1. Chapter 21. Wave Motion Physics, 6th Edition
Chapter 21. Wave Motion
Mechanical Waves
21-1. A transverse wave has a wavelength of 30 cm and vibrates with a frequency of 420 Hz.
What is the speed of this wave? [ λ = 30 cm = 0.30 m ]
v = fλ = (420 Hz)(0.30 m) ; v = 126 m/s
21-2. A person on a pier counts the slaps of a wave as the crests hit a post. If 80 slaps are heard
in one minute and a particular crest travels a distance of 8 m in 4 s, what is the length of a
single wave?
s 8m 80 waves
v= = = 2 m/s; f = = 1.33 Hz
t 4s 60 s
v 2 m/s
λ= = ; λ = 1.5 m
f 1.33 Hz
21-3. A transverse wave is pictured in Fig. 21-13. Find the amplitude, wavelength, period, and
speed of the wave if it has a frequency of 12 Hz. [ A = 12 cm, λ = 28 cm ]
From the figure: A = 12 cm, λ = 28 cm;
v = fλ = (12 Hz)(0.28 m); v = 3.36 m/s
1 1
T= = ; T = 0.0833 s
f 12 Hz
21-4. For the longitudinal wave in Fig. 21-13, find the amplitude, wavelength, period, and speed
of the wave if it has a frequency of 8 Hz. If the amplitude were doubled, would any of the
other factors change?
From figure: A = 12 cm and λ = 28 cm
1 1
v = fλ = (8 Hz)(0.28 m); v = 2.24 m/s; T= = ; T = 0.125 s
f 8 Hz
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2. Chapter 21. Wave Motion Physics, 6th Edition
21-5. A 500-g metal wire has a length of 50 cm and is under a tension of 80 N. What is the
speed of a transverse wave in the wire?
Fl (80 N)(0.50 m)
v= = ; v = 8.94 m/s
m 0.50 kg
21-6. If the wire in Problem 21-5 is cut in half, what will be its new mass? Show that the speed
of the wave is unchanged? Why?
0.500 kg Fl (80 N)(0.250 m)
m= ; m = 0.250 kg; v= = = 8.94 m/s
2 m 0.250 kg
The speed is the same, because linear density m/l is not changed.
21-7. A 3-m cord under a tension of 200 N sustains a transverse wave speed of 172 m/s. What
is the mass of the rope?
Fl Fl Fl (200 N)(3 m)
v= ; v2 = ; m= = ; m = 0.0203 kg
m m v2 (172 m/s) 2
21-8. A 200-g cord is stretched over a distance of 5.2 m and placed under a tension of 500 N.
Compute the speed of a transverse wave in the cord?
Fl (500 N)(5.2 m)
v= = ; v = 114 m/s
m 0.200 kg
21-9. What tension is needed to produce a wave speed of 12 m/s in a 900-g string that is 2 m
long?
mv 2 (0.900 kg)(12 m/s) 2
F= = ; F = 64.8 N.
l 2m
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3. Chapter 21. Wave Motion Physics, 6th Edition
21-10. A wooden float at the end of a fishing line makes eight complete oscillations in 10 s. If it
takes 3.60 s for a single wave to travel 11 m, what is the wavelength of the water waves?
8 waves 11 m
f = = 0.800 Hz; v= = 3.06 m/s;
10 s 3.6 s
v 3.06 m/s
λ= = ; λ = 3.82 m
f 0.800 Hz
*21-11. What frequency is required to cause a rope to vibrate with a wavelength of 20 cm when
it is under a tension of 200 N. Assume the linear density of the rope to be 0.008 kg/m.
v 1 F 1 (200 N)
f = ; f = = ; f = 791 Hz
λ λ λ (0.20 m) 0.008 kg/m
*21-12. A tension of 400 N causes a 300-g wire of length 1.6 m to vibrate with a frequency of
40 Hz. What is the wavelength of the transverse waves?
Fl (400 N)(1.6 m) v 46.2 m/s
v= = = 46.2 m/s; λ= = ; λ = 1.15 m
m 0.30 kg f 40 Hz
*21-13. A horizontal spring is jiggled back and forth at one end by a device that makes 80
oscillations in 12 s. What is the speed of the longitudinal waves if condensations are
separated by 15 cm as the wave progresses down the spring?
