20
TRAVELING WAVES
Conceptual Questions
20.1. va  vb  vc . Wave speed is independent of wave amplitude.
20.2. (a) v 
(c) v 
T
 /4
T

 2v
v 
2T

v  280 cm/s (b) v 
 2v
v   cm/s (d)   
T
v

4 2
v  100 cm/s
4m
  so the speed is unchanged: v  200 cm/s.
4L
20.3. The constant 1 mm displacement interval appears at later times. Therefore, the displacement at this point
reached 2 mm before it settled at 1 mm.
20.4. (a) The wave is traveling to the right. It reaches the 2 cm point at later times, 7 starting from the left.
(b) 200 cm/s. The leading edge reaches the 2 cm point at t  0.03 s. At t  0.01s it was 4 cm to the left, at 2 cm.
v
x
4 cm

 200 cm/s
t (0.03 s  0.01s)
20.5. a  b  c because v  f  constant for these frequencies, so large f implies small .
20.6.  1  2 0 f1  f0 . The frequency is unchanged because that is the frequency of the driving force that moves
successive oscillators. But v   f , so if the speed is doubled, so is the wavelength.
20.7. The amplitude of the wave is the maximum displacement, which is 4.0 cm. The wavelength is the distance
between two consecutive peaks, which gives   14 m  2 m  12 m. The frequency of the wave is
f 
v


24 m/s
 2.0 Hz
12 m
We solve for  0 from the initial conditions at x  0 m and t  0 s: 2  4sin(0   0 ) 
1

 sin( 0 )   0  .
2
6
20.8. The amplitude of the wave is the maximum displacement, which is 1.0 cm. The wavelength is the distance
between two consecutive peaks, which gives   2.0 m. The frequency of the wave is
f 
v


1.0 m/s
 0.50 Hz
2.0 m
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20-1
20-2
Chapter 20
 
1
  2
 0 
We solve for  0 from the initial conditions at x  0 m and t  1s: 1  1sin  2  0    0   
2
2
2

 


0   .
2
20.9. (a) 0; they are on the same wave front. (b) 4 rad; because they are two wave crests apart. (c)  rad; because F
is on a crest and E on an adjacent trough.
20.10. PC  PB  PA . Use P 
E
in each case.
t
PA 
2J
1W
2s
PB 
10 J
2W
5s
PC 
2J
2W
1s
20.11. For one professor I  (1.0  1012 W/m2 )  1052  1.5849  107 W/m2 ; for 100 professors I  1.5849 
105 W/m2. The new sound intensity level is
 1.5849  105 W/m2 
 72 dB
 1.0  1012 W/m2 


  (10 dB)log10 
Notice that for 10 times the intensity, the sound intensity level goes up by 10 dB; for 100 times the intensity, the
sound intensity level goes up by 20 dB.
20.12. The correct answer is D because the initial shift is to a lower frequency, which means the source is initially moving
away from you until t  2 s. Then the shift is to a higher frequency, which means the source is moving toward you.
Exercises and Problems
Section 20.1 The Wave Model
20.1. Model: The wave is a traveling wave on a stretched string.
Solve: The wave speed on a stretched string with linear density  is
vstring 
TS

 150 m/s 
75 N

   3.333  103 kg/m
For a wave speed of 180 m/s, the required tension will be
2
TS   vstring
 (3.333 103 kg/m)(180 m/s)2  110 N
20.2. Model: The wave is a traveling wave on a stretched string.
Solve: The wave speed on a stretched string with linear density  is vstring  TS/ . The wave speed if the tension is
doubled will be
vstring


TS
1
1

vstring 
(200 m/s)  141 m/s
2
2
2
20.3. Solve:
L  v t 
TS

L
t 
TS
T L
T
t  S t  L  S t 
m/L
m
m
TS
20 N
( t )2 
(50 ms)2  2.0 m
m
0.025 kg
Assess: 2.0 m seems like a reasonable length for a string.
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Traveling Waves
20-3
Section 20.2 One-Dimensional Waves
20.4. Model: This is a wave traveling at constant speed. The pulse moves 1 m to the right every second.
Visualize: The snapshot graph shows the wave at all points on the x-axis at t  0 s. The leading edge of the wave
will reach x  5.0 m1s later, at t  1s. The first part of the wave causes a sudden upward displacement at t  1s.
The flat top of the wave is 3 m wide, so it will take 3 s to pass the x  5.0 m point, keeping this point high until
t  4 s. The falling slope is 1 m wide, so it will take 1 s for the the displacement at x  5.0 m to drop back to zero.
The trailing edge of the pulse arrives at x  5.0 m at t  1s, which is 5 s after the figure given in the problem. The
displacement now becomes zero and stays zero for all later times.
20.5. Model: This is a wave traveling at constant speed. The pulse moves 1 m to the left every second.
Visualize: The snapshot graph shows the wave at all points on the x-axis at t  2 s The leading edge of the wave—
moving left—will reach x  0 m1s later, at t  3 s. The first part of the wave causes a sudden upward displacement
at t  3 s. The falling slope of the wave is 4 m wide, so it will take 4 s for the the displacement at x  0 m to
decrease from 1cm to  1cm. The trailing edge of the pulse arrives at x  0 m at t  7 s, which is 5 s after the
figure given in the problem. The displacement now becomes zero and stays zero for all later times.
20.6. Model: This is a wave traveling to the right at a constant speed of 1 m/s.
Visualize: This is the history graph of a wave at x  0 m. At t  1s, the time for which we’re to draw the snapshot
graph, the displacement at x  0 m is  1cm. The graph shows that the x  0 m point first “went high” at t  0 s,
and this leading edge will have moved 1 m to the right by the time of our snapshot graph at t  1s. The trailing edge
reaches x  0 m at t  6 s. That’s 5 s after our snapshot graph, so at t  1s the trailing edge must still be 5 m left of
x  0 m. Thus at t  1s the wave will stretch from x  5 to x  1 m.
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20-4
Chapter 20
20.7. Model: This is a wave traveling to the left at a constant speed of 1 m/s.
Visualize: This is the history graph of a wave at x  2 m. At t  0 s, the time for which we’re to draw the snapshot
graph, the displacement at x  2 m is 0 cm but is just beginning to rise. That is, the leading edge of a left-moving
pulse is just reaching x  2 m at t  0 s. The trailing edge reaches x  2 m at t  4 s. That’s 4 s after our snapshot
graph, so at t  0 s the trailing edge must still be 4 m right of x  2 m. Thus at t  0 s the wave will stretch from
x  2 m to x  6 m.
20.8. Visualize:
Figure EX20.8 shows a snapshot graph at t  0 s of a longitudinal wave. This diagram shows a row of particles with an
inter-particle separation of 1.0 cm at equilibrium. Because the longitudinal wave has a positive amplitude of 0.5 cm
between x  3 cm and x  8 cm, the particles at x  3, 4, 5, 6, 7, and 8 cm are displaced to the right by 0.5 cm.
20.9. Visualize:
We first draw the particles of the medium in the equilibrium positions, with an inter-particle spacing of 1.0 cm. Just
underneath, the positions of the particles as a longitudinal wave is passing through are shown at time t  0 s. It is
clear that relative to the equilibrium the particle positions are displaced negatively on the left side and positively on
the right side. For example, the particles at x  0 cm and x  1cm are at equilibrium, the particle at x  2 cm is
displaced left by 0.5 cm, the particle at x  3 cm is displaced left by 1.0 cm, the particle at x  4 cm is displaced left
by 0.5 cm, and the particle at x  5 cm is undisplaced. The behavior of particles for x  5 cm is opposite of that for
x  5 cm.
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Traveling Waves
20-5
Section 20.3 Sinusoidal Waves
20.10. Solve: (a) The wavelength is

