid compared to a gas.

*OQ13.10 Answer (c). The power carried by a wave is proportional to its frequency, wave speed, and the square of its amplitude. If the frequency does not change, the amplitude is increased by a factor of

2 . The wave speed does not change.

*OQ13.11 In order of decreasing size we have (b) > (d) > (a) > (c) > (e). In

′f = f (v + vo )[ ] (v − vs)[ ] we can consider the size of the fraction

(v + vo ) (v − vs) in each case, where the positive direction for the observer is toward the source, the positive direction for the source is toward the observer: (a) 343/343 = 1, (b) 343/(343 − 25) = 1.08, (c) 343/(343 + 25) = 0.932, (d) (343 + 25)/343 = 1.07, (e) (343 − 25)/343 = 0.927.

*OQ13.12 (a) through (d): Yes to all. The maximum element speed and the wave speed are related by vy, max = ω A = 2π fA = 2π fA/λ. Thus the amplitude or the wavelength of the wave can be adjusted to make either vy, max or v larger.

544 Mechanical Waves


*OQ13.13 (i) Answer (a). As the wave passes from the massive string to the less massive string, the wave speed will increase according to

v =

Tµ

.

(ii) Answer (c). The frequency will remain unchanged. The rate at which crests come up to the boundary is the same rate at which they leave the boundary.

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

*OQ13.14 Answer (d). We have fs = 1 000 Hz, v = 343 m/s, vo = −30 m/s, vs = 50 m/s. We find

′f =

f v + vo( )v − vs( ) =

1 000 Hz( ) 343 m/s( ) + −30 m/s( )[ ]343 m/s − 50 m/s

= 1 068 Hz

*OQ13.15 Answer (c). The ambulance driver, sitting at a fixed distance from the siren, hears the actual frequency emitted by the siren. However, the distance between you and the siren is decreasing, so you will detect a frequency higher than the actual 500 Hz.

*OQ13.16 (i) Answer (b). The frequency increases by a factor of 2 because the wave speed, which is dependent only on the medium through which the wave travels, remains constant.

(ii) Answer (c).

*OQ13.17 (i) Answer (c). Both observer and source have equal speeds in opposite directions relative to the medium, so in

′f = (v + vo )/(v − vs) we would have something like (343 − 25)f/(343 − 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. The wind “stretches” the wavelength out.

(iii) Answer (a).

ANSWERS TO CONCEPTUAL QUESTIONS

*CQ13.1 Our brave Siberian saw the first 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 compressional waves in the air and in the ground. The wave in the ground, which can be called either sound or a

Chapter 13 545


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 first 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, reflected sound, and perhaps the sound of the falling trees.

*CQ13.2 The type of wave you generate depends upon the direction of the disturbance (vibration) you generate and the direction of its travel (propagation).

(a) To use a spring (or slinky) to create a longitudinal wave, pull a few coils back and release.

(b) For a transverse wave, jostle the end coil side to side.

*CQ13.3 As the pulse moves down the string, the elements 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 definition.

*CQ13.4 The speed of sound in air is proportional to the square-root of the absolute temperature, T . The speed of sound is greater in warmer air, so the pulse from the camera would return sooner than it would on a cooler day from an object at the same distance. The camera would interpret an object as being closer than it actually is on a hot day.

*CQ13.5 It depends on from what the wave reflects. If reflecting from a less dense string, the reflected part of the wave will be right side up. A wave inverts when it reflects off a medium in which the wave speed is smaller.

*CQ13.6 No. The vertical speed of an element will be the same on any string because it depends only on frequency and amplitude: vy, max = ωA = 2π fA. The elements of strings with different wave speeds will have the same maximum vertical speed.

*CQ13.7 The speed of light is so high that the arrival of the flash is practically simultaneous with the lightning discharge. Thus, the delay between the flash and the arrival of the sound of thunder is the time sound takes to travel the distance separating the lightning from you. By counting the seconds between the flash and thunder and knowing the approximate speed of sound in air, you have a rough measure of the distance to the lightning bolt.

*CQ13.8 Both. There are actually two Doppler shifts. The first shift arises from



the source (you) moving toward the observer (the cliff). The second arises from the observer (you) moving toward the source (the cliff). If, instead of a cliff, there is a spacecraft moving toward you, then there are shifts due to moving source (you) and moving observer (the spacecraft) before reflection, and moving source (the spacecraft) and moving observer (you) after reflection.

*CQ13.9 Since the frequency is 3 cycles per second, the period is 1/3 second = 333 ms.

*CQ13.10 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 =

Tµ

increases with height.

*CQ13.11 A beam of radio waves of known frequency is sent toward a speeding car, which reflects the beam back to a detector in the police car. The amount the returning frequency has been shifted depends on the velocity of the oncoming car.

*CQ13.12 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.

*CQ13.13 (a) Let ∆t = ts − tp represent the difference in arrival times of the two waves at a station at distance d = vsts = vptp from the hypocenter.

Then d = Δt

1vs

−1vp

⎛

⎝⎜

⎞

⎠⎟

−1

.

(b) Knowing the distance from the first station places the hypocenter on a sphere around it. A measurement from a second station limits it to another sphere, which intersects with the first in a circle. Data from a third non-collinear station will generally limit the possibilities to a point.

Chapter 13 547


SOLUTIONS TO END-OF-CHAPTER PROBLEMS

Section 13.1 Propagation of a Disturbance P13.1 We obtain a function of the same shape by writing

y(x, t) = 6/[(x − x0)2 + 3]

where the center of the pulse is at x0 = 4.5t. Thus, we have

y =6

x − 4.50t( )2 + 3⎡⎣ ⎤⎦

Note that for y to stay constant as t increases, x must increase by 4.5t, as it should to describe the wave moving at 4.5 m/s.

P13.2 (a)

ANS FIG. P13.2 (a)

(b)

ANS FIG. P13.2 (b)

(c)

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.

(d) The wave has travelled d = vt = 2.40 m to the right.



Section 13.2 Analysis Model: Traveling Wave P13.3 (a) We note that sin (θ) = −sin (–θ) = sin (−θ + π), so the given wave

function can be written as

y(x, t) = (0.350) sin (–10πt + 3πx + π – π/4)

Comparing, 10π t − 3π x + π/4 = kx −ωt + φ. For constant phase, x must increase as t increases, so the wave travels in the positive x direction. Comparing the specific form to the general form, we find that

v =

ωk =

10π3π

= 3.33 m/s.