80 waves
f = = 6.67 Hz; v = fλ = (6.67 Hz)(0.15 m); v = 1.00 m/s
12 s
35

4. Chapter 21. Wave Motion Physics, 6th Edition
Energy of a Periodic Wave
21-14. A 2-m length of string has a mass of 300 g and vibrates with a frequency of 2 Hz and an
amplitude of 50 mm. If the tension in the rope is 48 N, how much power must be
delivered to the string?
m 0.30 kg F (48 N)
µ= = = 0.150 kg/m; v= = = 17.9 m/s
l 2m µ 0.15 kg/m
P = 2π2f2A2µ v = 2π2(2 Hz)2(0.05 m)2(0.15 kg/m)(17.9 m/s); P = 0.530 W
21-15. An 80-g string has a length of 40 m and vibrates with a frequency of 8 Hz and an
amplitude of 4 cm. Find the energy per unit of length passing along the string?
E 0.08 kg
= 2π 2 f 2 A2 µ = 2π 2 (8 Hz) 2 (0.04 m) 2 ; E/l = 4.04 x 10-3 J/m
l 40 m
21-16. If the wavelength of the transverse wave in Problem 21-11 is 1.6 m, what power is
supplied by the source?
E
P = 2π 2 f 2 A2 µ v =   v; v = f λ = (8 Hz)(1.6 m) = 12.8 m/s
l 
E
P =   v = (4.04 x 10-3 J/m)(12.8 m/s); P = 5.17 x 10-2 W
l 
*21-17. A 300-g string has a length of 2.50 m and vibrates with an amplitude of 8.00 mm. The
tension in the string is 46 N. What must be the frequency of the waves in order that the
average power be 90.0 W? [ P = 2π2f2A2µ v ]
m 0.300 kg Fl 46 N
µ= = = 0.120 kg/m; v= = = 19.6 m/s
l 2.50 m m 0.120 kg/m
P 90.0 W
f = = ; f = 174 Hz
2π A µ v
2 2
2π (0.008 m) (0.120 kg/m)(19.6 m/s)
2 2
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5. Chapter 21. Wave Motion Physics, 6th Edition
Standing Wave and Characteristic Frequencies
21-18. A string vibrates with a fundamental frequency of 200 Hz. What is the frequency of the
second harmonic and of the third overtone?
f n = nf1 ; f 2 = 2(200 Hz) ; f2 = 400 Hz
Third overtone is the fourth harmonic: f4 = 4(200 Hz); f4 = 800 Hz
21-19. If the fundamental frequency of a wave is 330 Hz, what is the frequency of the fifth
harmonic and the second overtone? fn = n f1 = n(330 Hz)
f n = nf1 ; f 2 = 5(330 Hz) ; f5 = 1650 Hz
Second overtone is the third harmonic: f3 = 3(330 Hz); f4 = 990 Hz
21-20. The linear density of a string is 0.00086 kg/m. What should be the tension in the rope in
order for a 2 m length of this string to vibrate at 600 Hz for its third harmonic?
3 F 9 F 4l 2 µ f 32
f3 = ; f 32 =  ; F=
2l µ 4l 2  µ  9
4(2 m) 2 (0.00086 kg/m)(600 Hz) 2
F= ; F = 550 N
9
21-21. A 10-g string, 4 m in length, has a tension of 64 N. What is the frequency of its
fundamental mode of vibration? What are the frequencies of the first and second
overtones?
1 Fl 1 (64 N)(4 m)
f1 = = ; f1 = 20 Hz
2L m 2(4 m) 0.0100 kg
First Overtone = f2 = 2(20 Hz) = 40 Hz
Second Overtone = f3 = 3(20 Hz) = 60 Hz
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6. Chapter 21. Wave Motion Physics, 6th Edition
21-22. The second harmonic of a vibrating string is 200 Hz. If the length of the string is 3 m and
its tension is 200 N, compute the linear density of the string.
2 F 1F F
f2 = ; f 22 =  ; µ=
2l µ l2  µ  f 22l 2
(200 N)
µ= ; µ = 5.56 x 10-4 kg/m
(200 Hz) 2 (3 m) 2
*21-23. A 0.500-g string is 4.3 m long and has a tension of 300 N. If it is fixed at each end and
vibrates in three segments, what is the frequency of the standing waves?