2
2

 4.2 m
k 1.5 rad/m
(b) The frequency is
f 
v


200 m/s
 48 Hz
4.19 m

2
 3.1 rad/m
2.0 m
20.11. Solve: (a) The wave number is
k
2

(b) The wave speed is

 30 rad/s 
v   f      (2.0 m) 
 9.5 m/s
 2 
 2 
20.12. Model: The wave is a traveling wave.
Solve: (a) A comparison of the wave equation with Equation 20.14 yields: A  3.5 cm, k  . rad/m,   rad/s,
and  0  0 rad. The frequency is
f 
 124 rad/s

 19.7 Hz  20 Hz
2
2

2
2

 2.33 m  2.3 m
k
2.7 rad/m
(b) The wavelength is
(c) The wave speed v   f  46 m/s.
20.13. Model: The wave is a traveling wave.
Solve: (a) A comparison of the wave equation with Equation 20.14 yields: A  5.2 cm, k   rad/m,    rad/s,
and 0  0 rad The frequency is
f 

2

72 rad/s
 115 Hz  11 Hz
2
(b) The wavelength is

2
2

 114 m  11 m
k
55 rad/m
(c) The wave speed v   f  13 m/s.
20.14. Solve: The amplitude of the wave is the maximum displacement, which is 6.0 cm. The period of the wave is
0.60 s, so the frequency f  1/T  1/0.60 s  1.67 Hz. The wavelength is

v
2 m/s

 1.2 m
f 1.667 Hz
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20-6
Chapter 20
Section 20.4 Waves in Two and Three Dimensions
20.15. Solve: According to Equation 20.28, the phase difference between two points on a wave is
2
   2   1 
(r  r )
 2 1
If 1   rad at r1  4.0 m, we can determine 2 at any r value at the same instant using this equation. At r2  3.5 m,
2  1 
At r2  4.5 m,  
3

2
2

(r2  r1)   rad 
2

(35 m  40 m)  rad
20 m
2
rad.
20.16. Solve: According to Equation 20.28, the phase difference between two points on a wave is
r 2
2
  2  1  2

(r2  r1)  (3 rad  0 rad) 
(80 cm  20 cm)    40 cm



20.17. Visualize: A phase difference of 2 rad corresponds to a distance of  . Set up a ratio.
Solve:
 5.5 rad 
 5.5 rad  v  5.5 rad  340 m/s
5.5 rad
x=
= 2.5 m
 = 
 =

2 rad
2

rad


 2 rad  f  2 rad  120 Hz
Assess: 2.5 m seems like a reasonable distance.
x

=
20.18. Visualize:
Solve: (a) Because the same wavefront simultaneously reaches listeners at x  7.0 m and x  3.0 m,
  0 rad 
2

(r2  r1)  r2  r1
Thus, the source is at x  2.0 m, so that it is equidistant from the two listeners.
(b) The third person is also 5.0 m away from the source. Her y-coordinate is thus y  (5 m)2  (2 m)2  46 m.
Section 20.5 Sound and Light
20.19. Solve: Two pulses of sound are detected because one pulse travels through the metal to the microphone while
the other travels through the air to the microphone. The time interval for the sound pulse traveling through the air is
x 400 m
tair 

 001166 s  1166 ms
vair 343 m/s
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Traveling Waves
20-7
Sound travels faster through solids than gases, so the pulse traveling through the metal will reach the microphone before
the pulse traveling through the air. Because the pulses are separated in time by 9.00 ms, the pulse traveling through the
metal takes tmetal  2.66 ms to travel the 4.00 m to the microphone. Thus, the speed of sound in the metal is
x
400 m

 1504 m/s  1500 m/s
tmetal 000266 s
vmetal 
20.20. Solve: (a) In aluminum, the speed of sound is 6420 m/s. The wavelength is thus equal to
v
6420 m/s

 3.21 103 m  3.21 mm  3.2 mm
f 2.0  106 Hz
(b) The speed of an electromagnetic wave is c. The frequency would be

c
f 


3.0  108 m/s
3.21 103 m
 9.3  1010 Hz
20.21. Solve: (a) The frequency is
f 
c


3.0 108 m/s
 1.5 109 Hz  1.5 GHz
0.20 m
(b) The speed of a sound wave in water is vwater  1480 m/s. The wavelength of the sound wave would be

vwater
1480 m/s

 9.87  107 m  990 nm
9
f
1.50 10 Hz
20.22. Model: Light is an electromagnetic wave that travels with a speed of 3  108 m/s.
Solve: (a) The frequency of the blue light is
f blue 
c


