Physics, 6th Edition
Chapter 21. Wave Motion
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? v=
s 8m = = 2 m/s; t 4s
λ=
f =
80 waves = 1.33 Hz 60 s
v 2 m/s = ; f 1.33 Hz
λ = 1.5 m
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); T=
1 1 = ; f 12 Hz
v = 3.36 m/s
T = 0.0833 s
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
v = fλ = (8 Hz)(0.28 m); v = 2.24 m/s;
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T=
1 1 = ; f 8 Hz
T = 0.125 s
Physics, 6th Edition
Chapter 21. Wave Motion
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) = ; m 0.50 kg
v=
v = 8.94 m/s
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? m=
0.500 kg ; 2
m = 0.250 kg;
v=
Fl (80 N)(0.250 m) = = 8.94 m/s 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? v=
Fl ; m
v2 =
Fl ; m
m=
Fl (200 N)(3 m) = ; v2 (172 m/s) 2
m = 0.0203 kg
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? v=
Fl (500 N)(5.2 m) = ; m 0.200 kg
v = 114 m/s
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? F=
mv 2 (0.900 kg)(12 m/s) 2 ; = l 2m
34
F = 64.8 N.
Physics, 6th Edition
Chapter 21. Wave Motion
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? f =
8 waves = 0.800 Hz; 10 s
λ=
v 3.06 m/s = ; f 0.800 Hz
v=
11 m = 3.06 m/s; 3.6 s
λ = 3.82 m
*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. f =
v ; λ
f =
1 λ
F 1 (200 N) = ; λ (0.20 m) 0.008 kg/m
f = 791 Hz
*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? v=
Fl (400 N)(1.6 m) = = 46.2 m/s; m 0.30 kg
λ=
v 46.2 m/s = ; f 40 Hz
λ = 1.15 m
*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? f =
80 waves = 6.67 Hz; 12 s
v = fλ = (6.67 Hz)(0.15 m);
35
v = 1.00 m/s
Physics, 6th Edition
Chapter 21. Wave Motion
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?
µ=
m 0.30 kg = = 0.150 kg/m; l 2m
v=
F (48 N) = = 17.9 m/s 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; l
v = f λ = (8 Hz)(1.6 m) = 12.8 m/s
E P = v = (4.04 x 10-3 J/m)(12.8 m/s); l
P = 5.17 x 10-2 W
*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?
µ=
[ P = 2π2f2A2µ v ]
m 0.300 kg = = 0.120 kg/m; l 2.50 m
v=
36
Fl 46 N = = 19.6 m/s m 0.120 kg/m
Physics, 6th Edition
Chapter 21. Wave Motion
f =
P 90.0 W = ; 2 2 2 2π A µ v 2π (0.008 m) (0.120 kg/m)(19.6 m/s) 2
f = 174 Hz
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? f n = nf1 ;
fn = n f1 = n(330 Hz)
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? f3 =
F=
3 2l
F ; µ
f 32 =
9 F ; 4l 2 µ
F=
4(2 m) 2 (0.00086 kg/m)(600 Hz) 2 ; 9
4l 2 µ f 32 9 F = 550 N
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? f1 =
1 2L
Fl 1 (64 N)(4 m) = ; m 2(4 m) 0.0100 kg
37
f1 = 20 Hz
Physics, 6th Edition
Chapter 21. Wave Motion First Overtone = f2 = 2(20 Hz) =
40 Hz
Second Overtone = f3 = 3(20 Hz) =
60 Hz
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. f2 =
µ=
2 2l
F ; µ
f 22 =
1F ; l2 µ
(200 N) ; (200 Hz) 2 (3 m) 2
µ=
F f 22l 2
µ = 5.56 x 10-4 kg/m
*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? f3 =
3 2l
Fl 3 (300 N)(4.3 m) = ; m (2)(4.3 m) 0.5 x 10-3 kg
f3 = 560 Hz
*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 = = ; f 2 2 f1 f2
f2 =
2(600 Hz) ; 5
f2 = 240 Hz
*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? v=
Fl (100 N)(8 m) = = 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 38
Physics, 6th Edition
Chapter 21. Wave Motion f =
v (81.65 m/s) = ; λ (16 m)
f1 = 5.10 Hz
Challenge Problems 21-26. A longitudinal wave of frequency 400 Hz has a velocity of 60 m/s. What is the wavelength?
