MR. SURRETTE VAN NUYS HIGH SCHOOL
CHAPTER 10: WAVES, SHM, AND SOUND
1. Ocean waves with a wavelength of 120 m are coming in at a rate of 8 per minute. What is their speed?
1A.
(1) v = f

(2) f = (8 cycles / 1 min) (1 min / 60 s)
(3) f = 0.13 cycles / sec
(4) v = ( 0.13 Hz )(120 m)
(5) v = 16.0 m/s
2. The speed of a sound wave in sea water is 1500 m/s. If this wave is transmitted at frequency 10 kHz, what is its wavelength (in cm)?
2A.
(1) v = f

(2)

= v / f
(3)

= (1500 m/s) / 10 x 10
3
Hz
(4)

= 0.15 m
(5)

= 15 cm
3. A piano string of mass 0.0057 kg/m is under a tension of 1425 N. Find the velocity with which a wave travels on the string.
3A.
(1) v = (F/

) 1/2
(2) v = (1425 N / 0.0057 kg/m)
1/2
(3) v = 500 m/s
4. Two pulses travel towards one another at 5 m/s. They are initially 10 m apart, or 5 m from their midpoint.
4a. How long will it take until the string looks like the sketch below (the vertical mark denotes the midpoint)?
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MR. SURRETTE VAN NUYS HIGH SCHOOL
A. One second . When the centers of the pulses coincide they cancel each other completely.
4b. How long will it take until the string looks like the sketch below?
A. 2 seconds . Each pulse moves 10 meters in 2 seconds.
5. A 0.35 kg object, suspended from a spring with a spring constant of k = 15 N/m, is moving in simple harmonic motion and has an amplitude of 0.08 m. What is its kinetic energy at the instant when its displacement is 0.04 m?
5A.
(1) Determine the velocity:
v = [(k/m)(A
2
– x
2
)]
1/2
(2) v = [(15 N/m / 0.35 kg)((0.08 m)
2
– (0.04 m)
2
)]
1/2
(3) v = 0.454 m/s
(4) Find K: K = ½ mv
2
(5) K = ½ (0.35 kg)(
0.454 m/s )
2
(6) K = 3.6 x 10 -2 J
6. A mass of 0.22 kg, hanging from a spring with a spring constant of 95 N/m, is set into an up and down simple harmonic motion. What is the potential energy stored in the spring when the mass is displaced 0.2 m?
6A.
(1) U s
= ½ kx 2
(2) U s
= ½ (95 N/m)(0.2 m)
2
(3) U s
= 1.9 J
7. A 0.32 kg block rests on a frictionless level surface and is attached to a horizontally aligned spring with a spring constant of 115 N/m. The block is initially displaced 5.5 cm from the equilibrium point and then released to set up a simple harmonic motion. What is the frequency?
7A.
(1) f = [1 / (2

)](k / m)
1/2
(2) f = [1 / (2

)][(115 N/m) / (0.32 kg)]
1/2
(3) f = 3.02 Hz
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MR. SURRETTE VAN NUYS HIGH SCHOOL
8. A 3.45 kg particle rests on a frictionless horizontal surface and is attached to the free end of a spring
( k = 510 N/m). The particle is pulled horizontally so that it stretches the spring 2.50 cm and is then released.
8a.
What is the period of the motion?
A.
(1) T s
= 2

(m/k) 1/2
(2) T s
= 2

(3.45 kg / 510 N/m)
1/2
(3) T s
= 0.52 s
8b. What is the frequency of the motion?
A.
(1) f = 1 / T
(2) 1 / 0.52 s
(3) f = 1.92 Hz
8c. What is the angular frequency of the motion?
(1)

= 2
 f
(2)

= 2

( 1.92 Hz )
(3)