3.0  108 m/s
450 109 m
 6.67  1014 Hz
(b) The frequency of the red light is
f red 
3.0  108 m/s
650 109 m
 4.62  1014 Hz
(c) Calculate the index of refraction,
material 
vacuum
n
n
vacuum 650 nm

 1.44
material 450 nm
20.23. Model: Radio waves are electromagnetic waves that travel with speed c.
Solve: (a) The wavelength is
c 30 108 m/s

 296 m
f
1013 MHz
(b) The speed of sound in air at 20C is 343 m/s. The frequency is
v
343 m/s
f  sound 
 116 Hz

296 m

20.24. Model: Microwaves are electromagnetic waves that travel with a speed of 3  108 m/s.
Solve: (a) The frequency of the microwave is
f microwaves 
c


30 108 m/s
2
 10  1010 Hz  10 GHz
30 10 m
(b) The refractive index of air is 1.0003, so the speed of microwaves in air is vair  c/1.00  c. The time for the
microwave signal to travel is
t
50 km
50 103 m

 0167 ms  017 ms
vair
(30  108 m/100)
Assess: A small time of 0.17 ms for the microwaves to cover a distance of 50 km shows that the electromagnetic
waves travel very fast.
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20-8
Chapter 20
20.25. Model: Light is an electromagnetic wave.
Solve: (a) The time light takes is
t
30 mm 30 102 3 m
30 102 3 m


 15  1011 s
8
vglass
c/n
(30 10 m/s)/150
(b) The thickness of water is
d  vwatert 
c
nwater
t
30 108 m/s
(15  1011 s)  34 mm
133
20.26. Model: Assume that the glass has index of refraction n  15 This means that vglass  c/n  2 108 m/s
Visualize: We apply v   f twice, once in air and then in the glass. The frequency will be the same in both cases.
Solve: (a) In the air
fair 
vair
air

30 108 m/s
 857  108 Hz  86  108 Hz
035 m
The frequency is the same in both media, so fglass  86  108 Hz.
(b) Now that we know fglass and vglass , we can find  glass 
 glass 
vglass
fglass

20  108 m/s
857  108 Hz
 23 cm
Assess: We get the same answer from  glass   air /nglass  35 cm/15  23 cm.
20.27. Solve: (a) The speed of light in a material is given by Equation 20.29:
n
The refractive index is
n
 vac
 mat
 vsolid  c
 solid
 vac
c
vmat
 vmat 
c
n
 (30 108 m/s)
420 nm
 188  108 m/s
670 nm
(b) The frequency is
f 
vsolid
 solid

188  108 m/s
 448  1014 Hz
420 nm
Section 20.6 Power, Intensity, and Decibels
20.28. Solve: The energy delivered to the eardrum in time t is E  Pt , where P is the power of the wave. The
intensity of the wave is I  P/a where a is the area of the ear drum. Putting the above information together, we have
E  Pt  ( Ia)t  I r 2t  (20 102 3 W/m2 ) (30 103 m)2 (60 s)  34 106 J
20.29. Solve: The energy delivered to an area a in time t is E  Pt , where the power P is related to the intensity I
as I  P/a. Thus, the energy received by your back is
E  Pt  Iat  (0.80)(1400 W/m2 )(0.30  50 m2 )(3600 s)  60  105 J
20.30. Solve: (a) The intensity of a uniform spherical source of power Psource a distance r away is I  Psource/4 r 2.
Thus, the intensity at the position of the microphone is
35 W
I50 m 
 11  102 3 W/m2
2
4 (50 m)
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Traveling Waves
20-9
(b) The sound energy impinging on the microphone per second is
P  Ia  (11  102 3 W/m2 )(10 102 4 m2 )  11102 7 W  11102 7 J/s
 Energy impinging on the microphone in 1 second  11102 7 J
20.31. Model: Assume the intensity scales inversely with the square of the distance.
 I 
Visualize:  = (10 dB)log10   .
 I0 
Solve: The intensity at 1.0 km is I  


30 m 2
I.
1000 m


 I 
  (10 dB)log10    (10 dB)log10 
I
 0




30 m 2
I
1000 m
I0





2
 I 
 30 m 
 (10 dB)log10 
  (10 dB)log10    30 db  140 dB = 110 dB
 1000 m 
 I0 
Assess: 110 dB is still loud, but not as damaging.
20.32. Solve: Because the sun radiates waves uniformly in all directions, the intensity I of the sun’s rays when they
impinge upon the earth is
I
Psun
4 r
2
 I earth 
Psun
2
4 rearth


4 1026 W
4 (1496 10 m)
11
2
 1400 W/m2

With rsun-Venus  m and rsun-Mars   m, the intensities of electromagnetic waves at these
planets are I venus  2700 W/m2 and I Mars  610 W/m2.
20.33. Visualize: Equation 20.35 gives the sound intensity level as
 I 