Therefore, the velocity is

3.33i( ) m/s .

(b) Substituting t = 0 and x = 0.100 m, we have

y 0.100 0( ) = 0.350 m( ) sin −0.300π +π4

⎛⎝⎜

⎞⎠⎟ = −0.054 8 m

= −5.48 cm

(c) k =

2πλ

= 3π : λ = 0.667 m ω = 2π f = 10π : f = 5.00 Hz

(d) vy =

∂y∂t

= 0.350( ) 10π( ) cos 10π t − 3π x +π4

⎛⎝⎜

⎞⎠⎟

vy , max = 10π( ) 0.350( ) = 11.0 m/s

P13.4 The linear wave equation is

∂2y∂x2 =

1v2

∂2y∂t2 .

If y = eb(x−vt),

Then

∂y∂t

= −bveb x−vt( ) and

∂y∂x

= beb x−vt( )

∂2y∂t2 = b2v2eb x−vt( ) and

∂2y∂x2 = b2eb x−vt( )

Therefore,

∂2y∂t2 = v2 ∂2y

∂x2 , demonstrating that eb(x−vt) is a solution.

P13.5 (a) ω = 2π f = 2π 5.00 s−1( ) = 31.4 rad s

(b) λ =

vf=

20.0 m/s5.00 s−1 = 4.00 m

Chapter 13 549


k =

2πλ

=2π

4.00 m= 1.57 rad/m

(c) In y = A sin kx −ω t + φ( ) we take A = 12.0 cm. At x = 0 and t = 0

we have y = 12.0 cm( ) sin φ. To make this fit y = 0, we take φ = 0. Then

y = 0.120 sin (1.57x − 31.4t), where x and y are in meters and t isin seconds

(d) The transverse velocity is

∂y∂t

= −Aω cos kx −ωt( ).

Its maximum magnitude is

Aω = 12.0 cm( ) 31.4 rad s( ) = 3.77 m s

(e) ay =

∂vy

∂t=

∂∂t

−Aω cos kx −ωt( )[ ] = −Aω 2 sin kx −ωt( )

The maximum value is Aω 2 = 0.120 m( ) 31.4 s−1( )2

= 118 m/s2

P13.6 Using data from the observations, we have λ = 1.20 m and

f =

8.00 crests12.0 s

=8.00 cycles

12.0 s=

8.0012.0

Hz

Therefore, v = λ f = 1.20 m( ) 8.00

12.0 Hz⎛

⎝⎜⎞⎠⎟ = 0.800 m/s .

P13.7 (a) A = ymax = 8.00 cm = 0.080 0 m k =

2πλ

=2π

0.800 m( ) = 2.50π m−1

ω = 2π f = 2π 3.00( ) = 6.00π rad s

Therefore, y = A sin kx +ωt( ).

Or [where y(0, t) = 0 at t = 0],

y = 0.080 0( ) sin 2.50πx + 6.00π t( ) .

(b) In general, y = 0.080 0 sin 2.50x + 6.00π t + φ( ).

Assuming y(x, 0) = 0 at x = 0.100,

then we require that 0 = 0.080 0 sin 0.250π + φ( )

or φ = −0.250π .

Therefore,



y = 0.080 0 sin 2.50πx + 6.00π t − 0.250π( ) , where x and y are inmeters and t is in seconds.

P13.8 Compare the specific equation to the general form:

y = (0.020 0 m) sin (2.11x − 3.62t) = y = A sin (kx − ω t + φ)

(a) A = 2.00 cm

(b) k = 2.11 rad m → λ =

2πk

= 2.98 m

(c) ω = 3.62 rad s → f =

ω2π

= 0.576 Hz

(d) v = fλ =

ω2π

2πk

=3.622.11

= 1.72 m s

P13.9 At time t, the motion at point A, where x = 0, is

yA = (15.0 cm) cos (−50.3t)

At point B, the motion is

yB = (15.0 cm) cos (15.7xB − 50.3t) = (15.0 cm) cos −50.3t ± π/3( ) which implies

15.7xB = 15.7 m−1( )xB = ±π 3→ xB = ±0.066 7 m = ±6.67 cm

P13.10 y = 0.120 m( ) sin

π8

x + 4π t⎛⎝⎜

⎞⎠⎟

(a) v =

∂y∂t

: v = 0.120( ) 4π( ) cosπ8

x + 4π t⎛⎝⎜

⎞⎠⎟

v 0.200 s, 1.60 m( ) = −1.51 m/s

(b) a =

∂v∂t

: a = −0.120 m( ) 4π( )2 sinπ8

x + 4π t⎛⎝⎜

⎞⎠⎟

a 0.200 s, 1.60 m( ) = 0

(c) k =

π8=

2πλ

: λ = 16.0 m

(d) ω = 4π =

2πT

: T = 0.500 s

Chapter 13 551


(e) v =

λT=

16.0 m0.500 s

= 32.0 m s

P13.11 f =

40.0 vibrations30.0 s

=43

Hz

v =

425 cm10.0 s

= 42.5 cm/s

λ =

vf=

42.5 cm s43 Hz

= 31.9 cm = 0.319 m

*P13.12 (a)

ANS FIG. P13.12(a)

(b) k =

2πλ

=2π

0.350 m= 18.0 rad m

(c) T =

1f=

112.0/s

= 0.083 3 s

(d) ω = 2π f = 2π 12.0 s = 75.4 rad s

(e) v = fλ = 12.0 s( ) 0.350 m( ) = 4.20 m s

(f) y = A sin kx +ωt + φ( ) specializes to

y = 0.200 m( ) sin 18.0x m + 75.4t s + φ( )

(g) At x = 0, t = 0 we require

−3.00 × 10−2 m = 0.200 m( ) sin +φ( )φ = −8.63° = −0.151 rad

so

y x, t( ) = 0.200 sin 18.0x + 75.4t − 0.151( ) , where x and y are inmeters and t is in seconds.



*P13.13 v = fλ = 4.00 Hz( ) 60.0 cm( ) = 240 cm s = 2.40 m s

P13.14 (a) Let us write the wave function as y x, t( ) = A sin kx +ωt + φ( ).

We have yi = y 0, 0( ) = A sin φ = 0.020 0 m

and vi = v 0, 0( ) = ∂y

∂t 0, 0

= Aω cos φ = −2.00 m/s.

Also, ω =

2πT

=2π

0.025 0 s= 80.0π s−1.

