Chapter 11
WAVES
Conceptual Questions
1. Wrapping a thick coil of copper wire around a piano string increases the string’s mass density and therefore
decreases the speed of waves traveling along it. The fundamental wavelength is fixed by the length of the string—
the decreased wave velocity must therefore be accompanied by a decrease in the frequency at which the string
vibrates.
2. Piano, guitar, and violin strings produce transverse waves when they are plucked or bowed—they do not produce
longitudinal sound waves. The transverse motion of the strings causes longitudinal sound waves to be produced in
the surrounding air.
3. The wavelength of the fundamental standing wave on a cello string depends only upon the length of the string.
The wavelength of the sound waves produced by the vibrating cello string is determined by the frequency of the
string’s vibration. This frequency is proportional to the string’s wave velocity, which depends upon both the mass
per unit length and the tension of the string.
4. (a) The wavelength of the fundamental will decrease.
(b) The frequency of the fundamental will increase.
(c) The wave velocity is constant, thus the time for a pulse to travel the length of the string will decrease.
(d) The maximum velocity of a point on the string is proportional to the frequency and will therefore increase.
(e) The maximum acceleration of a point on the string is proportional to the square of the frequency and will
therefore increase.
5. Words spoken by two people at the same time are comprehensible because sound waves travel through each
other—interfering while superimposed, but returning to their original waveform as they again separate.
6. (a) Increasing the tension of the string changes the speed of transverse waves and thus the frequency.
(b) Pressing her finger on the string changes the string length, which changes the wavelength and hence the
frequency.
(c) Each string has a different tension and mass per unit length; thus the speed of transverse waves differs from
string to string. The fundamental wavelength remains the same because the strings are all of the same length,
but the frequencies are different.
7. A transverse wave is produced by a disturbance that tends to shear the medium, separating layers at right angles to
the direction of the wave velocity.
511
Chapter 11: Waves
College Physics
8. In the figure below, the shape of a wave traveling to the right is depicted at two times separated by a short
interval. Arrows indicate the direction of the string’s movement.
v
9. Since transverse waves do not travel through the core while longitudinal waves do, some part of the core is a
molten, viscous liquid that cannot support the transmission of a transverse wave. A longitudinal wave can create
compressions and rarefactions in the liquid and travel on through.
10. The electrical signal mimicking the noise is modified electronically to produce a wave that is out of phase with the
noise. When played in the headphones it interferes destructively with the noise, thereby canceling it.
11. If the wires going to one speaker are reversed, the sound waves emitted will be out of phase with those from the
other speaker. For a listener situated midway between the speakers, the sounds from the two speakers will
interfere destructively, and the sound level will be noticeably reduced. If the listener moves slightly, the difference
in the path length from the two speakers changes. If this change in path length is small compared to the
wavelength, the waves still interfere destructively to a large extent. For this reason there is a noticeably weaker
bass (long wavelength) over a large area between the speakers, while high frequencies (short wavelengths) are
less affected.
Problems
1. Strategy Form a proportion with the intensities, treating the Sun as an isotropic source. Use Eq. (11-1).
Solution Find the intensity of the sunlight that reaches Jupiter.
P
2
2
r 
IJ
r 2
4 rJ 2
 1 

 E , so I J   E  I E  
(1400 W m 2 )  52 W m 2 .

P
2
IE
r
5.2


r
J


J
4 r 2
E
2. (a) Strategy It takes half the time for the sound to cross the valley as it does to make the round trip.
Solution The cliff is x  vt  (343 m s)(0.75 s)  260 m away.
(b) Strategy Treat the radio and echo as isotropic sources. Use Eq. (11-1) and form a proportion.
Solution Find the intensity of the music arriving at the cliff.
I cliff
I radio

P (4 rc 2 )
P (4 rr 2 )

r
, so I cliff   r
2
rc
 rc
rr 2
2
2

 1 
2
5
10 W m 2 .
I

 radio 
 (1.0  10 W m )  1.5  10
257.25



3. Strategy Form a proportion with the intensities, treating the jet airplane as an isotropic source. Use Eq. (11-1).
Solution Find the intensity of the sound waves at the ears of the person.
2
2
2
r 
I 2 P (4 r22 )  r1 
 5.0 m 
(1.0  102 W m 2 )  170 mW m 2 .

   , so I 2   1  I1  

I1 P (4 r12 )  r2 
 120 m 
 r2 
512
College Physics
Chapter 11: Waves
4. Strategy The power equals the intensity times the area.
Solution Find the power radiated by the jet airplane in the form of sound waves.
2
 5.0 m 
2
2
2
P  IA  
 (1.0  10 W m )4 (120 m)  31 kW
120
m


5. Strategy The power equals the intensity times the area.
Solution Find the rate at which the Sun emits electromagnetic waves.
P  IA  I (4 RE 2 )  4 (1.4  103 W m 2 )(1.50  1011 m)2  4.0  1026 W
6. Strategy Refer to the figure. Use the definition of average speed.
Solution
(a) vx 
x 1.80 m  1.50 m

 1.5 m s
t
0.20 s
(b) v y 
y 8.7 cm  4.5 cm

 21 cm s
t
0.20 s
7. Strategy Refer to the figure. Use the definition of average speed.
Solution
(a) Find the speed.
x 1.80 m  1.50 m
vx 

 1.5 m/s
t
0.20 s
Find the position.
xf  xi  vt  1.80 m  (1.5 m s)(3.00 s  0.20 s)  6.0 m
(b) tf 
xf  xi
4.00 m  1.80 m
 ti 
 0.20 s  1.7 s
vx
1.5 m s
8. Strategy Use Eq. (11-4).
Solution Find the linear mass density of the cord.
F
F
75 N

 3.8 g m .
v
, so  
2

v
(140 m s) 2
9. Strategy Use Eq. (11-4).
Solution Find the speed of the transverse waves on the string.
F
90.0 N
v

 168 m s

3.20  103 kg m
513
Chapter 11: Waves
College Physics
10. Strategy Use Eq. (11-2) to find the wave speeds for both strings. Use the speeds and the lengths of the strings to
find the additional time required for the slower wave to reach the end of its string.
Solution
FL
(180.0 N)(15.0 m)
v1  1 1 
 186 m s and v2 
0.0780 kg
m1
F2 L2
m2