 I0 
  (10 dB)log10 
where I0  10  1012 W/m2.
Solve:
(a)
 30 106 W/m2 
 I 
  65 dB
  (10 dB)log10 
12
W/m2 
 I0 
 10  10
  (10 dB)log10 
(b)
 30  102 W/m2 
 I 
  105 dB
  (10 dB)log10 
12
W/m2 
 I0 
 10 10
  (10 dB)log10 
Assess: As mentioned in the chapter, each factor of 10 in intensity changes the sound intensity level by 10 dB;
between the first and second parts of this problem the intensity changed by a factor of 104 , so we expect the sound
intensity level to change by 40 dB.
20.34. Visualize: We can solve Equation 20.35 for the sound intensity, finding I  I0  10 /10 dB.
Solve:
(a)
I  I0 10 /10 dB  (10 1012 W/m2 ) 1046  40 108 W/m2
(b)
I  I0 10 /10 dB  (10 1012 W/m2 )  10103  20 102 W/m2
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20-10
Chapter 20
Assess: Since the sound intensity levels in the two parts of this problem differ by 57 dB, we expect the sound
intensities to differ by a factor of 1057.
20.35. Model: Assume the pole is tall enough that we don’t have to worry about the ground absorbing or reflecting
sound.
Visualize: The area of a sphere of radius R is A  4 R2 . Also, I = P/A. We seek P when R  20 m.
Solve:
P = IA  ( I0 10 /10 dB )(4 R2 )  ( I0 1090 dB/10 dB )(4 (20 m)2 ) = 5.0 W
Assess: 5.0 W is a reasonable power output for a speaker.
Section 20.7 The Doppler Effect
20.36. Model: Your friend’s frequency is altered by the Doppler effect. The frequency of your friend’s note
increases as he races towards you (moving source and a stationary observer). The frequency of your note for your
approaching friend is also higher (stationary source and a moving observer).
Solve: (a) The frequency of your friend’s note as heard by you is
f
400 Hz
f  0 
 432 Hz
vS
250 m/s
1
1
v
340 m/s
(b) The frequency heard by your friend of your note is
 v 
 250 m/s 
f   f0 1  0   (400 Hz) 1 
  429 Hz
v
340 m/s 


20.37. Model: The frequency of the opera singer’s note is altered by the Doppler effect.
Solve: (a) Using 90 km/h   m/s, the frequency as her convertible approaches the stationary person is
f0
600 Hz

 650 Hz
1  vS/v 1  25 m/s
343 m/s
(b) The frequency as her convertible recedes from the stationary person is
f0
600 Hz
f 

 560 Hz
1  vS/v 1  25 m/s
343 m/s
f 
20.38. Model: The bat’s chirping frequency is altered by the Doppler effect. The frequency is increased as the bat
approaches and it decreases as the bat recedes away.
Solve: The bat must fly away from you, so that the chirp frequency observed by you is less than 25 kHz. From
Equation 20.38,
f0
25,000 Hz
f 
 20,000 Hz 
 vS  858 m/s  86 m/s
1  vS/v
 vS 
1 

 343 m/s 
Assess: This is a rather large speed: 85.8 m/s  180 mph. This is not possible for a bat.
20.39. Model: The mother hawk’s frequency is altered by the Doppler effect.
Solve: The frequency is f  as the hawk approaches you is
f0
800 Hz
 900 Hz 
 vS  381 m/s
vS
1  vS /v
1
343 m/s
Assess: The mother hawk’s speed of 38.1 m/s  80 mph is reasonable.
f 
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Traveling Waves
20-11
20.40. Visualize: The function D(x, t) represents a pulse that travels in the positive x-direction without changing shape.
Solve: (a)
(b) The leading edge of the pulse moves forward 3 m each second. Thus, the wave speed is 3 m/s.
(c) | x  3t | is a function of the form D( x  vt ), so the pulse moves to the right at v  3 m/s.
20.41. Solve: (a) We see from the history graph that the period T  0.20 s and the wave speed v  4.0 m/s. Thus,
the wavelength is

v
 vT  (40 m/s)(020 s)  080 m
f
(b) The phase constant 0 is obtained as follows:
D(0 m, 0 s)  Asin 0  2 mm  (2 mm)sin 0  sin 0  1  0  12  rad
(c) The displacement equation for the wave is
2 t

 2 x

 2 x
D( x,t )  Asin 
 2 ft  0   (20 mm)sin 

   (20 mm)sin(25 x  10 t  12  )
 

 080 m 020 s 2 
where x and t are in m and s, respectively.
20.42. Solve: (a) We can see from the graph that the wavelength is   2.0 m. We are given that the wave’s
frequency is f  5.0 Hz. Thus, the wave speed is v  f  10 m/s.
(b) The snapshot graph was made at t  0 s. Reading the graph at x  0 m, we see that the displacement is
D( x  0 m,t  0 s)  D(0 m,0 s)  05 mm  12 A
Thus
D(0 m, 0 s)  12 A  Asin 0  0  sin 1
 12   6 rad or 56 rad
Note that the value of D(0 m, 0 s) alone gives two possible values of the phase constant. One of the values will cause
the displacement to start at 0.5 mm and increase with distance—as the graph shows—while the other will cause the
displacement to start at 0.5 mm but decrease with distance. Which is which? The wave equation for t  0 s is
 2 x

D( x,t  0)  A sin 
 0 



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20-12
Chapter 20
If x is a point just to the right of the origin and is very small, the angle (2 x /  0 ) is just slightly bigger than the
angle 0  Now sin31  sin30, but sin151  sin150, so the value 0  16  rad is the phase constant for which the
displacement increases as x increases.
(c) The equation for a sinusoidal traveling wave can be written as
 x

 2 x


D( x, t )  A sin 
 2 ft  0   A sin 2   ft   0 





 

Substituting in the values found above,
  x
 
D( x,t )  (10 mm)sin 2 
 50 s 1 t   
 6
  20 m


20.43. Model: The wave is a traveling wave on a stretched string.
Solve: The wave speed on a string whose radius is R, length is L, and mass density is  is vstring  TS / with

m V  R 2 L


  R 2
L
L
L
If the string radius doubles, then

vstring

TS
 (2 R)
2

v0
2
20.44. Model: The wave pulse is a traveling wave on a stretched string.
Solve: While the tension TS is the same in both the strings, the wave speeds in the two strings are not. We have
v1 
TS
and
1
v2 
TS
2
 v12 1  v22 2  TS
Because v1  L1 /t1 and v2  L2 /t2 , and because the pulses are to reach the ends of the string simultaneously, the
above equation can be simplified to
L12 1
t
2