Use the identity sin2φ + cos2φ = 1 and the expressions for yi and vi:

A sin φ( )2

A2 +Aω cos φ( )2

A2ω 2 = 1

A sin φ( )2 +Aω cos φ( )2

ω 2 = A2

A2 = yi2 +

vi

ω⎛⎝⎜

⎞⎠⎟

2

= 0.020 0 m( )2 +−2.00 m/s80.0π s−1

⎛⎝⎜

⎞⎠⎟

2

A = 0.021 5 m

(b)

ωyi

vi

=ω A sin φ( )ωA cos φ

= tan φ → tan φ =80.0π 0.020 0( )

−2.00= −2.51

Your calculator’s answer φ = tan−1 (−2.51) = −1.19 rad is an angle in the fourth quadrant with a negative sine and positive cosine, just the reverse of what is required. Recall on the unit circle, an angle with a negative tangent can be in either the second or fourth quadrant. The sine is positive and the cosine is negative in the second quadrant. The angle in the second quadrant is

φ = π − 1.19 rad = 1.95 rad

(c) vy , max = Aω = 0.021 5 m( ) 80.0π s( ) = 5.41 m/s

(d) λ = vxT = 30.0 m s( ) 0.025 0 s( ) = 0.750 m

k =

2πλ

=2π

0.750 m= 8.38 m−1 , ω = 80.0π s−1

y x, t( ) = 0.021 5( ) sin 8.38x + 80.0π t + 1.95( )

Chapter 13 553


Section 13.3 The Speed of Transverse Waves on Strings P13.15 The total time interval is the sum of the two time intervals.

In each wire

Δt =

Lv= L

µT

Let A represent the cross-sectional area of one wire. The mass of one wire can be written both as m = ρV = ρAL and also as m = µL.

Then we have µ = ρA =

πρd2

4.

Thus, Δt = L

πρd2

4T⎛⎝⎜

⎞⎠⎟

1 2

.

For copper,

Δt = 20.0 m( )π( ) 8 920 kg/m3( ) 1.00 × 10−3 m( )2

4( ) 150 N( )⎡

⎣⎢⎢

⎤

⎦⎥⎥

1 2

= 0.137 s

For steel,

Δt = 30.0 m( )π( ) 7 860 kg/m3( ) 1.00 × 10−3 m( )2

4( ) 150 N( )⎡

⎣⎢⎢

⎤

⎦⎥⎥

1 2

= 0.192 s

The total time interval is 0.137 + 0.192 = 0.329 s

P13.16 The tension in the string is T = mg, where g is the acceleration of gravity on the Moon, about one-sixth that of Earth. From the data given, what is the acceleration of gravity on the Moon?

The wave speed is

v =

Tµ

=Mgm/L

=MgL

m=

Lt→

MgLm

=L2

t2 → g =mLMt2 .

g =

mLMt2 =

4.00 × 10−3 kg( ) 1.60 m( )3.00 kg( ) 26.1× 10−3 s( )2 = 3.13 m/s2

The calculated gravitational acceleration of the Moon is almost twice thatof the accepted value.



P13.17 The down and back distance is 4.00 m + 4.00 m = 8.00 m.

The speed is then v =

dtotal

t=

4 8.00 m( )0.800 s

= 40.0 m s =Tµ

.

Now, µ =

0.200 kg4.00 m

= 5.00 × 10−2 kg/m.

So T = µv2 = 5.00 × 10−2 kg/m( ) 40.0 m/s( )2 = 80.0 N .

P13.18 From the free-body diagram mg = 2T sin θ

T =

mg2 sin θ

The angle θ is found from

cos θ =

3L/8L/2

=34

∴θ = 41.4°

(a) v =

Tµ

v =

mg2µ sin 41.4°

=9.80 m/s2

2 8.00 × 10−3 kg/m( ) sin 41.4°

⎛

⎝⎜

⎞

⎠⎟ m

or

v = 30.4 ( ) m , where v is in meters per second and m is in kilograms.

(b) v = 60.0 = 30.4 m and

m = 3.89 kg

P13.19 The two wave speeds can be written as:

v1 = T1 µ and v2 = T2 µ

Since µ is constant, µ =

T2

v22 =

T1

v12 , and

T2 =

v2

v1

⎛⎝⎜

⎞⎠⎟

2

T1 =30.0 m s20.0 m s

⎛⎝⎜

⎞⎠⎟

2

6.00 N( ) = 13.5 N

*P13.20 v =

Tµ

=1 350 kg ⋅m s2

5.00 × 10−3 kg m= 520 m s

ANS. FIG. P13.18

Chapter 13 555


ANS. FIG. P13.23

Section 13.4 Reflection and Transmission P13.21 (a) If the end is fixed, there is inversion of the pulse upon reflection.

Thus, when they meet, they cancel and the amplitude is zero.

(b) If the end is free, there is no inversion on reflection. When they meet, the amplitude is 2A = 2 0.150 m( ) = 0.300 m.

Section 13.5 Rate of Energy Transfer by Sinusoidal Waves on Strings

P13.22 Originally,

P0 =12µω2A2v

P0 =12µω2A2 T

µ

P0 =12ω2A2 Tµ

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

P = 1

2ω2A2 T 2µ( ) = 2 1

2ω2A2 Tµ( ) = 2P0

P13.23 µ = 30.0 g m = 30.0× 10−3 kg m

λ = 1.50 m

f = 50.0 Hz: ω = 2π f = 314 s−1

2A = 0.150 m: A = 7.50 × 10−2 m

(a) y = A sin

2πλ

x −ωt⎛⎝⎜

⎞⎠⎟

y = 0.075( ) sin 4.19x − 314t( )

(b) P = 1

2µω2A2v = 1

230.0× 10−3( ) 314( )2 7.50× 10−2( )2 314

4.19⎛⎝⎜

⎞⎠⎟ W

P = 625 W



P13.24 f =

vλ=

30.0 m/s0.500 s

= 60.0 Hz ω = 2π f = 120π rad s

P =

12µω 2A2v =

12

0.1803.60

⎛⎝⎜

⎞⎠⎟ 120π( )2 0.100( )2 30.0( ) W = 1.07 kW

P13.25 Comparing the given wave function, y = (0.15) sin (0.80x − 50t), with the general wave function, y = A sin (kx − ωt), we have k = 0.80 rad/m, ω = 50 rad/s, and A = 0.15 m.