= 12.1 rad/s
8d. What is the displacement x of the particle from equilibrium at t = 0.720 s?
(1) x = Acos(
 t)
(2) x = (2.50 cm)cos[(12.1 rad/s)(0.720 s)]
(3) x = 2.47 cm
9. What is the period of a simple pendulum 52 cm long?
9A.
(1) T p
= 2

(L/g)
1/2
(2) T p
= 2

(0.52 m / 9.8 m/s
2
)
1/2
(3) T p
= 1.45 s
10. A mass of 0.20 kg is attached to a spring of force constant k = 400 N/m, stretched 5 cm from equilibrium, and released from rest.
10a. Compute the mechanical energy.
A.
Take the mass to be at the maximum point A , where the spring is stretched out to its maximum. At this point, its potential energy is 100% and kinetic energy is 0%.
(1) E = K + U
(2) E = 0 + U
(3) E = U
(4) U s
= ½ kx 2
(5) U s
= ½ (400 N/m)(0.05 m)
2
(6) U s
= 0.5 J
(7) E = 0.5 J
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MR. SURRETTE VAN NUYS HIGH SCHOOL
10b. Calculate the speed of the mass when it passes through the equilibrium position ( x = 0).
A.
(1) E = K + U
(2) E = K + 0
(3) E = K
(4) E = ½ mv 2
(5) v = [2E/m]
½
(6) v = [2( 0.50 J ) / (0.20 kg)]
1/2
(7) v = 2.24 m/s
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MR. SURRETTE VAN NUYS HIGH SCHOOL
CHAPTER 10: WAVES, SHM, AND SOUND
1. A standing wave is set up on a 1.5 m string fixed at both ends. The string vibrates in 4 distinct segments when driven by a 105 Hz source. What is the wave velocity in the string?
1A.
(1) f n
= nv / 2L
(2) v = 2Lf n
/ n
(3) v = 2(1.5 m)(105 Hz) / 4
(4) v = 78.8 m/s
2. If a sound source with a 1200 Hz frequency is at rest, and a listener moves at a speed of 34.5 m/s away from the source, what is the frequency heard by the listener? (The velocity of sound = 340 m/s)
2A.
(1) f ’ = f [(v +/- v o
) / (v +/- v s
)]
(2) f ’ = (1200 Hz) [(340 m/s – 34.5 m/s) / (340 m/s)]
(3) f ’ = 1078 Hz
3. An 85 Hz note is sounded on a piano. If the length of the piano wire is 1.75 m and its mass density is
4.5 g/m, what is the tension of the wire?
3A.
(1) Determine the wavelength:
 n
= 2L / n
(2)
 n
= 2(1.75 m) / 1
(3)
 n
= 3.5 m
(4) Determine the velocity: v = f

(5) v = (80 Hz)( 3.5 m )
(6) v = 280 m/s
(7) Determine the tension of the wire: v = (F/

) 1/2
(8) F =
 v 2
(9) F = (0.0045 kg/m)( 280 m/s )
2
(10) F = 352.8 N
4. A fireworks rocket explodes at a height of 120 m above the ground. An observer on the ground directly under the explosion experiences an average sound intensity of 6 x 10
-3
W/m
2
. What is the sound level in dB heard by the observer?
4A.
(1)

= 10 log (I / I o
)
(2)

= 10 log ( 6 x 10 -3 W/m 2 / 1 x 10 -12 W/m 2 )
(3)

= 97.8 dB
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MR. SURRETTE VAN NUYS HIGH SCHOOL
5. A clarinet behaves like a tube closed at one end. If its length is 80 cm, and the velocity of sound is
340 m/s, what is its fundamental frequency in Hz?
5A.
(1) f n
= n (v / 4L)
(2) f n
= (1) [(340 m/s) / (4)(0.8 m)]
(3) f n
= 106 Hz
6. A bat, flying at 5.0 m/s, emits a chirp at 40kHz. If this sound pulse is reflected by a wall, what is the frequency of the echo received by the bat?
6A.
(1) f ’ = f [(V +/- V o
) / (V +/- V s
)]
(2) f ’ = (40 kHz)[ (340 m/s + 5 m/s) / (340 m/s – 5 m/s)]
(3) f ’ = 4.12 x 10 4 Hz
7. Shortening a guitar string to one third its initial length will change its natural frequency by what factor?
7A .
(1) v = f