L22 2
t2

L1
2
40 g/m


 2  L1  2 L2
L2
1
20 g/m
Since L1  L2  4 m,
2L2  L2  4 m  L2  166 m  17 m
and
L1  2(166 m)  234 m  23 m
20.45. Solve: t is the time the sound wave takes to travel down to the bottom of the ocean and then up to the
ocean surface. The depth of the ocean is
2d  (vsound in water )t  d  (750 m/s)t
Using this relation and the data from Figure P20.45, we can generate the following table for the ocean depth (d ) at
various positions (x) of the ship.
x (km)
t (s)
d (km)
0
20
40
45
50
60
6
4
4
8
4
2
4.5
3.0
3.0
6.0
3.0
1.5
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Traveling Waves
20-13
20.46. Visualize:
Solve: The explosive’s sound travels down the lake and into the granite, and then it is reflected by the oil surface.
The echo time is thus equal to
techo  twater down  tgranite down  tgranite up  twater up
094 s 
dgranite
dgranite
500 m
500 m



 dgranite  790 m
1480 m/s 6000 m/s 6000 m/s 1480 m/s
20.47. Model: Assume a room temperature of 20C
Visualize:
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20-14
Chapter 20
Solve: The distance between the source and the left ear (E L ) is
dL  x2  ( y  01 m)2  [(50 m)cos45]2  [(50 m)sin 45  01 m]2  50712 m
Similarly dR  49298 m Thus,
dL  dR  d  01414 m
For the sound wave with a speed of 343 m/s, the difference in arrival times at your left and right ears is
d
01414 m
t 

 410 s
343 m/s 343 m/s
20.48. Model: The laser beam is an electromagnetic wave that travels with the speed of light.
Solve: The speed of light in the liquid is
vliquid 
30 102 2 m
138 102 9 s
 2174  108 m/s
The liquid’s index of refraction is
n
c
vliquid

30  108
2174  108
 138
Thus the wavelength of the laser beam in the liquid is
liquid 
vac
n

633 nm
 459 nm
138
20.49. Model: v   f applies.
Solve: (a) The frequency must remain the same since the harmonic oscillators in one medium excite the oscillators
in the second medium. So f water  440 Hz.
(b) The table in the chapter gives vwater  1480 m/s.
v 1480 m/s

 3.4 m
f
440 Hz
Assess: This is a reasonable wavelength for a sound wave.

20.50. Solve: The difference in the arrival times for the P and S waves is
t  tS  tP 
d d
1
1


6
  120 s  d 

  d  123  10 m  1230 km
vS vP
 4500 m/s 8000 m/s 
Assess: d is approximately one-fifth of the radius of the earth and is reasonable.
20.51. Model: This is a sinusoidal wave.
Solve: (a) The equation is of the form D( y, t )  Asin(ky   t  0 ), so the wave is traveling along the y-axis.
Because it is  t rather than  t the wave is traveling in the negative y-direction.
(b) Sound is a longitudinal wave, meaning that the medium is displaced parallel to the direction of travel. So the air
molecules are oscillating back and forth along the y-axis.
(c) The wave number is k  896 m1, so the wavelength is
2
2


 0701 m
k 896 m2 1
The angular frequency is   3140 s2 1, so the wave’s frequency is
f 
 3140 s2 1

 500 Hz
2
2
Thus, the wave speed v   f  (0.70 m)(500 Hz)  350 m/s. The period T  1/f  0.00200 s  2.00 ms.
Assess: The wave is a sound wave with speed v  350 m/s. This is greater than the room-temperature speed of 343 m/s,
so the air temperature must be greater than 20
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Traveling Waves
20-15
20.52. Model: This is a sinusoidal wave.
Solve: (a) The displacement of a wave traveling in the positive x-direction with wave speed v must be of the form
D(x, t)  D(x vt). Since the variables x and t in the given wave equation appear together as x  vt, the wave is
traveling toward the left, that is, in the x direction.
(b) The speed of the wave is

2 /020 s
v 
 12 m/s
k 2 rad/24 m
The frequency is
 2 rad/020 s
f 

 50 Hz
2
2
The wave number is
2 rad
k
 26 rad/m
24 m
(c) The displacement is
  020 m 050 s 
D(020 m, 050 s)  (30 cm)sin 2 

 1   15 cm
  24 m 020 s  
20.53. Model: This is a sinusoidal wave traveling on a stretched string in the x direction.
Solve: (a) From the displacement equation of the wave, A  2.0 cm, k  12.57 rad/m, and   638 rad/s. Using the
equation for the wave speed in a stretched string,
vstring 
2
2
 
 638 rad/s 
2
 TS  vstring
     (500  103 kg/m3 ) 
  126 N

k
 1257 rad/m 
TS
(b) The maximum displacement is the amplitude Dmax ( x,t )  200 cm.
(c) From Equation 20.17,
v y max   A  (638 rad/s)(20 102 m)  128 m/s
20.54. Solve: The wave number and frequency are calculated as follows:
2 2 rad
k

 4 rad/m    vk  (40 m/s)(4 rad/m)  16 rad/s
 050 m
Thus, the displacement equation for the wave is
D( y,t )  (50 cm)sin[(4 rad/m) y  (16 rad/s)t ]
Assess: The positive sign in the sine function’s argument indicates motion along the y direction.
20.55. Solve: The angular frequency and wave number are calculated as follows:
 400 rad/s
  2 f  2 (200 Hz)  400 rad/s  k  
  rad/m
v
400 m/s
The displacement equation for the wave is
D( x,t )  (0010 mm)sin[( rad/m) x  (400 rad/s)t  12  rad]
Assess: Note the negative sign with  t in the sine function’s argument. This indicates motion along the x
direction.
20.56. Model: We have a sinusoidal traveling wave on a stretched string.
Solve: (a) The wave speed on a string and the wavelength are calculated as follows:
v
TS


20 N
v 1000 m/s
 100 m/s    
 10 m
0002 kg/m
f
100 Hz
(b) The amplitude is determined by the oscillator at the end of the string and is A  1.0 mm. The phase constant can
be obtained from Equation 20.15 as follows:
D(0 m, 0 s)  Asin 0  10 mm  (10 mm)sin 0  0  