(a) v = fλ = ω

2π2πk

= ωk= 50.0

0.800 m s = 62.5 m s

(b) λ = 2π

k= 2π

0.800 m = 7.85 m

(c) f = 50.0

2π= 7.96 Hz

(d) P = 1

2µω2A2v = 1

212.0× 10−3( ) 50.0( )2 0.150( )2 62.5( ) W = 21.1 W

P13.26 A = 5.00× 10−2 m µ = 4.00× 10−2 kg m

P = 300 W T = 100 N

Therefore, v =

Tµ

= 50.0 m/s.

P =12µω 2A2v:

ω 2 =2P

µA2v=

2 300 W( )4.00 × 10−2 kg/m( ) 5.00 × 10−2 m( )2

50.0 m/s( )

ω = 346 rad s

f = ω2π

= 55.1 Hz

Section 13.6 Sound Waves P13.27 The sound pulse must travel 150 m before reflection and 150 m after

reflection. We have d = vt:

t = d

v= 300 m

1 533 m s= 0.196 s

Chapter 13 557


P13.28 (a) Since vlight >> vsound, and assuming that the speed of sound is constant through the air between the lightning strike and the observer, we have

d ≈ 343 m s( ) 16.2 s( ) = 5.56 km

(b)

No, we do not need to know the value of the speed of light.The speed of light is much greater than the speed of sound,so the time interval required for the light to reach you isnegligible compared to the time interval for the sound.

P13.29 (a) λ = v

f= 343 m/s

1 480 s−1 = 0.232 m

(b) ′λ = v

′f= 343 m/s

1 397 s−1 = 0.246 m ,

Δλ = ′λ − λ = 13.8 mm

P13.30 λ = v

f= 340 m/s

60.0 × 103 s−1 = 5.67 mm

P13.31 We write the pressure variation as ∆P = ∆Pmax sin (kx − ω t). Note that

k = 2π

λ= 2π

0.100 m( )= 62.8 m−1

and ω =

2π vλ

=2π 343 m s( )

0.100 m( ) = 2.16 × 104 s−1.

Therefore,

ΔP = 0.200 sin 62.8x − 2.16 × 104t⎡⎣ ⎤⎦

where ∆P is in Pa, x is in meters, and t is in seconds.

P13.32 The sound speed is

v = 331 m/s + 0.600 m/s ⋅ °C( ) 26.0°C( ) = 347 m s

(a) Let t represent the time for the echo to return. Then

d = 1

2vt = 1

2347 m/s ( ) 24.0 × 10−3 s( ) = 4.16 m

(b) Let Δt represent the duration of the pulse:

Δt = 10λ

v= 10λ

fλ= 10

f= 10

22.0 × 106 s−1 = 0.455 µs



(c) L = 10λ = 10v

f= 10 347 m s( )

22.0 × 106 s–1 = 0.158 mm

P13.33 (a) If f = 2.40 MHz, λ =

vf=

1 500 m/s2.40 × 106 s−1 = 0.625 mm

(b) If f = 1.00 MHz, λ =

vf=

1 500 m/s106 s−1 = 1.50 mm

If f = 20.0 MHz, λ =

1 500 m/s2 × 107 s−1 = 75.0 µm

P13.34 We use ΔPmax = ρvω smax = ρv

2π vλ

⎛⎝⎜

⎞⎠⎟ smax:

λmin =

2πρv2smax

ΔPmax

=2π 1.20 kg/m3( ) 343 m/s( )2 5.50 × 10−6 m( )

0.840 Pa= 5.81 m

P13.35 (a) A = 2.00 µm

(b) λ = 2π

15.7= 0.400 m = 40.0 cm

(c) v = ω

k= 858

15.7= 54.6 m s

(d) s = 2.00 cos 15.7( ) 0.050 0( ) − 858( ) 3.00 × 10−3( )⎡⎣ ⎤⎦ = −0.433 µm

(e) vmax = Aω = 2.00 µm( ) 858 s−1( ) = 1.72 mm s

P13.36 ΔPmax = ρω vsmax = 1.20 kg m3( ) 2π 2 000 s−1( )[ ] 343 m s( ) 2.00 × 10−8 m( )

ΔPmax = 0.103 Pa

*P13.37 ΔPmax = ρvω smax

smax =ΔPmax

ρvω=

4.00 × 10−3 N m2( )1.20 kg m3( ) 343 m s( ) 2π( ) 10.0 × 103 s−1( )

= 1.55 × 10−10 m

*P13.38 It is easiest to solve part (b) first:

(b) The distance the sound travels to the plane is

ds = h2 + h

2⎛⎝⎜

⎞⎠⎟

2

= h 52

Chapter 13 559


The sound travels this distance in 2.00 s, so

ds =

h 52

= 343 m s( ) 2.00 s( ) = 686 m

giving the altitude of the plane as h =

2 686 m( )5

= 614 m

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

v 2.00 s( ) = h

2= 307 m

Thus, the speed of the plane is:

v = 307 m

2.00 s= 153 m s

Section 13.7 The Doppler Effect

P13.39 (a) ′f =

f v + vo( )v − vs( )

′f = 2 500

343+ 25.0( )343 − 40.0( )

= 3.04 kHz

(b) ′f = 2 500

343+ −25.0( )343 − (−40.0)

⎛

⎝⎜

⎞

⎠⎟ = 2.08 kHz

(c) ′f = 2 500

343+ −25.0( )343 − 40.0

⎛

⎝⎜

⎞

⎠⎟ = 2.62 kHz while police car overtakes

′f = 2 500 343+ 25.0

343 − −40.0( )⎛

⎝⎜⎜

⎞

⎠⎟⎟ = 2.40 kHz after police car passes

P13.40 (a) ω = 2π f = 2π 115 min−1

60.0 s min⎛⎝⎜

⎞⎠⎟= 12.0 rad/s

vmax =ωA = 12.0 rad/s( ) 1.80 × 10−3 m( ) = 0.021 7 m/s



(b) The heart wall is a moving observer:

Δ ′f = ′f − f = f v + vO

v− f

⎛⎝⎜

⎞⎠⎟− f = f vO

v⎛⎝⎜

⎞⎠⎟

= 2 000 000 Hz( ) 0.021 71 500

⎛

⎝⎜

⎞

⎠⎟ = 28.9 Hz

(c) Now, the heart wall is a moving source:

Δf ″ = f ′v

v − vs

⎛

⎝⎜⎞

⎠⎟− f = f

v + vo

v⎛⎝⎜

⎞⎠⎟

vv − vs

⎛

⎝⎜⎞

⎠⎟− f

Δf ″ = fv v + vo( )v v − vs( )

⎛

⎝⎜

⎞

⎠⎟ −

v v − vs( )v v − vs( )

⎛

⎝⎜

⎞

⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥= f

vo + vs

v − vs

⎡

⎣⎢

⎤

⎦⎥

Since the velocities of the source and the observer in these expressions are both referring to the movement of the heart wall, and the velocity of the sound wave is much greater than those velocities, we may approximate:

Δf ″ ≅ f2vs

v⎡⎣⎢

⎤⎦⎥

Δf ″ ≅ 2.00 × 106 Hz( ) 0.043 41 500

⎡⎣⎢

⎤⎦⎥= 57.9 Hz

P13.41 Let va represent the magnitude of the velocity of the ambulance.