(2) f = v /

(3) f ’ = v / [(1/3) 
(4) f ’ = 3(v / 
 f ’ = 3 f
8. A pipe 60 cm long is open at both ends. The speed of sound in air is 340 m/s.
8a.
What is the wavelength of the standing wave mode of lowest frequency (“the fundamental mode”)?
A.
(1) L = ½ 
1
(2)

1
= 2L
(3)

1
= 2(60 cm)
(4)

1
= 120 cm
8b. What is the frequency of the fundamental?
A.
(1) v = f

(2) f = v /

(3) f = 340 m/s / 1.2 m
(4) f = 283.3 Hz
8c. What are the frequencies of the next two (consecutive) allowed modes?
A. The next two frequencies are two times and three times the lowest: 566.7 Hz and 849.9 Hz .
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MR. SURRETTE VAN NUYS HIGH SCHOOL
9. The same pipe is now closed at one end. Repeat steps a, b, and c for this new pipe.
9a. Wavelength of the fundamental?
A.
(1) L = ¼ 
1
(2)

1
= 4L
(3)

1
= 4(60 cm)
(4)

1
= 240 cm
9b. Frequency of the fundamental?
A.
(1) v = f

(2) f = v /

(3) f = 340 m/s / 2.4 m
(4) f = 141.7 Hz
9c. What are the frequencies of the next two (consecutive) allowed modes?
A. f
3
= 425 Hz (n = 3) f
5
= 708.5 Hz (n = 5)
10. An ambulance has a siren of frequency 2315 Hz.
10a. Assuming a speed of sound in air of 340 m/s and no motion of the ambulance relative to the air, the siren produces sounds of wavelength:
A.
(1) v = f

(2)

= v / f
(3)

= 340 m/s / 2315 cycles/sec
(4)

= 0.15 m
10b. If you are at rest and the ambulance is approaching you at 40 m/s relative to the air, the wavelength of sound you observe is:
(a) equal to 15 cm
(b) more than 15 cm
(c) less than 15 cm
A. (c) The wave fronts are closer together in front of the ambulance so the wavelength you observe is less than 15 cm .
10c. What is the actual value of the wavelength you observe?
A.
(1) f ’ = f [v / (v – v s
)]
(2) f ’ = (2315 Hz) [(340 m/s) / (340 m/s – 40 m/s)]
(3) f ’ = (2315 Hz) [ (340 m/s) / (300 m/s)]
(4) f ’ = 2624 Hz
(5) v = f

(6)

= v / f
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MR. SURRETTE VAN NUYS HIGH SCHOOL
10c. (continued…)
(7)

= 340 m/s / 2624 Hz
(8)

= 0.13 m = 13 cm
10d. What is the frequency you observe?
A.
(1) v = f

(2) f = v /

(3) f = 340 m/s / 0.13 m
(4) f = 2615 Hz
10e. The ambulance is still heading towards you at the same speed. If you now start moving towards the oncoming ambulance at 25 m/s relative to the air, the value of the frequency you hear becomes:
A.
(1) f ’ = f [(v + v o
) / (v – v s)
]
(2) f ’ = (2315 Hz) [ (340 m/s + 25 m/s) / (340 m/s – 40 m/s) ]
(3) f ’ = 2817 Hz
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MR. SURRETTE VAN NUYS HIGH SCHOOL
CHAPTER 10: WAVES, SHM, AND SOUND
1. A standing wave is set up on a 4.0 m string fixed at both ends. The string vibrates in 7 distinct segments when driven by a 120 Hz source. What is the wave velocity in the string?
1A.
(1) f n
= nv / 2L
(2) v = 2Lf n
/ n
(3) v = 2(4.0 m)(120 Hz) / 7
(4) v = 137.1 m/s
2. A middle C (frequency 256 Hz) is sounded on a piano. If the length of the piano wire is 1.75 m and its mass density is 4.2 g/m, what is the tension of the wire?
2A.
(1) Determine the wavelength:
 n
= 2L / n
(2)
 n
= 2(1.75 m) / 1
(3)
 n
= 3.5 m
(4) Determine the velocity: v = f