(160.0 N)(15.0 m)
 203 m s.
0.0580 kg
The pulse moves faster on the second string. Find the time difference.
1 1 


1
1
x x
x  vt , so t1  t2 

 x     (15.0 m) 

  6.9 ms .
v1 v2
 186.05 m s 203.42 m s 
 v1 v2 
11. Strategy Use Eq. (11-2) for the speed of the transverse waves.
Solution The weight of the string divided by the load is
0.25 N
 2.5  104  0.025%.
3
1.00  10 N
The weight of the string is negligible since the result will be limited to two significant figures (by 0.25 N).
Find the time it takes the wave pulse to travel to the upper end of the string.
y
L
mL
mgL
(0.25 N)(10.0 m)
t 




 16 ms
v
F
Fg
FL m
(1.00  103 N)(9.80 N kg)
12. Strategy Use Eq. (11-6).
Solution Find the speed of the wave.
v  f   (500.0 Hz)(0.500 m)  250 m s
13. Strategy Use Eq. (11-5).
Solution Find the wavelength.
  vT  (75.0 m s)(5.00  103 s)  0.375 m
14. Strategy Use Eq. (11-6).
Solution Find the frequency.
v
120 m s
f  
 400 Hz
 30.0  102 m
15. Strategy Use Eq. (11-6).
Solution Find the frequencies.
(a)
f 
(b) f 
v

v


340 m s
 340 Hz
1.0 m

3.0  108 m s
 3.0  108 Hz
1.0 m
514
College Physics
Chapter 11: Waves
16. Strategy Use Eq. (11-6).
Solution Find the range of visible electromagnetic waves.
v 3.0  108 m s
v 3.0  108 m s
f1 

 7.5  1014 Hz and f 2 

 4.3  1014 Hz.
1 4.0  107 m
2 7.0  107 m
The frequency range is 4.3  1014 Hz to 7.5  1014 Hz .
17. Strategy Use Eq. (11-6).
Solution Find the frequency with which the buoy bobs up and down.
v 2.5 m s
f  
 0.33 Hz
7.5 m

18. Strategy and Solution If another swimmer were 9.6 m away from you, your motions would be the same (in
phase). For periodic motion, the motion half a wavelength from any point along a line parallel to the motion of the
waves is opposite to the motion at that point, so the other swimmer should be half a wavelength away, or
9.6 m
 4.8 m .
2
19. Strategy Use Eq. (11-7).
Solution Compute the wave speed.

5.0 rad s
v 
 0.83 cm s
k 6.0 rad cm
20. Strategy Read the amplitude from the equation of the wave. Use Eq. (11-7) to find the wavelength.
Solution
 
y ( x, t )  (3.5 cm) sin 
 x  (66 cm s)t   A sin(kx  t ), so we have
 3.0 cm

(a) A  3.5 cm
(b)  
2
2
 
k
 6.0 cm
3.0 cm
21. Strategy The wave on the string is of the form y ( x, t )  A sin(t  kx). Use the equation, the given information,
Eq. (11-7), and the relationship between period and angular frequency to find the amplitude, wavelength, period,
and wave speed.
Solution
(a) A  4.0 mm
(b)  
(c) T 
2
2

 1.0 m
k
6.0 m 1
2


2
6.0  102 s 1
 0.010 s
515
Chapter 11: Waves
(d) v 

k

College Physics
6.0  102 s1
6.0 m 1
 100 m s
(e) Since the signs of  t and kx are opposite, the wave travels in the +x-direction (to the right).
22. (a) Strategy Use Eq. (10-21).
Solution Find the maximum transverse speed of a point on the string.
vm   A  (130 rad s)(0.0220 m)  2.9 m s
(b) Strategy Use Eq. (10-22).
Solution Find the maximum transverse acceleration of a point on the string.
am   2 A  (130 rad s)2 (0.0220 m)  370 m s 2
(c) Strategy Use Eq. (11-7).
Solution Find the wave speed.
 130 rad s
 8.7 m s
v 
k 15 rad m
(d) Strategy Consider why transverse speed and wave speed are different.
Solution The answer to part (c) is different from the answer to part (a) because
the motion of the particles on the string is not the same as the motion of the wave along the string .
23. Strategy The equation for a transverse sinusoidal wave moving in the negative x-direction can be written in the
form y ( x, t )  A sin(kx  t ). Use Eq. (11-7) to find the angular frequency and the wavenumber.
Solution A  0.120 m,   0.300 m, v  6.40 m s , and y ( x, t )  A sin( t  kx ).

2 v


2 (6.40 m s)
2
2
 134 s 1 and k 

 20.9 m 1.
0.300 m
 0.300 m
Thus, the equation is
y ( x, t )  (0.120 m) sin[(134 s1 )t  (20.9 m 1 ) x ] .
24. Strategy The equation for a transverse sinusoidal wave moving in the positive x-direction and in the negative
y-direction in the next instant of time according to the situation given in the problem statement can be written in
the form y ( x, t )  A sin(kx  t ). Use Eqs. (10-21) and (11-7).
Solution Find the wave speed v by finding the maximum speed of a point on the string.
v  5.00vm  5.00 A
Find the wave number.


1
1
k 


 8.00 rad m
v 5.00 A 5.00 A 5.00(0.0250 m)
The equation for the transverse sinusoidal wave is
y ( x, t )  (2.50 cm) sin[(8.00 rad m) x  (2.90 rad s)t ] .
516
College Physics
Chapter 11: Waves
25. Strategy The maximum y-value is the amplitude. Find the horizontal distance for which the wave repeats to find
the wavelength. The wave speed is the distance the wave travels divided by the time interval it took to travel that
distance. Use Eq. (11-6) to find the frequency of the wave. The period is the reciprocal of the frequency.
Solution
(a) ymax  2.6 cm, so A  2.6 cm .
(b)   x  16 m  2 m  14 m
(c) v 
(d) f 
(e) T 
x 7.5 m  5.5 m

 20 m s
t
0.10 s
v


20 m s
 1.4 Hz
14 m
1
14 m

 0.70 s
f 20 m s
26. (a) Strategy The maximum y-value is the amplitude. Find the horizontal distance for which the wave repeats to
find the wavelength. The wave speed is the distance the wave travels divided by the time interval it took to
travel that distance.
Solution Plot y ( x, t )  (4.0 cm) sin[(378 rad s)t  (314 rad cm) x] at the two given times.
y (cm)
4
1
t = 480
s
t=0
2
0
0.01
0.02
0.03 x (cm)
2
4
ymax  4.0 cm, so A  4.0 cm ;   x  0.020 cm  0  0.020 cm ;
v
x 0.0025 cm  0

 1.2 cm s
t
1 480 s
(b) Strategy Set x  0; then the function to plot is y ( x, t )  (4.0 cm) sin[(378 rad s)t ]. The period of the
vibration is equal to the time interval for which the wave repeats. Multiply the wave speed by the period of
the vibration and compare to the wavelength found in part (a).
Solution
y (cm)
4
x=0
2
0
2
π
378
π
189
t (s)
4
T

 
s  16.6 ms and vT  (1.2 cm s) 
189
 189

s   0.020 cm  .