As it approaches you hear the frequency ′f = v v

v – va

⎛⎝⎜

⎞⎠⎟ f = 560 Hz.

The negative sign appears because the source is moving toward the observer. The opposite sign with source velocity magnitude describes the ambulance moving away. As the ambulance recedes, the Doppler-shifted frequency is

′′f = v

v + va

⎛⎝⎜

⎞⎠⎟ f = 480 Hz .

Solving the second of these equations for f and substituting into the other gives

′f = v

v – va

⎛⎝⎜

⎞⎠⎟

v + va

v⎛⎝⎜

⎞⎠⎟ ′′f or ′f v − ′f va = v ′′f + va ′′f

so the speed of the source is

va =

v ′f – ′′f( )′f + ′′f =

(343 m/s)(560 Hz – 480 Hz)560 Hz + 480 Hz = 26.4 m/s

Chapter 13 561


P13.42 The maximum speed of the speaker is described by

12

mvmax2 = 1

2kA2

vmax =km

A = 20.0 N/m5.00 kg

0.500 m( ) = 1.00 m s

The frequencies heard by the stationary observer range from

′fmin = f

vv + vmax

⎛⎝⎜

⎞⎠⎟

to ′fmax = f

vv − vmax

⎛⎝⎜

⎞⎠⎟

where v is the speed of sound.

′fmin = 440 Hz343 m/s

343 m/s + 1.00 m s⎛⎝⎜

⎞⎠⎟= 439 Hz

′fmax = 440 Hz343 m s

343 m/s − 1.00 m/s⎛⎝

⎞⎠ = 441 Hz

P13.43 (a) Sound moves upwind with speed (343–15) m/s = 328 m/s. Crests pass a stationary upwind point at frequency 900 Hz. Then

λ = v

f= 328 m/s

900 s−1 = 0.364 m

(b) By similar logic, λ = v

f= 343 + 15( ) m/s

900 s−1 = 0.398 m

(c) The source is moving through the air at 15 m/s toward the observer. The observer is stationary relative to the air.

′f = f

v + vo

v − vs

⎛⎝⎜

⎞⎠⎟= 900 Hz( ) 343 m/s + 0

343 m/s − 15.0 m/s⎛⎝

⎞⎠ = 941 Hz

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

′f = fv + vo

v − vs

⎛⎝⎜

⎞⎠⎟= 900 Hz( ) 343 m/s + 30.0 m/s

343 m/s − −15.0 m/s( )⎛⎝⎜

⎞⎠⎟

= 900 Hz( ) 373 m/s358 m/s

⎛⎝

⎞⎠ = 938 Hz



P13.44 The apparent frequency drops because of the Doppler effect. Using a T subscript for the situation when the athlete moves toward the horn, and A for movement away from the horn, we have,

′fA

′fT

=

v + vOA

v − vS

⎛

⎝⎜

⎞

⎠⎟ f

v + vOT

v − vS

⎛

⎝⎜

⎞

⎠⎟ f

= v + vOA

v + vOT

= v + −vO( )v + +vO( )

= v − vO

v + vO

where v0 is the constant speed of the athlete. Setting this ratio equal to 5/6, we have

56 = v − vO

v + vO

→ 5v + 5vO = 6v − 6vO → 11vO = v

Solving for the speed of the athlete,

vO = v

11 = 343 m/s

11 = 31.2 m/s

This is much faster than a human athlete can run.

P13.45 We first determine how fast the tuning fork is falling to emit sound with apparent frequency 485 Hz. Call the magnitude of its velocity vfall. The tuning fork source is moving away from the listener, so vs = –vfall.

Therefore, we use the equation ′f = v

v + vfall

⎛⎝⎜

⎞⎠⎟

f

Solving for vfall gives

v + vfall

v=

ff '

and vfall = vff '− 1

⎛⎝⎜

⎞⎠⎟

.

Substituting, we have vfall =

512 Hz485 Hz

− 1⎛⎝⎜

⎞⎠⎟ 343 m/s( ) = 19.1 m/s .

The time interval required for the tuning fork to reach this speed, from the particle under constant acceleration model, is given by

vy = 0 + ayt as t = vy/ay = (19.1 m/s)/(9.80 m/s2 ) = 1.95 s

The distance that the fork has fallen is

Δy = 0 +

12

ayt2 =

12

(9.80 m/s2 )(1.95 s)2 = 18.6 m

At this moment, the fork would appear to ring at 485 Hz to a stationary observer just above the fork. However, some additional time is required for the waves to reach the point of release. The fork is moving down, but the sound it radiates still travels away from its

Chapter 13 563


instantaneous position at 343 m/s. From the traveling wave model, the time interval it takes to return to the listener is

Δt = Δy/v = 18.6 m/(343 m/s) = 0.054 2 s

Over a total time t + ∆t = 1.95 s + 0.054 2 s = 2.00 s, the fork falls a total distance

dtotal =

12

gttotal fall2 = 19.7 m

*P13.46 (a) Equation 13.31, ′f = f v + vo

v − vs

⎛

⎝⎜

⎞

⎠⎟ , applies to an observer on B

because B is receiving sound from source A.

(b) The sign of vs should be positive because the source is moving

toward the observer, resulting in an increase in frequency.

(c) The sign of vo should be negative because the observer is

moving away from the source, resulting in a decrease in frequency.

(d) The speed of sound should be that of the medium of seawater,

1 533 m/s .

(e)

fo= fsv + vo

v − vs

⎛⎝⎜

⎞⎠⎟= 5.27 × 103 Hz( ) 1 533 m/s( ) + −3.00 m/s( )

1 533 m/s( ) − +11.0 m/s( )⎡

⎣⎢

⎤

⎦⎥

= 5.30 × 103 Hz

Section 13.3 Context Connection: Seismic Waves P13.47 The distance the waves have traveled is d = (7.80 km/s)t =

(4.50 km/s)(t + 17.3 s), where t is the travel time for the faster wave.