3 Fl 3 (300 N)(4.3 m)
f3 = = ; f3 = 560 Hz
2l m (2)(4.3 m) 0.5 x 10-3 kg
*21-24. A string vibrates with standing waves in five loops when the frequency is 600 Hz. What
frequency will cause the string to vibrate in only two loops?
f 5 5 f1 600 Hz 2(600 Hz)
= = ; f2 = ; f2 = 240 Hz
f 2 2 f1 f2 5
*21-25. A 120-g wire fixed at both ends is 8 m long and has a tension of 100 N. What is the
longest possible wavelength for a standing wave? What is the frequency?
Fl (100 N)(8 m)
v= = = 81.65 m/s
m 0.120 kg
The longest standing wave occurs for the fundamental, when λ = 2l
λ = 2(8 m)= 16 m; λ = 16 m
The velocity of the wave is v = fλ, so that
v (81.65 m/s)
f = = ; f1 = 5.10 Hz
λ (16 m)
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7. Chapter 21. Wave Motion Physics, 6th Edition
Challenge Problems
21-26. A longitudinal wave of frequency 400 Hz has a velocity of 60 m/s. What is the
wavelength?
v 60 m/s
λ= = ; λ = 0.150 m
f 400 Hz
21-27. What is the speed of a transverse wave in a rope of length 2.00 m and mass 80 g under a
tension of 400 N?
Fl (400 N)(2 m)
v= = ; v = 100 m/s
m 0.080 kg
21-28. A transverse wave travels at a speed of 8.00 m/s. A particular particle on the string
moves from its highest point to its lowest point in a time of 0.03 s. What is the
wavelength? [ From high to low is a time of ½T, so that T = 2(0.03 s) = 0.06 s. ]
1 1 v 8 m/s
f = = = 16.7 Hz; λ= = ; λ = 0.480 m
T 0.06 s f 16.7 Hz
21-29. A bass guitar string 750 mm long is stretched with sufficient force to produce a funda-
mental vibration of 220 Hz. What is the velocity of the transverse waves in this string?
The fundamental occurs when λ = 2l = 2(0.750 m); λ = 1.50 m
v = fλ = (220 Hz)(1.50 m); v = 330 m/s
*21-30. A 5-kg mass is hung from the ceiling by a 30-g wire 1.8 m long. What is the
fundamental frequency of vibration for this wire? [ F = mg = (5 kg)(9.8 m/s2) ]
1 Fl 1 (49 N)(1.8 m)
f1 = = ; f1 = 15.1 Hz
2l m 2(1.8 m) 0.030 kg
39

8. Chapter 21. Wave Motion Physics, 6th Edition
*21-31. A steel guy wire supporting a pole is 18.9 m long and 9.5 mm in diameter. It has a linear
density of 0.474 kg/m. When it is struck at one end by a hammer, the pulse returns in
0.3 s. What is the tension in the wire?
2l 2(18.9 m) Fl
v= = = 126 m/s ; v=
t 0.3 s m
F
v2 = ; F = µ v 2 = (0.474 kg/m)(126 m/s) 2 ; F = 7530 N
µ
*21-32. A 30-m wire weighing 400 N is stretched with a tension of 1800 N. How much time is
required for a pulse to make a round trip if it is struck at one end?
400 N Fl (1800 N)(30 m)
m= = 40.8 kg; v= = ; v = 36.4 m/s
9.8 m/s 2 m 40.8 kg
2l 2(30 m)
t= = ; t = 1.65 s
v 36.4 m/s
*21-33. Transverse waves have a speed of 20 m/s on a string whose tension is 8 N. What tension is
required to give a wave speed of 30 m/s for the same string?
Velocity is proportional to the square root of the tension, therefore:
v1 F1 v12 F1 F1v2 (8 N)(30 m/s) 2
2
= ; 2
= ; F2 = = ; F2 = 18.0 N
v2 F2 v2 F2 v12 (20 m/s) 2
*21-34. The fundamental frequency for a given string is 80 Hz. If the mass of the string is doubled
but other factors remain constant, what is the new fundamental frequency?
n Fl f1 1 f1 80 Hz
fn = ; = m
= 2; f2 = =
2L m f2 1
2m 2 1.414
f2 = 56.6 Hz
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9. Chapter 21. Wave Motion Physics, 6th Edition
Critical Thinking Questions
*21-35. In a laboratory experiment, an electromagnetic vibrator is used as a source of standing
waves in a string. A one meter length of the string is determined to have a mass of 0.6 g.