517
Chapter 11: Waves
College Physics
27. Strategy Use Eqs. (10-21) and (10-22). Plot the graphs.
Solution Find the maximum speed and maximum acceleration of a point on the string.
vm   A  (4.0 rad s)(0.0050 m)  0.063 m s and am   2 A  (4.0 rad s)2 (0.0050 m)  0.79 m s 2 .
Plot the graphs.
y(0, t )  (0.0050 m) cos[(4.0 s 1)t ]
y (m)
0.0050
x=0
0
0.25
0.50 t (s)
0.0050
v y leads y by 14 cycle; v y (0, t )  (0.063 m/s) sin[(4.0 s1)t ]
vy (m/s)
0.063
0
x=0
0.50 t (s)
0.25
0.063
a y leads v y by 14 cycle; a y (0, t )  (0.79 m/s 2 ) cos[(4.0 s 1)t ]
ay (m/s2)
x=0
0.79
0
0.50 t (s)
0.25
0.79
28. Strategy Use Eqs. (10-21) and (10-22). v y leads y by 1/4 cycle, so v y is a cosine function. Plot the graphs.
Solution y ( x, t )  (1.2 mm) sin[(2.0 s 1 )t  (0.50 m 1 ) x]; calculate the maximum speed.
vm   A  (2.0 s 1 )(1.2 mm)  7.5 mm/s, so v y (0, t )  (7.5 mm s) cos(2.0 s 1 )t at x  0.
y (mm)
vy (mm/s)
1.2
7.5
x=0
x=0
1.0
0
0.5
0
t (s)
1.2
7.5
518
0.5
1.0
t (s)
College Physics
Chapter 11: Waves
29. (a) Strategy Substitute t = 0, 0.96 s, and 1.92 s into y ( x, t )  (0.80 mm) sin[( 5.0 cm 1 ) x  ( 6.0 s 1 )t ] and
graph the resulting equations.
Solution The three equations are:
y ( x, 0)  (0.80 mm) sin[( 5.0 cm 1 ) x], y ( x, 0.96 s)  (0.80 mm) sin[( 5.0 cm 1 ) x  0.50], and
y ( x, 1.92 s)  (0.80 mm) sin[( 5.0 cm 1 ) x  1.0].
Find the wavelength.
2
2


 10 cm
 (5.0 cm)
k
The amplitude of the wave is A = 0.80 mm. The first graph (solid) begins at the origin. The second graph
(dashed) is shifted to the right by (5.0 cm  0.50)   0.80 cm. The third graph (dotted) is shifted to the right
twice as far as the second graph, or 1.6 cm.
The graphs are shown:
y (mm)
t = 0.96 s
0.80
t = 1.92 s
10
0
x (cm)
5.0
t=0
0.80
(b) Strategy Substitute t = 0, 0.96 s, and 1.96 s into y ( x, t )  (0.50 mm) sin[( 5.0 cm 1 ) x  ( 6.0 s 1 )t ] and
graph the resulting equations.
Solution The three equations are:
y ( x, 0)  (0.50 mm) sin[( 5.0 cm 1 ) x], y ( x, 0.96 s)  (0.50 mm) sin[( 5.0 cm 1 ) x  0.50], and
y ( x, 1.92 s)  (0.50 mm) sin[( 5.0 cm 1 ) x  1.0].
Find the wavelength.
2
2


 10 cm
 (5.0 cm)
k
The amplitude of the wave is A = 0.50 mm. The first graph (solid) begins at the origin. The second graph
(dashed) is shifted to the left by (5.0 cm  0.50)   0.80 cm. The third graph (dotted) is shifted to the left
twice as far as the second graph, or 1.6 cm. The graphs are shown:
y (mm)
t = 0.96 s
0.50
t=0
5.0
t = 1.92 s
0
10
x (cm)
0.50
(c) Strategy Refer to the results of parts (a) and (b).
Solution The graphs obtained in part (a) move to the right as time progresses, so
y ( x, t )  (0.80 mm)sin (kx  t ) represents a wave traveling in the +x-direction. The graphs obtained in part
(b) move to the left as time progresses, so y ( x, t )  (0.50 mm)sin (kx  t ) represents a wave traveling in the
–x-direction.
519
Chapter 11: Waves
College Physics
30. (a) Strategy and Solution Since the argument of the cosine function is  t  kx (both terms are positive), the
wave is moving to the left.
(b) Strategy The maximum y-value is the amplitude. The wave repeats every 4.0 cm, so the wavelength is
0.040 m. Use Eq. (11-7) to find the angular frequency and wavenumber.
Solution
ymax  2.0 mm, so
A  2.0 mm ;  
2
2

 160 rad m
k
 0.040 m
2 v


2 (10.0 m s)
 1600 rad s ;
0.040 m
(c) Strategy Choose the point (x, y) = (0, 0). 0  A cos[t  k (0)], so 0  cos t. cos t  0 when t  n 2,
where n is an odd integer. The smallest nonnegative n is 1 and it will give the smallest nonnegative time. Use
Eq. (11-6).
Solution
t