Then, 7.80 − 4.50( ) km s( )t = 4.50 km s( ) 17.3 s( )

or t =

4.50 km s( ) 17.3 s( )7.80 − 4.50( ) km s

= 23.6 s

and the distance is d = 7.80 km s( ) 23.6 s( ) = 184 km

P13.48 (a) The

longitudinal P wave travels a shorter distance and is

moving faster, so it will arrive at point B first.

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



a distance of 2Rsin 30.0° = 2 6.37 × 106 m( )sin 30.0° = 6.37 × 106 m

at a speed of 7 800 m/s.

Therefore, it takes ΔtP =

6.37 × 106 m7 800 m/s

817 s.

The Rayleigh wave that travels along the Earth’s surface must travel a distance of

s = Rθ = R

π3

rad⎛⎝⎜

⎞⎠⎟ = 6.67 × 106 m

at a speed of 4 500 m/s.

Therefore, it takes ΔtS =

6.67 × 106 m4 500 m/s

1 482 s.

The time difference is ΔT = ΔtS − ΔtP = 666 s = 11.1 min.

Additional Problems P13.49 Let M = mass of block, m = mass of string. For the block, ∑F = ma

implies T = mvb

2

r= mω2r. The speed of a wave on the string is then

v =

Tµ

=Mω 2rm/r

= rω Mm

the travel time of the wave on the string is given by

Δt =

rv=

1ω

mM

and the angle through which the block rotates is

Δθ =ω Δt =

mM

=0.003 2 kg0.450 kg

= 0.084 3 rad

P13.50 Assuming the incline to be frictionless and taking the positive x direction to be up the incline:

∑Fx = T − Mg sin θ = 0

or the tension in the string is T = Mg sin θ.

The speed of transverse waves in the string is then

Chapter 13 565


v =

Tµ

=Mg sin θ

m/L=

MgL sin θm

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

Δt =Lv= L

mMgL sin θ

=mL

Mg sin θ

*P13.51 The block-cord-Earth system is isolated, so energy is conserved as the block moves down distance x:

ΔK +ΔU = 0 →

K +Ug +Us( )top= K +Ug +Us( )bottom

0+ Mgx + 0+ 0 = 0+ 0+ 12

kx2

x = 2Mgk

(a) T = kx = 2Mg = 2 2.00 kg( ) 9.80 m s2( ) = 39.2 N

(b) L = L0 + x = L0 +

2Mgk

L = 0.500 m+ 39.2 N

100 N m= 0.892 m

(c) v = T

µ= TL

m

v =39.2 N × 0.892 m

5.0 × 10−3 kg

v = 83.6 m/s

P13.52 The block-cord-Earth system is isolated, so energy is conserved as the block moves down distance x:

ΔK + ΔU = 0→

K +Ug +Us( )top

= K +Ug +Us( )bottom

0 + Mgx + 0 + 0 = 0 + 0 +12

kx2

Mgx = 1

2kx2



(a) T = kx = 2Mg

(b) L = L0 + x = L0 +

2Mgk

(c)

v =

Tµ

=TLm

=2Mg

mL0 +

2Mgk

⎛⎝⎜

⎞⎠⎟

P13.53 (a) λ = v

f= 343 m s

1 000 s−1 = 0.343 m

(b) ′λ = v

′f= v

fv − vS

v⎛⎝⎜

⎞⎠⎟ =

343 − 40.0( ) m s1 000 s−1 = 0.303 m

(c) ′′λ = v

′′f= v

fv + vS

v⎛⎝⎜

⎞⎠⎟ =

343+ 40.0( ) m s1 000 s−1 = 0.383 m

(d) ′f = f

v − vO

v − vS

⎛⎝⎜

⎞⎠⎟= 1 000 Hz( ) 343 − 30.0( ) m s

343 − 40.0( ) m s= 1.03 kHz

P13.54 Assume a typical distance between adjacent people ∼ 1 m.

Then the wave speed is v =

ΔxΔt

~1 m0.1 s

~ 10 m/s.

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

T = 2π r

v~

2π 102( )10 m s

= 63 s ~ 1 min

P13.55 (a) P x( ) = 1

2µω2A2v = 1

2µω2A0

2e−2bx ωk

⎛⎝⎜

⎞⎠⎟ =

µω3

2kA0

2e−2bx

(b) P 0( ) = µω3

2kA0

2

(c)

P x( )P 0( )

= e−2bx

P13.56 (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

Chapter 13 567


swallows it up, F∑ = ma becomes kdx = adm or

kdmdx

= a. But

dmdx

= µ so a = k

µ. Also,

a = dv

dt= v

t when vi = 0. But L = vt, so

a = v2

L. Equating the two expressions for a, we have

kµ= v2

L or

v = kLµ

(b) Using the expression from part (a):

v = kL

µ= kL2

m= 100 N/m( ) 2.00 m( )2

0.400 kg= 31.6 m/s

P13.57 Sound takes this time to reach the man: Δts =

d − hv

. The minimum time

interval between when a warning is shouted and when the man responds to the warning is Δtmin = Δts +Δt .

Since the whole time interval to fall is given by

Δy = d − h( ) = 1

2gΔt f

2 → Δt f =

2 d − h( )g

The warning needs to come at least

ΔT = Δt f − Δt − Δts =

2 d − h( )g

− Δt − d − h

v

into the fall, when the pot is at the position

y f = yi + vyiΔT −12

gΔT 2

y f = 20.0 m −12

9.80 m/s2( )

×2 20.0 m − 1.75 m( )

g− 0.300 s −

20.0 m − 1.75 m343 m/s

⎛

⎝⎜

⎞

⎠⎟

2

y f = 7.82 m above the ground.

*P13.58 Sound takes this time to reach the man: Δts =

d − hv

. The minimum

time interval between when a warning is shouted and when the man responds to the warning is Δtmin = Δts + Δt.



Since the whole time interval to fall is given by

Δy = d − h( ) = 1

2gΔt f

2 → Δt f =

2 d − h( )g

The warning needs to come at least

ΔT = Δt f − Δt − Δts =

2 d − h( )g

− Δt − d − h

v

into the fall, when the pot is at the position

y f = yi + vyiΔT −12

gΔT 2

y f = d −12

g2 d − h( )

g− Δt −

d − hv

⎛

⎝⎜⎜

⎞

⎠⎟⎟

2

above the ground.