2
rad
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20-16
Chapter 20
(c) The wave (as distinct from the oscillator) is described by D( x,t )  Asin(kx  t  0 ). In this equation the wave
number and angular frequency are
2
2
k

 2 rad/m
  vk  (1000 m/s)(2 rad/m)  200 rad/s
 10 m
Thus, the wave’s displacement equation is
D( x,t )  (10 mm)sin[(2 rad/m) x  (200 rad/s)t  12  rad]
(d) The displacement is
D(050 m, 0.015 s)  (1.0 mm)sin[(2 rad/s)(0.50 m)  (200 rad/s)(0.015 s)  12  ]  1.0 mm
20.57. Model: We have a wave traveling to the right on a string.
Visualize:
Solve: The snapshot of the wave as it travels to the right for an infinitesimally small time t shows that the velocity
at point 1 is downward, at point 3 is upward, and at point 2 is zero. Furthermore, the speed at points 1 and 3 is the
maximum speed given by Equation 20.17: v1  v3   A. The frequency of the wave is
2 (45 m/s)
 300 rad/s   A  (300 rad/s)(20 102 2 m)  19 m/s
030 m
Thus, v1  19 m/s, v2  0 m/s, and v3  19 m/s.
  2 f  2
v


20.58. Model: The wave pulse is a traveling wave on a stretched string. The two masses hanging from the steel
wire are in static equilibrium.
Visualize:
Solve: The wave speed along the wire is
vwire 
40 m
 1667 m/s
0024 s
Using Equation 20.2,
vwire  1667 m/s 
T1


T1
 T1  2084 N
(0060 kg/80 m)
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Traveling Waves
20-17
Because point 1 is in static equilibrium, with Fnet  0,
T1
 2721 N
cos 40
( Fnet ) x  T1  T2 cos 40  T2 
( Fnet ) y  T2 sin 40  w  0 N  w  mg  T2 sin 40  m 
(2721 N)sin 40
98 m/s 2
 178 kg
20.59. Solve: The wave speeds along the two metal wires are
v1 
T
1

2250 N
 500 m/s
0009 kg/m
v2 
T
2

2250 N
 300 m/s
0025 kg/m
The wavelengths along the two wires are
v 500 m/s 1
v
300 m/s 1
1  1 
 m
2  2 
 m
f 1500 Hz 3
f 1500 Hz 5
Thus, the number of wavelengths over two sections of the wire are
10 m 10 m
10 m 10 m

3

5
1
1 m
1
2
m
3
5




The number of complete cycles of the wave in the 2.00-m-long wire is 8.
20.60. Model: The object is in static equilibrium.  is the linear density of the string, not a coefficient of friction.
Visualize: Use tilted axes with the x-direction along the string. The tension in the string is TS.
Solve: From a free-body diagram we see that
 Fx = TS  Mg sin  0  TS  Mg sin
v
TS


Mg sin 

20.61. Model: The wave is traveling on a stretched string.
Solve: The wave speed on the string is
v
TS


50 N
 100 m/s
0005 kg/m
The speed of the particle on the string, however, is given by Equation 20.17. The maximum speed is calculated as
follows:
v
 100 m/s 
v y   A cos(kx   t   0 )  v y max   A  2 fA  2 A  2 
 (0030 m)  94 m/s

 20 m 
20.62. Model: A sinusoidal wave is traveling along a stretched string.
Solve: From Equation 20.17 and Equation 20.20, v y max   A and a y max   2 A. These two equations can be
combined to give
a y max 200 m/s2
v y max

20 m/s


 100 rad/s  f 
 159 Hz  16 Hz  A 

 20 cm
v y max
20 m/s
2

100 rad/s
20.63. Solve: (a) At a distance r from the bulb, the 5 watts of visible light have spread out to cover the surface of a
sphere of radius r. The surface area of a sphere is a a  4 r 2 . Thus, the intensity at a distance of 2 m is
P
P
50 W


 0095 W/m2
a 4 r 2 4 (20 m)2
Note that the presence of the wall has nothing to do with the intensity. The wall allows you to see the light, but the
light wave has the same intensity at all points 2 m from the bulb, whether it is striking a surface or moving through
empty space.
I
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20-18
Chapter 20
(b) Unlike the light from a light bulb, a laser beam does not spread out. We ignore the small diffraction spread of the
beam. The laser beam creates a dot of light on the wall that is 2 mm in diameter. The full 5 watts of light is
concentrated in this dot of area a   r 2   (0001 m)2  314 106 m2. The intensity is
P
5W

 16 MW/m2
a 314 106 m2
Although the power of the light source is the same in both cases, the laser produces light on the wall whose intensity
is over 16 million times that of the light bulb.
I
20.64. Model: The radio wave is an electromagnetic wave.
Solve: At a distance r, the 25 kW power station spreads out waves to cover the surface of a sphere of radius r. The
surface area of a sphere is 4 r 2 . Thus, the intensity of the radio waves is
I
Psource
4 r 2

25 103 W
4 (10 103 m) 2
 20  105 W/m2
20.65. Visualize: To find the power of a laser pulse, we need the energy it contains, U , and the time duration of
the pulse, t  Then to find the intensity, we need the area of the pulse. Its radius is 050 mm
Solve: (a) Using P  U/Δt , we find the following:
P  (10 103 J)/(15 109 s)  667 104 W
(b) Then from I  P/a, we obtain
I
(667  104 W)
 (50 104 m)2
 85  1010 W/m2
20.66. Model: We have a traveling wave radiated by the tornado siren.
Solve: (a) The power of the source is calculated as follows:
P
Psource
I50m  010 W/m2  source

 Psource  (010 W/m2 )4 (50 m)2  (1000 ) W
2
4 r
4 (50 m)2
The intensity at 1000 m is
Psource
(1000 ) W
I1000 m 