2


4 f

1

0.040 m


 1.0 ms
4 f 4v 4 10.0 m s 
The period is T 
1  0.040 m
 
 4.0 ms, so the three times are 1.0 ms, 5.0 ms, and 9.0 ms.
f v 10.0 m s
31. Strategy Compute the positions of the peaks of each pulse for the given times. Then use the principle of
superposition to graph the shape of the cord for each time.
Solution
t (s)
Short Pulse Position
Tall Pulse Position
0.15
10 cm  (40 cm s)(0.15 s)  16 cm
30 cm  (40 cm s)(0.15 s)  24 cm
0.25
10 cm  (40 cm s)(0.25 s)  20 cm
30 cm  (40 cm s)(0.25 s)  20 cm
0.30
10 cm  (40 cm s)(0.30 s)  22 cm
30 cm  (40 cm s)(0.30 s)  18 cm
y (cm)
1.5
y (cm)
1.5
1.0
1.0
t = 0.25 s
t = 0.15 s
0.5
0
0.5
10
20
30
40 x (cm)
20
30
40 x (cm)
0
y (cm)
1.5
t = 0.30 s
1.0
0.5
0
10
520
10
20
30
40 x (cm)
College Physics
Chapter 11: Waves
32. Strategy Compute the positions of the peaks of each pulse for the given times. Then use the principle of
superposition to graph the shape of the cord for each time.
Solution
t (s)
Positive Pulse Position
Negative Pulse Position
0.60
2.0 m  (2.5 m s)(0.60 s)  3.5 m
6.0 m  (2.5 m s)(0.60 s)  4.5 m
0.80
2.0 m  (2.5 m s)(0.80 s)  4.0 m
6.0 m  (2.5 m s)(0.80 s)  4.0 m
0.90
2.0 m  (2.5 m s)(0.90 s)  4.3 m
6.0 m  (2.5 m s)(0.90 s)  3.8 m
y (cm)
10.0
y (cm)
10.0
t = 0.60 s
5.0
0
2
4
6
t = 0.80 s
5.0
8
0
x (m)
5.0
5.0
10.0
10.0
2
4
6
8
x (m)
y (cm)
10.0
t = 0.90 s
5.0
0
2
6
8
x (m)
5.0
10.0
33. Strategy Sketch the sine waves. Use the principle of superposition to find the amplitudes. Let y1  A sin( t  kx)
   
   
and y2  A sin( t  kx   ) and use the trigonometric identity sin   sin   2sin 
 cos  2  .
2




Solution
(a)
y (cm)
4
2
0
2
60°
180°
300°
420°
θ
4
Use the principle of superposition.
 



y  y1  y2  A sin( t  kx )  A sin( t  kx   )  2 A sin   t  kx   cos  A sin   t  kx  
2
2
2



60.0
 6.9 cm .
where A  2 A cos  2(4.0 cm) cos
2
2
521
Chapter 11: Waves
(b)
College Physics
y (cm)
4
2
0
2
60°
180°
300°
420°
θ
4
A  2(4.0 cm) cos
90.0
 5.7 cm
2
34. Strategy Let y1  A sin( t  kx) and y2  A sin( t  kx   ) and use the trigonometric identity
   
   
sin   sin   2sin 
 cos  2  . Use the principle of superposition.
2




Solution Find the traveling wave.





y  y1  y2  A sin( t  kx)  A sin( t  kx   )  2 A sin   t  kx   cos  A sin   t  kx  
2
2
2



where A  2 A cos
2 A cos

2

2
 A. Find  .
 A, so   2 cos 1
1
 120 .
2
35. Strategy Use the principle of superposition and the trigonometric identity
   
   
sin   sin   2sin 
 cos  2  .
2




Solution Find the traveling sine wave.





y  y1  y2  A sin( t  kx)  A sin( t  kx   )  2 A sin   t  kx   cos  A sin   t  kx  
2
2
2


where A  2 A cos
cos

2


2
 6.69 cm. Find  .
A
A
6.69 cm
 2 cos 1
 96.0 .
, so   2 cos 1
2A
2A
2(5.00 cm)
36. Strategy f  v  and the frequency is the same in both mediums.
Solution Find the wavelength of the light in water.
va vw
v

, so w  w a  0.750(0.500  106 m)  375 nm .
va
a w
37. Strategy Refer to the figure. Use x  vx t and the principle of superposition.
Solution The pulse moves 1.80 m  1.50 m = 0.30 m in 0.20 s. So, the speed of the wave is
0.30 m
v
 1.5 m s. When the pulse reaches the right endpoint, it is reflected and inverted. When exactly half
0.20 s
of the pulse has been reflected and inverted, the superposition of the incident and reflected waves results in the
x 4.0 m  1.5 m
cancellation of the waves ( y1  y2  0). Thus, the string looks flat at t  
 1.7 s .
v
1.5 m s
522
College Physics
Chapter 11: Waves
38. Strategy Refer to the figure. Use x  vx t.
Solution The pulse moves 1.80 m  1.50 m = 0.30 m in 0.20 s. So, the speed of the wave is
0.30 m
v
 1.5 m s. The pulse travels to the right until it reaches the endpoint; it is then reflected and inverted.
0.20 s
It then travels to the left until it hits the left endpoint; it is again reflected and inverted. The pulse travels to the
right until it reaches x = 1.5 m. The total distance traveled is 2.5 m + 4.0 m + 1.5 m = 8.0 m. The elapsed time is
x
8.0 m
t 
 5.3 s .
v 1.5 m s
39. Strategy The waves are coherent. Use the principle of superposition.
Solution
(a) The resulting wave will have its largest amplitude if the waves interfere constructively. The phase difference
is 0 , and the amplitude is A1  A2  5.0 cm  3.0 cm  8.0 cm .
(b) The resulting wave will have its smallest amplitude if the waves interfere destructively. The phase difference
is 180 , and the amplitude is A1  A2  5.0 cm  3.0 cm  2.0 cm .
(c) 8.0 cm : 2.0 cm  4 :1
40. Strategy The waves are coherent. Use the principle of superposition. Intensity is proportional to the square of the
amplitude.
Solution
(a) The resulting wave will have its highest intensity when the waves interfere constructively. The phase
difference is 0 , and the amplitude is A  A1  A2  6.0 cm  3.0 cm  9.0 cm .
(b) The resulting wave will have its lowest intensity when the waves interfere destructively. The phase difference
is 180 , and the amplitude is A  A1  A2  6.0 cm  3.0 cm  3.0 cm .
(c) Form a proportion.
2
2
I1  A1 
 9.0 cm 
  
  9.0
I 2  A2 
 3.0 cm 
The ratio is 9 :1 .
523
Chapter 11: Waves
College Physics
41. Strategy Intensity is proportional to the amplitude squared. For constructive interference, the amplitude of the
superposition is the sum of the original amplitudes.
Solution Find A1/A2.
A1
I
25
5.0
 1 

A2
I2
15
3.0
Find the amplitude of the superposition.