P13.59 The trucks form a train analogous to a wave train of crests with speed

v = 19.7 m/s and unshifted frequency f =

23.00 min

= 0.667 min−1.

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

′f = fv + vo

v⎛⎝⎜

⎞⎠⎟ = 0.667 min−1( ) 19.7 + −4.47( )

19.7⎛⎝⎜

⎞⎠⎟

= 0.515 min

(b) ′′f = f

v + ′vo

v⎛⎝⎜

⎞⎠⎟ = 0.667 min−1( ) 19.7 + −1.56( )

19.7⎛⎝⎜

⎞⎠⎟= 0.614 min

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

P13.60 Imagine a short transverse pulse traveling from the bottom to the top of the rope. When the pulse is at position x above the lower end of the

rope, the wave speed of the pulse is given by v = T

µ, where T = µxg

is the tension required to support the weight of the rope below position x.

Therefore, v = gx.

But v = dx

dt, so that

dt = dx

gx.

Chapter 13 569


and t =

dx

gx0

L

∫ =1

g

x12 0

L

≈ 2Lg

P13.61 v = 2d

t gives

d = vt

2= 1

26.50 × 103 m s( ) 1.85 s( ) = 6.01 km

P13.62 Refer to Problem 60. At distance x from the bottom, the tension is

T =

mxgL

⎛⎝⎜

⎞⎠⎟+ Mg, so the wave speed is:

v =Tµ

=TLm

= xg +MgL

m⎛⎝⎜

⎞⎠⎟=

dxdt

→ dt =dx

xg + MgLm

⎛⎝⎜

⎞⎠⎟

(a) Then

t = dt

0

t

∫ = xg +MgL

m⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

−1 2

dx0

L

∫

gives

t =1g

xg + MgL m( )⎡⎣ ⎤⎦1 2

12

x=0

x=L

t =

2g

Lg +MgL

m⎛⎝⎜

⎞⎠⎟

1 2

−MgL

m⎛⎝⎜

⎞⎠⎟

1 2⎡

⎣⎢

⎤

⎦⎥

t = 2L

mgM + m − M( )

(b) When M = 0,

t = 2 L

gm − 0

m

⎛

⎝⎜

⎞

⎠⎟ = 2 L

g

(c) As m → 0 we expand

M + m = M 1+mM

⎛⎝⎜

⎞⎠⎟

1 2

= M 1+

12

mM

−18

m2

M2 +…⎛

⎝⎜⎞

⎠⎟



to obtain t = 2

Lmg

M + 12

mM

⎛⎝⎜

⎞⎠⎟− 1

8m2 M3 2( ) +…− M

⎛⎝⎜

⎞⎠⎟

t ≈ 2

Lg

12

mM

⎛⎝⎜

⎞⎠⎟= mL

Mg

where we neglect terms

18

m2

M3 2

⎛⎝⎜

⎞⎠⎟

and higher because terms with

m2 and higher powers are very small.

P13.63 (a) ′f = f v

v − vdiver( )

so 1− vdiver

v= f

′f⇒ vdiver = v 1− f

′f

⎛

⎝⎜

⎞

⎠⎟

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

we find

vdiver = 343 m/s( ) 1− 1 800 Hz

2 150 Hz⎛⎝⎜

⎞⎠⎟ = 55.8 m/s

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

′′f = ′f

v + vdiver( )v

⇒ ′′f = f vv − vdiver( )

⎡

⎣⎢

⎤

⎦⎥

v + vdiver( )v

so ′′f = 1 800 Hz( ) 343 m/s + 55.8 m/s( )

343 m/s − 55.8 m/s( ) = 2 500 Hz

P13.64 v =

4 450 × 103 m5.88 h

1 h3 600 s

⎛⎝⎜

⎞⎠⎟ = 210 m/s

davg =

v2

g=

210 m/s( )2

9.80 m/s2 = 4 500 m

This is greater than the average ocean depth, about 4 280 m.

*P13.65 The transverse wave velocity in the string is vtrans =

Tµ

,

where T is the tension in the cord, and µ is the mass per unit length of the cord. The tension T is generated by the centripetal force holding the mass and cord in uniform circular motion at the angular velocity ω,

Chapter 13 571


thus:

T = Fc = M v2

r= Mω2r

where we note that M is the mass of the block.

The mass density of the cord is µ = m

r; thus, the transverse wave

velocity is

vtrans =Tµ

=Mω 2r( )

mr

⎛⎝⎜

⎞⎠⎟

=Mω 2r2( )

m( ) =ωrMm

Now the transverse wave travels a distance r (the length of the cord) at a uniform velocity vtrans , thus, distance = r = vtrans

t, and therefore,

t =r

vtrans

=r

ωrMm

⎛

⎝⎜

⎞

⎠⎟

=1ω

mM

which we may solve numerically:

t =

1ω

mM

=1

10.0 rad/s( )0.00 320 kg

0.450 kg= 8.43 × 10−3 s

[See Note to P13.66.]

*P13.66 The transverse wave velocity in the string is vtrans =

Tµ

,

where T is the tension in the cord, and µ is the mass per unit length of the cord. The tension T is generated by the centripetal force holding the mass and cord in uniform circular motion at the angular velocity, ω, thus:

T = Fc = M v2

r= Mω2r

where we note that M is the mass of the block, and the mass density of

the cord is µ =

mr

.

Thus transverse wave velocity is



vtrans =Tµ

=Mω 2r( )

mr

⎛⎝⎜

⎞⎠⎟

=Mω 2r2( )

m( ) =ωrMm

Now the transverse wave travels a distance r (the length of the cord) at a uniform velocity vtrans ; thus, distance = r = vtranst, and therefore,

t =r

vtrans

=r

ωrMm

⎛

⎝⎜

⎞

⎠⎟

=1ω

mM

[Note: To solve this problem without integration of the mass density µ over the length of the cord to include the cord’s own mass as a contribution to its own tension, and thus to a nonuniform tension along the length of the cord (and thus also to a nonuniform wave velocity along the cord), we must assume that the mass of the cord m is very small compared to the mass of the block M. In such a case, the mass of the cord does not contribute to the centripetal force, or as a result, to the tension on the cord itself. The only role the cord’s mass will then play is in generating the linear density in the transverse wave velocity equation. To be forced to include mass of the cord in the centripetal force calculation is a significantly more difficult problem and is not attempted here.]