(5) v = (256 Hz)( 3.5 m )
(6) v = 896 m/s
(7) Determine the tension of the wire: v = (F/

) 1/2
(8) F =
 v
2
(9) F = (0.0042 kg/m)( 896 m/s ) 2
(10) F = 3372 N
3. A fireworks rocket explodes at a height of 82 m above the ground. An observer on the ground directly under the explosion experiences an average sound intensity of 9 x 10
-2
W/m
2
. What is the sound level in dB heard by the observer?
3A.
(1) B = 10 log (I / I o
)
(2) B = 10 log ( 9 x 10
-2
(3)

= 109.5 dB
W/m
2
/ 1 x 10
-12
W/m
2
)
4. A clarinet behaves like a tube closed at one end. If its length is 75 cm, and the velocity of sound is
340 m/s, what is its fundamental frequency in Hz?
4A.
(1) f n
= n (v / 4L)
(2) f n
= (1) [(340 m/s) / (4)(0.75 m)]
(3) f n
= 113.3 Hz
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MR. SURRETTE VAN NUYS HIGH SCHOOL
5. A pipe 215 cm long is open at both ends. The speed of sound in air is 340 m/s.
5a. What is the wavelength of the standing wave mode of lowest frequency (“the fundamental mode”)?
A. Since antinodes are located at both ends of the pipe, one wavelength is twice the pipelength, or 430 cm .
5b. What is the frequency of the fundamental?
A.
(1) v = f

(2) f = v /

(3) f = 340 m/s / 4.3 m
(4) f = 79.1 Hz
5c. What is the frequency of the next allowed mode?
A.
Additional modes are produced by adding one-half wavelength sections of a standing wave to the original fundamental mode. The next frequency is two times the lowest: 158.2 Hz .
6. An ambulance has a siren of frequency 1798 Hz.
6a. Assuming a speed of sound in air of 340 m/s and no motion of the ambulance relative to the air, the siren produces sounds of wavelength:
A.
(1) v = f

(2)

= v / f
(3)

= 340 m/s / 1798 cycles/sec
(4)

= 0.189 m = 18.9 cm
6b. You are now at rest and the ambulance is approaching you at 32 m/s relative to the air. What is the value of the wavelength you observe?
A.
(1) f ’ = f [v / (v – v s
)]
(2) f ’ = (1798 Hz)[ (340 m/s) / (340 m/s – 32 m/s)]
(3) f ’ = (1798 Hz) [ (340 m/s) / (308 m/s)]
(4) f ’ = 1985 Hz
(5) v = f

(6)

= v / f
(7)

= 340 m/s / 1985 Hz
(8)