5.0
5.0 
A  A1  A2  A2
 A2  A2  1 

3.0
3.0 

Find the intensity of the superposition.
2
2


I
A
5.0
5.0 
5.0 
2
2

 1
, so I  1 
 I 2   1 
 (15 mW m )  79 mW m .
I 2 A2
3.0
3.0
3.0




42. Strategy Intensity is proportional to the amplitude squared. For destructive interference, the amplitude of the
superposition is the absolute value of the difference of the original amplitudes.
Solution Find A1/A2.
A1
I
25
 1 
A2
I2
28
Find the amplitude of the superposition.

25 
A  A1  A2  A2 1 

28 

Find the intensity of the superposition.
2
2


I
A
25
25 
25 
3
2
2
, so I  1 

 1
 I 2  1 
 (28 10 W m )  80 W m .
I 2 A2
28
28
28




43. Strategy Intensity is proportional to the amplitude squared. For constructive interference, the amplitude of the
superposition is the sum of the original amplitudes. For destructive interference, the amplitude of the
superposition is the absolute value of the difference of the original amplitudes. For incoherent waves, the
intensities add.
Solution
(a) Find A1/A2.
A1
I
0.040 2.0
 1 

A2
I2
0.090 3.0
Find the amplitude of the superposition.
 2.0 
 2.0 
A  A1  A2  A2 
  A2  A2 1 

3.0


 3.0 
Find the intensity of the superposition.
2
2
I
A
2.0
 2.0 
 2.0 
2
2

 1
, so I  1 
 I 2  1 
 (0.090 W m )  0.25 W m .
3.0
I 2 A2
 3.0 
 3.0 
524
College Physics
Chapter 11: Waves
(b) Find the amplitude of the superposition.
 2.0 
A  A1  A2  A2  1 

 3.0 
Find the intensity of the superposition.
2
2
 2.0 
 2.0 
2
2
I  1 
 I 2  1 
 (0.090 W m )  0.010 W m
 3.0 
 3.0 
(c) Find the intensity of the superposition.
I  I1  I 2  0.040 W m 2  0.090 W m 2  0.130 W m 2
44. Strategy The intensity minimums imply that the distance 37.1 m  25.8 m  11.3 m is equal to a whole number
of wavelengths m plus one-half wavelength. Determine the number (or numbers) of wavelengths m that gives a
frequency (or frequencies) between 100 Hz and 150 Hz. Use Eq. (11-6).
Solution The number of wavelengths m is related to the distance between intensity minimums by
1
v 
1  343 m s

 (30.35 Hz)m  15.18 Hz.
 m     11.3 m. In terms of frequency, we have f    m  
2
2  11.3 m




Substitute 100 Hz and 150 Hz for f and solve for m to find the range of possible values.
(30.35 Hz)m  15.18 Hz  100 Hz, so m  2.8 and (30.35 Hz)m  15.18 Hz  150 Hz, so m  4.4.
The two possible values of m are 3 and 4. Try them both.
1  343 m s
1  343 m s


f (3)   3  
 106 Hz and f (4)   4  
 137 Hz.
2  11.3 m
2  11.3 m


Both values are within the range of allowed frequencies, so the possible frequencies of the sound waves coming
from the speakers are 106 Hz and 137 Hz.
45. Strategy Use Eqs. (11-2) and (11-13).
Solution
2
 f 
T1
T
1 T1L
v
TL

, so  2   2 .
and v 
. f1 
2L
m
2L m
4 Lm
T1
 f1 
Calculate the percentage reduction in the tension.
f1 
  T 100%  1   f
f
T1  f2
T1  T2
1
 100% 
T1
T1
2
1




f
 1
2
2
2

1  0.040  
  100%  1  
   100%  7.8%
1

 
 

46. Strategy According to Eq. (11-13), the fundamental frequency of a guitar string is directly proportional to the
speed of a wave on the string. According to Eq. (11-2), the speed of a wave on the string in directly proportional
to the square root of the tension in the string.
Solution f  v and v  T , so f  T , where T is the tension. Therefore, since 1.15  1.07,
the frequency increases by 7% .
47. Strategy Nodes are separated by a distance of  /2. Use Eq. (11-7).
Solution Find the distance between adjacent nodes.
 2


 0.016 m
distance  
2 2k 2.0  102 rad m
525
Chapter 11: Waves
College Physics
48. Strategy The frequencies are given by f n  nv (2 L). The speed of the transverse waves is related to the tension
by v  T  .
Solution
(a) Find the speed of the transverse waves.
v
 f1 , so v  2 Lf1  2(1.50 m)(450.0 Hz)  1350 m s .
2L
(b) Find the tension.
v
1 T

, so T  4 L2 f12  4(25.0  106 kg m)(1.50 m)2 (450.0 Hz)2  45.6 N .
f1 
2L 2L 
(c) The frequencies are the same for both mediums, but the wavelength depends upon the wave speed.
v 340 m s
 0.76 m.
f  f1  450.0 Hz, so   
f 450.0 Hz
The wavelength and frequency are 0.76 m and 450.0 Hz , respectively.
49. Strategy The frequencies are given by f n  nv (2 L). The speed of the transverse waves is related to the tension
by v  T  .
Solution
(a) Find the frequency of the fundamental oscillation.
v
1 T
1
12 N


 33 Hz
f1 
2 L 2 L  2(1.5 m) 1.2  103 kg m
(b) Find the tension.
4 L2 f32 4(1.2  103 kg m)(1.5 m) 2 (0.50  103 Hz) 2
3v
3 T
f3 
, so T 


 300 N .
2L 2L 
9
9
50. Strategy Use Newton’s second law and Eqs. (11-4) and (11-13).
Solution Find the tension in the string.
Fy  T  mg  0, so T  mg .
T
T
Find the speed of waves on the string.
T
mg
v