P13.67 (a) If the velocity of the insect is vx ,

40.4 kHz = 40.0 kHz( ) 343 m/s + 5.00 m/s( ) 343 m/s − vx( )

343 m/s − 5.00 m/s( ) 343 m/s + vx( )

Solving, vx = 3.29 m/s .

(b) Therefore,

the bat is gaining on its prey at 1.71 m s .

P13.68 Since cos2θ + sin2θ = 1 , sinθ = ± 1− cos2θ (each sign applying half the time),

ΔP = ΔPmax sin kx −ωt( ) = ±ρvω smax 1− cos2 kx −ωt( )

Therefore,

ΔP = ± ρvω smax2 − smax

2 cos2 kx −ωt( ) = ± ρvω smax2 − s2

P13.69 (a) µ x( ) is a linear function, so it is of the form µ x( ) = mx + b

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

Chapter 13 573


so m =

µL − µ0

L.

Then µ x( ) = µL − µ0( )x

L+ µ0 .

(b) Imagine the crest of a short transverse pulse traveling from one end of the string to the other. Consider the pulse to be at position

x. From v =

dxdt

, the time interval required to move from x to x +

dx is

dxv

. The time interval required to move from 0 to L is

Δt =dxv

0

L

∫ =dx

T /µ0

L

∫ =1T

µ x( ) dx0

L

∫Δt =

1T

µL − µ0( )xL

+ µ0

⎛⎝⎜

⎞⎠⎟

1 2µL − µ0

L⎛⎝⎜

⎞⎠⎟ dx

LµL − µ0

⎛⎝⎜

⎞⎠⎟

0

L

∫Δt =

1T

LµL − µ0

⎛⎝⎜

⎞⎠⎟

µL − µ0( )xL

+ µ0

⎛⎝⎜

⎞⎠⎟

3 2132( )

0

L

Δt =2L

3 T µL − µ0( ) µL3 2 − µ0

3 2( )

P13.70 (a) We have ′f = fv

v − u and

′′f = fv

v − −u( ). We then have

′f − ′′f = fv

1v − u

− 1v + u( )

Δf = fv v + u − v + u( )

v2 − u2 = 2uvfv2 1− u2

v2( ) =2 u

v

1− u2

v2

f

(b) 130 km/h = 36.1 m/s

Δf = 2 36.1 m/s( ) 400 Hz( )

340 m/s( ) 1− 36.1 m/s( )2

340 m/s( )2⎡⎣ ⎤⎦

= 85.9 Hz

P13.71 (a) If vO = 0 m/s, then ′f =

vv − vS cos θS

f .

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



cosθS =

45

so ′f =

343 m/s343 m/s − 0.800 25.0 m/s( ) 500 Hz( ) ,

or ′f = 531 Hz .

(b) 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

′fmax =v

v − vS cos 0°f =

343 m/s343 m/s − 25.0 m/s

500 Hz( )

= 539 Hz

to

′fmin =v

v − vS cos 180°f =

343 m/s343 m/s + 25.0 m/s

500 Hz( )

= 466 Hz

(c) Now vO = +40.0 m/s, and the train is 40.0 m from the intersection, and the car is 30.0 m from the intersection, so

cos θO =

35

′f =

343 m/s + 0.600(40.0 m/s)343 m/s − 0.800(25.0 m/s)

(500 Hz) = 568 Hz

Chapter 13 575


ANSWERS TO EVEN-NUMBERED PROBLEMS P13.2 (a) See ANS. FIG P13.2(a); (b) See ANS. FIG P13.2(b); (c) 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; (d) The wave has traveled d = vt = 2.40 m to the right.

P13.4 See P13.4 for full solution.

P13.6 0.800 m/s

P13.8 (a) 2.00 cm; (b) 2.98 m; (c) 0.576 Hz; (d) 1.72 m/s

P13.10 (a) –1.51 m/s; (b) 0; (c) 16.0 m (d) 0.500 s; (e) 32.0 m/s

P13.12 (a) See ANS FIG P13.12(a); (b) 18.0 rad/m; (c) 0.083 3 s; (d) 75.4 rad/s; (e) 4.20 m/s; (f) y = 0.200 m( ) sin 18.0x / m + 75.4t / s + φ( ) ; (g) y(x, t) = 0.200 sin (18.0x + 75.4t − 0.151), where x and y are in meters and t is in seconds.

P13.14 (a) 0.021 5 m; (b) 1.95 rad; (c) 5.41 m/s; (d) y x, t( ) = 0.021 5( ) sin 8.38x + 80.0πt + 1.95( )

P13.16 The calculated gravitational acceleration of the Moon is almost twice that of the accepted value.

P13.18 (a) v = 30.4( ) m where v is in meters per second and m is in kilograms;

(b) m = 3.89 kg

P13.20 520 m/s

P13.22 2P0

P13.24 1.07 kW

P13.26 55.1 Hz

P13.28 (a) 5.56 km; (b) No, we do not need to know the value of the speed of light. The speed of light is much greater than the speed of sound, so the time interval required for the light to reach you is negligible compared to the time interval for the sound.

P13.30 5.67 mm

P13.32 (a) 4.16 m; (b) 0.455 μs; (c) 0.158 mm

P13.34 5.81 m

P13.36 0.103 Pa

P13.38 (a) 153 m/s; (b) 614 m

P13.40 (a) 0.021 7 m/s (b) 28.9 Hz; (c) 57.9 Hz



P13.42 (a) fmin = 439 Hz, fmax = 441 Hz

P13.44 This is much faster than a human athlete can run.

P13.46 (a) B; (b) positive; (c) negative; (d) 1 533 m/s; (e) 5.30 × 103 Hz

P13.48 (a) longitudinal P wave; (b) 666 s

P13.50

mLMg sin θ

P13.52 (a) 2 Mg; (b) L0 +

2 Mgk

; (c)

2 Mgk

L0 +2 Mg

k⎛⎝⎜

⎞⎠⎟

P13.54 ~1 min

P13.56 (a)

kLµ

; (b) 31.6 m/s

P13.58 d −

12

g2 d − h( )

g− Δt −

d − hv

⎛

⎝⎜

⎞

⎠⎟

2

above the ground

P13.60 2 L

g

P13.62 (a) t = 2 L

gM +m − M( ) ; (b)

2 L

g; (c)

mLMg

P13.64 This is greater than the average ocean depth, about 4 280 m.

P13.66

1ω

mM

P13.68 See P13.68 for complete solution.

P13.70 (a)

2 uv

1− u2

v2

f ; (b) 85.9 Hz

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