 250  W/m2
2
4 (1000 m)
4 (1000 m)2
(b) The maximum distance is calculated as follows:
P
(1000 ) W
I  source
 10  106 W/m2 
 r  16 km
4 r 2
4 r 2
20.67. Model: Assume the saw is far enough off the ground that we don’t have to worry about reflected sound.
Visualize: First note that 1  2  20dB  I1/I 2  10 10  100 (a change of 10 dB corresponds to a change in
intensity by a factor of 10). Then use I1 A1  P and then P  I 2 A2  A2  P/I 2 , and finally solve for R2  A2/4 .
Solve: Put all of the above together.
P
A2
I2
R2 


4
4
I1 A1
I1
I2
I2
4

(4 R1 )
2
4
 R1
I1
 R1 100  (50 m)(10)  50 m
I2
Assess: The scaling laws help and the answer is reasonable.
20.68. Model: Assume the two loudspeakers broadcast the same power and that the platforms are high enough off
the ground that we don’t have to worry about reflected sound.
Visualize: Call the distance between the loudspeakers d. Call the intensity halfway between the speakers (at d/2) I1
and the sound intensity level there 1( 75 dB); call them I 2 and  2 at 1/4 the distance from one pole and 3/4 the
distance from the other pole on the line between them. We seek  2 .
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Traveling Waves
20-19
We first apply a general approach for different sound intensity levels:

I 
 I 
 I /I 
I 
   2   1  (10dB) log10  2   log10  1    (10 dB)log10  2 0   (10 dB)log10  2 
 I1 
 I0 
 I 0  
 I1 /I 0 

Solve: Recall that for the general case of spherical symmetry I  P/A, where P is the power emitted by the source
and A  4 R2 is the area of the sphere. Now we find the ratio of the intensities I 2 /I1 and then plug it in the formula
above and add it to 75 dB.
P
P
2P
I1 


2
2
4 (d/ 2)
4 (d/2)
d2
I2 
P
4 (d/4)
2

P
4 (3d/4)
2

4P
d
2

4P
9 d
2

(36  4) P
9 d
2

40P
9 d
2

20
I1
9
I 
 20 
  (10 dB)log10  2   (10 dB)log10    348 dB
 9 
 I1 
2  1    75 dB  348 dB  78 dB
Assess: An increase of about 3dB corresponds to a doubling of the intensity. 20/9 is close to double.
20.69. Solve: If we solve the equation for I, we have:
I  I0 10(  /10 dB)
Now plugging in 60 dB for  , we get I  106 W/m2 and plugging in 61 dB for  , we get I  13 106 W/m2  The
ratio of the latter to the former is 13
20.70. Visualize: Take the log10 of both sides of I  cPsourcer  x.
log I = log(cPsource )  log(r  x ) = log(cPsource )  x log r
Solve: Now use the equation for sound level intensity.
 I 
  (10dB)log    (10dB)[log I  log I 0 ]
 I0 
Insert the expression for log I from above.
  10dBlog(cPsource )  x log r  log I0 
Divide both sides by 10 dB and note log1012  12.

10 dB
= log(cPsource )  x log r  (12)
Group constant terms.

10 dB
=  x log r  (log(cPsource )  12)
This equation leads us to believe that a graph of

10 dB
vs. log r would produce a straight line whose slope is  x and
whose intercept is log(cPsource )  12.
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20-20
Chapter 20
The spreadsheet shows that the linear fit is excellent and that the slope is 1.5. So we conclude that x  1.5.
Assess: We expected a number less than 2. Also, the intercept of 10 tells us that log(cPsource )  2.
20.71. Model: As suggested, model the bald head as a hemisphere with radius R  0080m. This means the
surface area of the bald head (hemisphere) is A  2 R2  00402m2.
Visualize: We are given   93 dB and E  010 J. We also know that I  I0 10 /10dB and P  IA. Also recall
P  Et.
Solve: Put all of the above together to find t.
E E
E
010 J
t 



 1250 s  21 min
P
IA ( I 0 10 /10dB )(2 R2 ) (1012 W/m2 1093 )(00402 m2 )
Assess: 21 min seems like quite a while to deliver 0.10 J of energy, but sounds waves don’t carry a lot of energy
unless the intensity is high.
20.72. Model: The sound generator’s frequency is altered by the Doppler effect. The frequency increases as the
generator approaches the student, and it decreases as the generator recedes from the student.
Solve: The generator’s speed is
 100

vS  r  r (2 f )  (10 m)2 
rev/s   1047 m/s
 60

The frequency of the approaching generator is
f0
600 Hz
f1 

 619 Hz  620 Hz
1  vS/v 1  1047 m/s
343 m/s
Doppler effect for the receding generator, on the other hand, is
f0
600 Hz
f 

 582 Hz  580 Hz
1  vS/v 1  1047 m/s
343 m/s
Thus, the highest and the lowest frequencies heard by the student are 620 Hz and 580 Hz.
20.73. Solve: We will closely follow the details of section 20.7 in the textbook. Figure 20.29 shows that the wave
crests are stretched out behind the source. The wavelength detected by Pablo is   13 d , where d is the distance the
wave has moved plus the distance the source has moved at time t  3T. These distances are xwave  vt  3vT and
xsource  vSt  3vST . The wavelength of the wave emitted by a receding source is thus
d xwave  xsource 3vT  3vST


 (v  vS )T
3
3
3
The frequency detected in Pablo’s direction is thus
v
v
f0
f 


 (v  vs )T 1  vS /v
 
20.74. Model: We are looking at the Doppler effect for the light of an approaching source.
Solve: (a) The time is
t
54  106 km
3  105 km/s
(b) Using Equation 20.40, the observed wavelength is