2.00 m

mg
Find the fundamental frequency.
f1 
v
1 mg
1
(2.20 kg)(9.80 m s 2 )


 616 Hz
2L 2L 
2(2.00 m)
3.55  106 kg m
526
College Physics
Chapter 11: Waves
51. Strategy The frequencies are given by f n  nv (2 L). The speed of the transverse waves is related to the mass per
unit length by v  T  .
Solution Find the mass per unit length of the guitar string.
v
T
1 T
82 N
f1 
, so   2 2 

 4.5  104 kg m .
2
2
2L 2L 
4L f1
4(0.65 m) (329.63 Hz)
52. Strategy The fundamental frequency is given by f n  nv (2L) with n = 1. Use the definition of average speed.
Solution Find the fundamental frequency.
2.0 m
f1 
v
 0.050 s  10 Hz
2 L 2(2.0 m)
53. (a) Strategy and Solution All frequencies higher than the fundamental are integral multiples of the
fundamental. Since there are no other frequencies between the two given, the fundamental is the difference
between those two. Thus, the fundamental frequency is 1040 Hz  780 Hz  260 Hz .
(b) Strategy Use Eqs. (11-2) and (11-13).
Solution Find the total mass of the string.
v
F
F
1 FL
1200 N



 2.8 g .
f1 
, so m 
2
2L 2L m
4mL
4 f1 L 4(260 Hz)2 (1.6 m)
54. Strategy The weight of the weight equals the tension in the string. Use Eqs. (11-4) and (11-13) to find the
required weight of the weight.
Solution Find the weight F.
fn 
F
nv
n

2L 2L
4 L2 f n 2
n2
F



n2 F
4  L2
, so
4(0.120  103 kg m)(0.42 m) 2 (110 Hz) 2
n2
 1.0 n 2 N, where n  1, 2, 3, ... .
55. Strategy Use Eqs. (11-2) and (11-13).
Solution
nv
TL
fn 
and v 
. Find the total mass of the wire.
m
2L
f1 
v
1 TL


2L 2L m
T
T
300.0 N
, so m 

 0.050 kg .
2
4 Lm
4 Lf1
4(2.0 m)(27.5 Hz)2
527
Chapter 11: Waves
College Physics
56. (a) Strategy Replace each quantity with its SI units.
Solution Show that
FL
has units
m
FL m has units of speed.
Nm

kg
kg  m s 2  m
 m s , which are the units of speed.
kg
(b) Strategy Set [ F ]a [ L]b [m]c equal to m s , the units of speed, and determine a, b, and c.
Solution Show that no combination of L, m, and F other than
FL m has units of speed.
[ F ]a [ L]b [m]c  m  s 1
N a mb kg c 
kg a  m a
s
2a
mb kg c 
kg a  c m a b s2 a  kg 0  m1  s 1
Equate exponents.
1
1
1
a + c = 0, so a = c. a + b = 1, so b = 1  a = 1 + c. 2a = 1, so a  , c   , and b  .
2
2
2
1/ 2 1/ 2 1/ 2
The only combination of F, L, and m that gives units of speed is F L m
 FL / m ; therefore, the
speed of transverse waves on the string can only be
FL / m times some dimensionless constant.
57. Strategy The wave speed for the 1.0-Hz waves is twice that for the 2.0-Hz waves, so it takes the 1.0-Hz waves
120 s to reach you. (120 s + 120 s = 240 s is the time it takes the 2.0-Hz waves to reach you; twice as long.)
Solution Compute the distance to the boat.
x  vt  (1.56 m s)(120 s)  190 m
58. Strategy The wave is harmonic; y( x, t )  (1.2 cm)sin[(0.50 rad/s)t  (1.00 rad/m)x]. Use Eqs. (10-21) and
(10-22). Plot the graphs.
Solution Find the maximum velocity and maximum acceleration for a point on the string.
vm   A  (0.50 rad s)(1.2 cm)  1.9 cm s and am   2 A  (0.50 rad s) 2 (1.2 cm)  3.0 cm s 2 .
v y leads y by a quarter cycle, so v y is a cosine function.
v y  (1.9 cm s) cos[(0.50 rad s)t  (1.00 rad m) x]
a y leads v y by a quarter cycle, so a y leads y by a half cycle; a y is a negative sine function.
a y  (3.0 cm s 2 ) sin[(0.50 rad s)t  (1.00 rad m) x]
The period is T 
y (cm)
1.2
0
1.2
2


2
 4.0 s.
0.50 rad s
vy (cm/s)
1.9
x=0
3
6
t (s)
ay (cm/s2)
3.0
x=0
0
3
6
t (s)
0
3.0
1.9
528
x=0
3
6
t (s)
College Physics
Chapter 11: Waves
59. Strategy Use Eq. (11-6).
Solution Find the wavelength of the radio waves.
v 3.0  108 m s
 3.3 m
 
f
90  106 Hz
60. (a) Strategy In a longitudinal wave, the motion of particles in the medium is along the same line as the direction
of propagation of the wave.
Solution
A given particle will oscillate sinusoidally in the  y -direction about its equilibrium position with an
amplitude of 5.0 cm and a period of 1/8 s.
[T   v  10 cm (80 cm s )  1 8 s].
(b) Strategy In a transverse wave, the motion of particles in the medium is perpendicular to the direction of
propagation of the wave.
Solution If the wave under consideration were transverse,
the particles would oscillate along a direction perpendicular to the y -axis .
61. Strategy Speed is inversely proportional to the time of travel. Form a proportion and use x  vt.
Solution Relate the speeds to the times of travel.
vP 10.0 km s 5.00 tS



vS
8.0 km s
4.0 tP
Find the time for the S wave to travel from the source to the detector.
2.0 s
 4.0 
t  tS  tP  tS  tS 
  tS (1  0.80), so tS  0.20  10 s.
5.00


Calculate the distance between the source and the detector.
d  vStS  (8.0 km s)(10 s)  80 km
62. Strategy Refer to the figure to determine the frequency; count the number of cycles during the indicated time
period. Use Eq. (11-6) and the definition of average speed.
Solution Estimate the wavelength of the seismic waves.
180,000 m

v
s
 4.030.0
 2.4 km
cycles
f
2.6 s 1.0 s
529
Chapter 11: Waves
College Physics
63. Strategy According to Eq. (11-13), the frequency of the string is inversely proportional to the length of the string.
Solution Form a proportion.
1
L
f
f  , so 2  1 .
L
L1
f2
Relate the distance between frets to the frequencies and the length of the string.