λ=
v 60 m/s = ; f 400 Hz
λ = 0.150 m
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? v=
Fl (400 N)(2 m) = ; m 0.080 kg
v = 100 m/s
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. ] f =
1 1 = = 16.7 Hz; T 0.06 s
λ=
v 8 m/s = ; f 16.7 Hz
λ = 0.480 m
21-29. A bass guitar string 750 mm long is stretched with sufficient force to produce a fundamental 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); v = fλ = (220 Hz)(1.50 m);
λ = 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) ]
39
Physics, 6th Edition
Chapter 21. Wave Motion
f1 =
1 2l
Fl 1 (49 N)(1.8 m) = ; m 2(1.8 m) 0.030 kg
f1 = 15.1 Hz
*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? v=
v2 =
F ; µ
2l 2(18.9 m) = = 126 m/s ; t 0.3 s
v=
F = µ v 2 = (0.474 kg/m)(126 m/s) 2 ;
Fl m F = 7530 N
*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? m=
400 N = 40.8 kg; 9.8 m/s 2 t=
v=
Fl (1800 N)(30 m) = ; m 40.8 kg
2l 2(30 m) = ; v 36.4 m/s
v = 36.4 m/s
t = 1.65 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 = v2
F1 ; F2
v12 F1 = ; v22 F2
F1v22 (8 N)(30 m/s) 2 F2 = 2 = ; F2 = 18.0 N v1 (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?
40
Physics, 6th Edition
Chapter 21. Wave Motion
fn =
n 2L
Fl ; m
f1 = f2
1 1
m
= 2;
f2 =
2m
f1 80 Hz = 2 1.414
f2 = 56.6 Hz
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? f3 =
f3 =
3 2l
1m
Fl 3 mg (1 m) = m 2(1 m) 6 x 10-4 kg
3 (0.392 kg)(9.8 m/s 2 )(1 m) ; 2 6 x 10-4 kg
0.392 kg
f3 = 120 Hz
Now using the same vibrator frequency, what mass will produce four loops (n = 4). Fl = 120 Hz; m
16 Fl 4 2 = 1.44 x 10 Hz 2 4(1 m ) m
f4 =
4 2L
F=
4(1.44 x 106 Hz 2 )(6 x 10-4 kg)(1 m 2 ) ; 16(1 m)
F = 2.16 N
m=
W 2.16 N = g 9.8 m/s 2
m = 22.0 g
m = 0.2204 kg
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Physics, 6th Edition
Chapter 21. Wave Motion
*21-36. To understand the parameters that affect wave velocity in a vibrating string suppose that v=
F = 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 F2 v22 = =; F1 v12
2 F1 v22 = ; F1 (100 m/s) 2
v2 = 2 x 104 m 2 /s 2 ;
v2 = 141 m/s
l 2 v22 = =; l 1 v12
2l 1 v22 = ; l 1 (100 m/s) 2
v2 = 2 x 104 m 2 /s 2 ;
v2 = 141 m/s
F2 = F1
m1 v22 = ; 2m1 (100 m/s) 2
v2 = 5 x 103 m 2 /s 2 ;
v2 = 70.7 m/s
1
m2
1
m1
=;
*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 AB2 mv 2π 2 f A2 (2 AA ) 2 mv 4 = = = PA 2π 2 f A2 AA2 mv 2π 2 f A2 AA2 mv 1 PB = 4PA = 4(2 mW);
42
PB = 8 mW
Physics, 6th Edition
Chapter 21. Wave Motion
*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)? f2 = f1
F2 F1
=
F2 ; F1
F2 f 22 (256 Hz) 2 = = ; F1 f12 (253 Hz) 2
F2 = 1.024 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? f3 =
3 2l
F 3 (200 N) = ; µ 2(1.2 m) (0.0004 kg/m)
Since f3 = 3f1, Since f5 = 5f1,;
f1 =
884 Hz ; 3
f5 = 5(295 Hz)
43
f3 = 884 Hz
f1 = 295 Hz f1 = 1470 Hz