 180 s  30 min
1  vs /c
1  0.1c/c
0 
(540 nm)  (0.9045)(540 nm)  488 nm  490 nm
1  vs /c
1  0.1c/c
Assess: 490 nm is slightly blue shifted from green.
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Traveling Waves
20-21
20.75. Model: The Doppler effect for light of a receding source yields an increased wavelength.
Solve: Because the measured wavelengths are 0.5% longer, that is,   1.0050 , the distant galaxy is receding away
from the earth. Using Equation 20.40,
  10050 
1  vs /c
1  vs /c
0  (1005)2 
 vs  00050 c  15  106 m/s
1  vs /c
1  vs /c
20.76. Model: The Doppler effect for light of an approaching source leads to a decreased wavelength.
Solve: The red wavelength (0  650 nm) is Doppler shifted to green (  540 nm) due to the approaching light
source. In relativity theory, the distinction between the motion of the source and the motion of the observer
disappears. What matters is the relative approaching or receding motion between the source and the observer. Thus,
we can use Equation 20.40 as follows:
1  vs /c
1  vs /c
  0
 540 nm  (650 nm)
1  vs /c
1  vs /c
 vs  55  104 km/s  20  108 km/h
The fine will be
 1$ 
(20 108 km/h  50 km/h) 
  $200 million
 1 km/h 
Assess: The police officer knew his physics.
20.77. Solve: The time for the wave to travel from California to the South Pacific is
t
d 800  106 m

 5405.4 s
v
1480 m/s
d
 148028 m/s
t
Since the 4.0 m/s increase in velocity is due to an increase of 1C, an increase of 0.28 m/s occurs due to a
temperature increase of
 1C 

 (028 m/s)  007C
 40 m/s 
Thus, a temperature increase of approximately 0.07C can be detected by the researchers.
A time decrease to 5404.4 s implies the speed has changed to v 
20.78. Model: As the guitar string is stretched its linear density will decrease, but only slightly, so we will assume
that   1.3g/m will apply through the problem.
Visualize: The fact that we have a length L, a tension force, and are looking for a change in length suggests this is a
Young’s modulus problem.
Solve: First solve for the tension force from v =
TS

. This tension force becomes the F in the Young’s modulus
equation. From the chapter on elasticity we have
L =
LF
YA
where Y  20 1010 N/m2 is Young’s modulus for steel from Table 15.3, F  TS  v2 , and A   R2 is the crosssectional area of the string.
LF L(v 2  ) (0.75 m)(250 m/s)2 (0.0013 kg/m)
L =
=
=
= 1.8 mm
YA Y ( R 2 )
(20 1010 N/m2 ) (0.00023 m)2
Assess: That f  196 Hz when the string is properly tuned is correct but irrelevant information. That frequency,
however, illustrates that only 65 cm of the string is really vibrating, the other 10 cm are wrapped around the tuning
screw.
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20-22
Chapter 20
20.79. Model: The wave pulse is a traveling wave on a stretched wire.
Visualize:
Solve: (a) At a distance y above the lower end of the rope, the point P is in static equilibrium. The upward tension in
the rope must balance the weight of the rope that hangs below this point. Thus, at this point
T  w  Mg  ( y) g
where   m/L is the linear density of the entire rope. Using Equation 20.2, we get
v
T


 yg
 gy

(b) The time to travel a distance dy at y, where the wave speed is v  gy , is
dy
dy

v
gy
Finding the time for a pulse to travel the length of the rope requires integrating from one end of the rope to the other:
dt 
T
L
L
dy
1 
2
t  Ñ
dt  Ñ

2 y 0  
gy
g


g
0
0
L  t  2
L
g
20.80. Visualize:
Solve: (a) Using the graph, the refractive index n as a function of distance x can be mathematically expressed as
n n
n  n1  2 1 x
L
At position x, the light speed is v  c/n. The time for the light to travel a distance dx at x is
dx n
1
n n 
dt 
 dx   n1  2 1 x  dx
v c
c
L

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Traveling Waves
20-23
To find the total time for the light to cover a thickness L of a glass we integrate as follows:
T
L
T Ñ
dt 
0
L
L
2
1 
n2  n1 
n1
(n2  n1)
n1
 n2  n1  L  n1  n2 
n

x
dx

dx

x
dx

L


1




L
cÑ
L
cÑ
cL Ñ
c

 cL  2  2c 
0
0
0
(b) Substituting the given values into this equation,
(150  160)
T
 0010 m  517 1011 s
8
2(30 10 m/s)
20.81. Model: Use the shallow wave equation v  gd .
Visualize: Say the wave is moving to the right from an initial depth of 5.0 m at 100 m from shore to a depth of 0.0 m
at the shore.
Solve: We want T =  dt where dt  dx/v  dx/ gd .
T   dt  
100
100
dx

gd  0
dx

 5

g
(100  x) 
 100

100
10
20
 2 100  x  

0  10  29 s
0
5g 
5g
0
10
5g
100
0
dx
100  x
Assess: 29 s seems like a reasonable time for a shallow wave to travel 100 m.
20.82. Model: Assume v
v0 .
Visualize: Apply the Doppler effect twice, first with the observer (the object) approaching the source, and second
with the source (the object now reflecting the sound) approaching the observer.
Solve:
(a) With the object (observer) approaching we have
 v 
f   1  0  f0
v

Plug that as the new
f 0 into the equation for the receding source (now the object).
f  
f
1
v0
v
1   f

v0
0
v
v0
1 v
=
v  v0
f0
v  v0
The shift is now
 v  v0 
 v  v0  v  v0 
v0
v
f  f   f0  
 1 f 0 = 
 2 f0 0
 f0 = 2 f0
v

v
v

v
v

v
v
0
0
0




1
1
(b) Bernoulli’s equation for fluid dynamics is p1   v12   gy1 = p2  v22   gy2 . In this case y1  y2 , and the
2
2
speed of the blood in the heart is v1  0, so
1
p =   v22
2
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20-24
Chapter 20
The blood in this part is the object in part (a), so v0 there is now v2 =
f
v, where f0 = 2.5 MHz, f = 6000 Hz, and
2 f0
the speed of sound in blood is the same as the speed of sound in water: v = 1480 m/s.
2
2
 6000 Hz

1  f 
1
p =   
v  =  (1060 kg/m3 ) 
(1480 m/s)  = 1672 Pa
2  2 f0 
2
2(2.5
MHz)


Convert to mm of Hg.
1672 Pa =12.54 mm of Hg  13 mm of Hg
Assess: 13 mm of Hg seems like a reasonable pressure difference.
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