f
f 
L  L1  L2  L1  1 L1  L1 1  1 
f2
f
2

1 

First fret: L  (64.8 cm) 1 
  3.64 cm
1.0595


1 

Second fret: 3.64 cm  (64.8 cm  3.64 cm) 1 
  7.07 cm
1.0595


1 

Third fret: 7.074 cm  (64.8 cm  7.074 cm) 1 
  10.32 cm
 1.0595 
64. Strategy Use Eq. (11-13).
Solution Find the lowest standing wave frequencies in each situation.
(a) Since f n 
v
nv
and f1 
, f n  nf1  n(300.0 Hz).
2L
2L
n
nf1
2
600.0 Hz
3
900.0 Hz
4
1.200 kHz
(b) The lowest frequency is now f 2 , and only even harmonics are allowed (always a node at the center).
n
nf1
2
600.0 Hz
4
1.200 kHz
6
1.800 kHz
8
2.400 kHz
(c) The effective length of the string is now half of the original length.
nv
nv
nv
fn 
, so f n 

 2 f n.
2L
2( L 2) L
n
f n  2 f n
1
600.0 Hz
2
1.200 kHz
3
1.800 kHz
4
2.400 kHz
530
College Physics
Chapter 11: Waves
65. Strategy The fundamental frequency depends on the wavelength of the fundamental and the wave speed. The
wavelength is determined by the length of the wire. The wave speed depends on the tension and the linear mass
density. So we need to find out how those three quantities (length, tension, linear mass density) change when the
wire is cut in half and then find the new fundamental frequency. The only numerical value we know is the original
frequency, so we work by proportions. According to Eq. (11-4), the speed of waves on a wire is directly
proportional to the square root of the tension. According to Eq. (11-13), the frequency of the waves on a wire is
directly proportional to the speed of the waves. Therefore, the frequency is directly proportional to the square root
of the tension in a wire. Use Newton’s second law.
Solution
T1
For a wire fixed at both ends, the wavelength of the
T2
T2
fundamental is 2L. When the wire is cut in half, then,
1
the wavelength is cut in half: f  i . The linear
2
mg
mass density does not change (half the length and half
mg
the mass), but the tension does (two wires supporting
the sign instead of one).
For the single wire supporting the sign (see FBD): Fy  Ti  mg  0, so Ti  mg .
For the two wires: Fy  2Tf  mg  0, so Tf 
1
1
mg  T .
2
2 i
The frequency is f = v/λ and the wave speed is v =
f f i

fi f
F /  , so f 
1
F


. Form a proportion:
Tf i
1
 2 1  2, so f f  2 f1  2  660 Hz = 930 Hz
T1 f
2
66. (a) Strategy For a harmonic wave moving to the left (x-direction),
2 
 2
y  A sin(t  kx)  A sin 
t
x .
T
 

Solution Write an equation for the surface seismic waves.
2
 2

y( x, t )  (0.020 m) sin 
t
x  , so y( x, t )  (0.020 m) sin[(1.6 rad/s)t  (0.0016 rad/m) x] .
3
4.0
s
4.0  10 m 

(b) Strategy Use Eq. (10-21).
Solution Find the maximum speed of the ground as the waves move by.
2 A 2 (0.020 m)
vm   A 

 0.031 m s
T
4.0 s
(c) Strategy Use Eq. (11-6).
Solution Find the wave speed.
 4.0 km
vf  
 1.0 km s
T
4.0 s
67. (a) Strategy Use Hooke’s law.
Solution Explain why the tension in the spring is approximately proportional to the length.
Hooke’s law: T  k ( x  x0 )  kx for x  x0.
531
Chapter 11: Waves
College Physics
(b) Strategy Use Eq. (11-2) and the result of part (a).
Solution Find the time it takes the wave to travel the length of the spring.
v
TL

m
kL2
L
since T  kx  kL, and L  vt , so

m
t
kL2
or t 
m
m
.
k
Since neither m nor k change, the increase in length does not affect the time of travel, so t  4.00 s .
68. Strategy Use dimensional analysis.
Solution  has units m. g has units m s 2 .   g has units m 2 s 2 .
 g has units m s. So, v   g .
69. Strategy Use dimensional analysis.
Solution  has units N m  kg s 2 .  has units kg m3 .  has units m.  (   ) has units
kg s 2
m2
. [ (   )]1/ 2 has units m s. So, v 

m  kg m3 s 2

. Since v depends upon  , surface waves are

dispersive.
70. Strategy Use the properties of traveling waves to answer the questions concerning the seismic wave described by
the equation y ( x, t )  (7.00 cm) cos[(6.00 rad cm) x  (20.0 rad s)t ].
Solution
(a) Because kx and t have the same sign, the wave is moving to the left .
(b) The particles in the medium move a distance from their equilibrium positions equal to the amplitude of the
wave. Thus, they move 7.00 cm .
(c) The frequency of this wave is f 
 20.0 s 1