One end of the string is connected to the tip of the vibrator and the other passes over a
pulley 1 m away and is attached to a weight hanger. A mass of 392 g hanging from the
free end causes the string to vibrate in three segments. What is the frequency of the
vibrator? What new mass attached to the free end will cause the string to vibrate in four
loops? What is the fundamental frequency? 1m
3 Fl 3 mg (1 m)
f3 = =
2l m 2(1 m) 6 x 10-4 kg
3 (0.392 kg)(9.8 m/s 2 )(1 m) 0.392 kg
f3 = ; f3 = 120 Hz
2 6 x 10-4 kg
Now using the same vibrator frequency, what mass will produce four loops (n = 4).
4 Fl 16  Fl 
f4 = = 120 Hz;   = 1.44 x 10 Hz
4 2
2L m 4(1 m 2 )  m 
4(1.44 x 106 Hz 2 )(6 x 10-4 kg)(1 m 2 )
F= ; F = 2.16 N
16(1 m)
W 2.16 N
m= = m = 0.2204 kg m = 22.0 g
g 9.8 m/s 2
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10. Chapter 21. Wave Motion Physics, 6th Edition
*21-36. To understand the parameters that affect wave velocity in a vibrating string suppose that
F
v= = 100 m / s
m
What is the new wave speed v’ for each of the following changes: (a) F’ = 2F, (b) m’ =
2m, (c) ’ = 2  ?
Note the following proportions from the formula: v2 ∝ F, v2 ∝ , and v2 ∝ 1/m
2 2
F2 v2 2 F1 v2
= 2 =; = ; v2 = 2 x 104 m 2 /s 2 ; v2 = 141 m/s
F1 v1 F1 (100 m/s) 2
2 2
 2 v2 2 1 v2
= 2 =; = ; v2 = 2 x 104 m 2 /s 2 ; v2 = 141 m/s
1 v1 1 (100 m/s) 2
2
F2 1
m2 m1 v2
= =; = ; v2 = 5 x 103 m 2 /s 2 ; v2 = 70.7 m/s
F1 1
m1 2m1 (100 m/s) 2
*21-37. A power of 2 mW generates waves down rope A, and another power source generates
waves down an identical rope B. The waves in each rope are of the same frequency f and
velocity v. If the amplitude in rope B is twice that of rope A, what power is supplied to
rope B? [ fB = fA; vB = vA; AB = 2AA; P1 = 0.002 W; P = 2π2f2A2µv; PB = ? ]
PB 2π 2 f B2 AB mv 2π 2 f A2 (2 AA ) 2 mv 4
2
= = =
PA 2π 2 f A2 AA mv
2
2π 2 f A2 AA mv
2
1
PB = 4PA = 4(2 mW); PB = 8 mW
*21-38. The fundamental frequency of a piano wire is 253 Hz. By what fraction must the
tension in the wire be increased in order that the frequency be the desired “C” note (256
Hz)?
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11. Chapter 21. Wave Motion Physics, 6th Edition
f2 F2 F2 F2 f 22 (256 Hz) 2 F2
= = ; = = ; = 1.024
f1 F1 F1 F1 f12 (253 Hz) 2 F1
The tension must be increased by a factor of 0.024
*21-39. A variable oscillator allows a laboratory student to adjust the frequency of a source to
produce standing waves in a vibrating string. A 1.20-m length of string (µ = 0.400 g/m) is
placed under a tension of 200 N. What frequency is necessary to produce three standing
loops in the vibrating string? What is the fundamental frequency? What frequency will
produce five loops?
3 F 3 (200 N)
f3 = = ; f3 = 884 Hz
2l µ 2(1.2 m) (0.0004 kg/m)
884 Hz
Since f3 = 3f1, f1 = ; f1 = 295 Hz
3
Since f5 = 5f1,; f5 = 5(295 Hz) f1 = 1470 Hz
43