= 0.171 m = 17.1 cm
6c. What is the frequency you observe?
A.
(1) v = f

(2) f = v /

(3) f = 340 m/s / 0.171 m
(4) f = 1985 Hz
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MR. SURRETTE VAN NUYS HIGH SCHOOL
6d. The ambulance is still heading towards you at the same speed. If you now start moving towards the oncoming ambulance at 30 m/s relative to the air, the value of the frequency you hear becomes:
A.
(1) f ’ = f [(v + v o
) / (v – v s)
]
(2) f ’ = (1798 Hz)[ (340 m/s + 30 m/s) / (340 m/s – 32 m/s)]
(3) f ’ = (1798 Hz) [ (370 m/s) / (308 m/s)]
(4) f ’ = 2160 Hz
7. A 0.35 kg object, suspended from a spring with a spring constant of k = 20 N/m, is moving in simple harmonic motion and has an amplitude of 0.07 m. What is its kinetic energy at the instant when its displacement is 0.04 m?
7A.
(1) Determine the velocity: v = [(k/m)(A 2 – x 2 )] 1/2
(2) v = [(20 N/m / 0.35 kg)((0.07 m)
2
– (0.04 m)
2
)]
1/2
(3) v = 0.434 m/s
(4) Find K: K = ½ mv 2
(5) K = ½ (0.35 kg)(
0.434 m/s )
2
(6) K = 3.3 x 10 -2 J
8. A mass of 0.22 kg, hanging from a spring with a spring constant of 85 N/m, is set into an up and down simple harmonic motion. What is the potential energy stored in the spring when the mass is displaced 0.14 m?
8A.
(1) U s
= ½ kx 2
(2) U s
= ½ (85 N/m)(0.14 m)
2
(3) U s
= 8.3 x 10 -2 J
9. Ocean waves with a wavelength of 82 m are coming in at a rate of 7 per minute. What is their speed?
9A.
(1) v = f

(2) f = (7 cycles / 1 min) (1 min / 60 s)
(3) f = 0.12 Hz
(4) v = ( 0.12 Hz )(82 m)
(5) v = 9.6 m/s
10. The speed of a sound wave in sea water is 1500 m/s. If this wave is transmitted at frequency 23 kHz, what is its wavelength (in cm)?
10A.
(1) v = f

(2)

= v / f
(3)

= (1500 m/s) / 23 x 10 3 Hz
(4)

= 0.065 m = 6.5 cm
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MR. SURRETTE VAN NUYS HIGH SCHOOL
11. A 2.75 kg particle rests on a frictionless horizontal surface and is attached to the free end of a spring
(k = 510 N/m). The particle is pulled horizontally so that it stretches the spring 3.75 cm and is then released.
11a.
What is the period of the motion?
A.
(1) T = 2

(m/k) 1/2
(2) T = 2

(2.75 kg / 510 N/m)
1/2
(3) T = 0.46 s
11b. What is the frequency of the motion?
A.
(1) f = 1 / T
(2) 1 / 0.46 s
(3) f = 2.17 Hz
11c. What is the angular frequency of the motion?
A.
(1)

= 2
 f
(2)

= 2

( 2.17 Hz )
(3)

= 13.6 rad/s
11d. What is the displacement x of the particle from equilibrium at t = 1.72 s?
A.
(1) x = Acos(
 t)
(2) x = (3.75 cm)cos[( 13.6 rad/s )(1.72 s)]
(3) x = 3.44 cm
12. What is the period of a simple pendulum 35 cm long?
12A.
(1) T p
= 2

(L/g) 1/2
(2) T p
= 2

(0.35 m / 9.8 m/s
2
)
1/2
(3) T p
= 1.19 s
13. A mass of 0.15 kg is attached to a spring of force constant k = 400 N/m, stretched 4 cm from equilibrium, and released from rest.
13a. Compute the mechanical energy.
A.
Take the mass to be at the maximum point A , where the spring is stretched out to its maximum. At this point, its potential energy is 100% and kinetic energy is 0%.
(1) E = K + U
(2) E = 0 + U
(3) E = U
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MR. SURRETTE VAN NUYS HIGH SCHOOL
13a. (continued…)
(4) U s
= ½ kx
2
(5) U s
= ½ (400 N/m)(0.04 m) 2
(6) U s
= 0.32 J
(7) E = 0.32 J
13b. Calculate the speed of the mass when it passes through the equilibrium position (x = 0).
A.
(1) E = K + U
(2) E = K + 0
(3) E = K
(4)
E = ½ mv 2
(5) v = [2E/m]
½
(6) v = [2( 0.32 J ) / (0.15 kg)]
1/2
(7) v = 2.07 m/s
PHYSICS PAGE 13