 10.0 Hz .
2
2
(d) The wavelength of this wave is  
(e) The wave speed is v 

k

2
2

 0.333 cm .
k
6.00 cm 1
20.0 s 1
6.00 cm 1
 3.33 cm s .
(f) A particle that is at y  7.00 cm and x  0 when t  0
oscillates sinusoidally along the y -axis about y = 0 with an amplitude of 7.00 cm .
(g) Since the motion of particles in the medium is perpendicular to the direction of propagation of the wave, the
wave is a transverse wave.
71. Strategy and Solution Refer to the figure.
(a) Since the wave is moving to the left, the peak of the pulse has not yet reached the position of point A.
So, point A is moving upward.
532
College Physics
Chapter 11: Waves
(b) The peak has passed point B, so point B is moving downward.
(c) The slope of the string is larger at point A than at point B, so the speed of the string segment is larger at
point A.
72. Strategy The position of the particle will follow the shape of the wave. The velocity is a bit more complicated.
As the wave passes the point under consideration, the particle moves upward rapidly until it reaches the “top” of
the wave. When it reaches the top, its velocity is instantaneously zero. The point then moves downward at an
average velocity less than that with which it rose. The velocity is equal to the slope of the position graph for any
time t.
Solution The plot of the position as a function of time:
x
t
The plot of the velocity as a function of time:
vx
t
73. Strategy Destructive interference occurs when the path length difference of the two sound waves is an odd
multiple of half of the wavelength.
Solution The wavelength is   v / f  (340 m/s)/(680 Hz)  0.50 m. The largest possible path length difference
is equal to the distance between the speakers, 1.5 m.  / 2  0.25 m, so the path length differences that cause
destructive interference are 0.25 m, 0.75 m, and 1.25 m. Let the speakers lie along the x-axis at x  0.75 m.
Then the path length difference is zero along the y-axis and 1.5 m along the x-axis. As the listener walks along the
circle of radius 1 m, the path length difference varies from 0 to 1.5 m. The path length difference equals 0.25 m,
0.75 m, and 1.25 m once for each quadrant of the circle (three occurrences of destructive interference). There are
four quadrants, so the listener observes destructive interference at 12 points along the circle.
74. (a) Strategy Substitute t = 0, 1.0 s, and 2.0 s into


y ( x, t )  (5.0 cm) sin[( 5.0 cm 1 ) x  ( 6.0 rad s)t ]  sin[( 5.0 cm 1 ) x  ( 6.0 rad s)t ]
and graph the resulting equations. Use Eq. (11-7) to find the wavelength.
Solution The three equations are:


y ( x, 0)  (5.0 cm) sin[( 5.0 cm 1 ) x]  sin[( 5.0 cm 1 ) x]  (10 cm) sin[( 5.0 cm 1 ) x],


y ( x, 2.0 s)  (5.0 cm) sin[( 5.0 cm 1 ) x  1.05]  sin[( 5.0 cm 1 ) x  1.05] .
y ( x, 1.0 s)  (5.0 cm) sin[( 5.0 cm 1 ) x  0.52]  sin[( 5.0 cm 1 ) x  0.52] , and
Find the wavelength.
2
2

 10 cm

 (5.0 cm)
k
The amplitude of the first (t = 0) wave is A = 10 cm. The second wave (t = 1.0 s) has a smaller amplitude than
533
Chapter 11: Waves
College Physics
the first, due to the opposite phase shifts. The third wave is even smaller in amplitude. The graphs are shown.
y (cm)
10
t=0
t = 1.0 s
0
5.0
10
x (cm)
t = 2.0 s
10
(b) Strategy and Solution The wave doesn’t “travel” anywhere—it just oscillates up and down—so this is a
standing wave.
75. Strategy Refer to Problem 78. The function is y  A[sin(kx  t )  sin(kx  t )]. Use the trigonometric identity
sin   sin   2sin[(   ) 2]cos[(   ) 2] and the principle of superposition.
Solution Use the identity.
y  A[sin(kx  t )  sin(kx   t )]  A sin( t  kx)  A sin( t  kx)
 kx  t  kx  t 
 kx  t  (kx   t ) 
 2 A sin 
 cos 
  2 A sin( kx) cos( t )  [2 A cos(t )]sin(kx)  A sin(kx)
2
2




Therefore, A  2 A cos(t ). Prove this holds for each of the amplitudes of the graphs from Problem 79.


y ( x, t )  (5.0 cm) sin[( 5.0 cm 1 ) x  ( 6.0 rad s)t ]  sin[( 5.0 cm 1 ) x  ( 6.0 rad s)t ]
1
 2(5.0 cm) sin[( 5.0 cm ) x]cos[( 6.0 rad s)t ]
 (10 cm) sin[( 5.0 cm 1 ) x]cos[( 6.0 rad s)t ]
So the maximum amplitude is 10 cm for x = 2.5 cm.
At t = 1.0 s,
y (2.5 cm, 1.0 s)  (10 cm) sin[( 5.0 cm 1 )(2.5 cm)]cos[( 6.0 rad s)(1.0 s)]  (10 cm)(1) cos( 6.0)  8.7 cm.
At t = 2.0 s,
y (2.5 cm, 2.0 s)  (10 cm) sin[( 5.0 cm 1 )(2.5 cm)]cos[( 6.0 rad s)(2.0 s)]  (10 cm)(1) cos( 3.0)  5.0 cm.
The values correspond nicely with those shown in the graphs in the solution for Problem 79.
Using the original function, we have
y (2.5 cm, 1.0 s)  (5.0 cm){sin[( 5.0 cm 1 )(2.5 cm)  ( 6.0 rad s)(1.0 s)]
 sin[( 5.0 cm 1 )(2.5 cm)  ( 6.0 rad s)(1.0 s)]}  8.7 cm
and
y (2.5 cm, 2.0 s)  (5.0 cm){sin[( 5.0 cm 1 )(2.5 cm)  ( 6.0 rad s)(2.0 s)]
 sin[( 5.0 cm 1 )(2.5 cm)  ( 6.0 rad s)(2.0 s)]}  5.0 cm.
Therefore, the amplitudes of the graphs of Problem 79 satisfy the equation A  2 A cos(t ), where A is the
amplitude of the wave plotted and A is 5.0 cm.
534
College Physics
Chapter 11: Waves
76. Strategy Use the principle of superposition.
Solution x  1.80 m  1.50 m  0.30 m in t  0.20 s, so v  0.30 m (0.20 s)  1.5 m s.
Find the position of the peak at t = 2.2 s.
xpeak  1.5 m  (1.5 m s)(2.2 s)  4.8 m (3.2 m; 4.8 m  4.0 m = 0.8 m to the left)
The peak of the pulse is now inverted due to reflection.
y (cm)
10
5
0
x = 4.0 m
x=0
5
10
x = 3.2 m
77. Strategy Use the principle of superposition.
Solution x  1.80 m  1.50 m  0.30 m in t  0.20 s, so v  0.30 m (0.20 s)  1.5 m s.
Find the position of the peak at t = 1.6 s.
xpeak  xi  vt  1.5 m  (1.5 m s)(1.6 s)  3.9 m
The peak of the pulse is nearly to the end of the string. The reflected pulse is below the string, so most of the
height of the original pulse is cancelled.
y (cm)
10
5
0
x = 3.9 m
x=0
x = 4.0 m
535