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17: Sound

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    17.1. Sound and light both travel at definite speeds, and the speed of sound is slower than the speed of light. The first shell is probably very close by, so the speed difference is not noticeable. The second shell is farther away, so the light arrives at your eyes noticeably sooner than the sound wave arrives at your ears.

    17.2. 10 dB: rustle of leaves; 50 dB: average office; 100 dB: noisy factory

    17.3. Amplitude is directly proportional to the experience of loudness. As amplitude increases, loudness increases.

    17.4. In the example, the two speakers were producing sound at a single frequency. Music has various frequencies and wavelengths.

    17.5. Regular headphones only block sound waves with a physical barrier. Noise-canceling headphones use destructive interference to reduce the loudness of outside sounds.

    17.6. When the tube resonates at its natural frequency, the wave’s node is located at the closed end of the tube, and the antinode is located at the open end. The length of the tube is equal to one-fourth of the wavelength of this wave. Thus, if we know the wavelength of the wave, we can determine the length of the tube.

    17.7. Compare their sizes. High-pitch instruments are generally smaller than low-pitch instruments because they generate a smaller wavelength.

    17.8. An easy way to understand this event is to use a graph, as shown below. It appears that beats are produced, but with a more complex pattern of interference.

    17.9. If I am driving and I hear Doppler shift in an ambulance siren, I would be able to tell when it was getting closer and also if it has passed by. This would help me to know whether I needed to pull over and let the ambulance through.

    Conceptual Questions

    1. Sound is a disturbance of matter (a pressure wave) that is transmitted from its source outward. Hearing is the human perception of sound.

    3. Consider a sound wave moving through air. The pressure of the air is the equilibrium condition, it is the change in pressure that produces the sound wave.

    5. The frequency does not change as the sound wave moves from one medium to another. Since the speed changes and the frequency does not, the wavelength must change. This is similar to the driving force of a harmonic oscillator or a wave on the string.

    7. The transducer sends out a sound wave, which reflects off the object in question and measures the time it takes for the sound wave to return. Since the speed of sound is constant, the distance to the object can found by multiplying the velocity of sound by half the time interval measured.

    9. The ear plugs reduce the intensity of the sound both in water and on land, but Navy researchers have found that sound under water is heard through vibrations mastoid, which is the bone behind the ear.

    11. The fundamental wavelength of a tube open at each end is 2L, where the wavelength of a tube open at one end and closed at one end is 4L. The tube open at one end has the lower fundamental frequency, assuming the speed of sound is the same in both tubes.

    13. The wavelength in each is twice the length of the tube. The frequency depends on the wavelength and the speed of the sound waves. The frequency in room B is higher because the speed of sound is higher where the temperature is higher.

    15. When resonating at the fundamental frequency, the wavelength for pipe C is 4L, and for pipes A and B is 2L. The frequency is equal to f = \(\frac{v}{\lambda}\). Pipe C has the lowest frequency and pipes A and B have equal frequencies, higher than the one in pipe C.

    17. Since the boundary conditions are both symmetric, the frequencies are fn = \(\frac{nv}{2L}\). Since the speed is the same in each, the frequencies are the same. If the wave speed were doubled in the string, the frequencies in the string would be twice the frequencies in the tube.

    19. The frequency of the unknown fork is 255 Hz. No, if only the 250 Hz fork is used, listening to the beat frequency could only limit the possible frequencies to 245 Hz or 255 Hz.

    21. The beat frequency is 0.7 Hz.

    23. Observer 1 will observe the highest frequency. Observer 2 will observe the lowest frequency. Observer 3 will hear a higher frequency than the source frequency, but lower than the frequency observed by observer 1, as the source approaches and a lower frequency than the source frequency, but higher than the frequency observed by observer 1, as the source moves away from observer 3.

    25. Doppler radar can not only detect the distance to a storm, but also the speed and direction at which the storm is traveling.

    27. The speed of sound decreases as the temperature decreases. The Mach number is equal to M = \(\frac{v_{s}}{v}\), so the plane should slow down.

    Problems

    29. smax = 4.00 nm, \(\lambda\) = 1.72 m, f = 200 Hz, v = 343.17 m/s

    31. a. \(\lambda\) = 68.60 \(\mu\)m

    b. \(\lambda\) = 360.00 \(\mu\)m

    33. a. k = 183.09 m−1

    b. \(\Delta\)P = −1.11 Pa

    35. s1 = 7.00 nm, s2 = 3.00 nm, kx1 + \(\phi\) = 0 rad; kx2 + \(\phi\) = 1.128 rad; k(x2 − x1) = 1.128 rad, k = 5.64 m−1; \(\lambda\) = 1.11 m, f = 306.31 Hz

    37. k = 5.28 x 10m; s(x, t) = 4.50 nm cos(5.28 x 103 m−1 x − 2\(\pi\)(5.00 MHz)t)

    39. \(\lambda\) = 3.43 mm

    41. \(\lambda\) = 6.00 m; smax = 2.00 mm; v = 600 m/s; T = 0.01 s

    43. (a) f = 100 Hz, (b) \(\lambda\) = 3.43 m

    45. f = 3400 Hz

    47. a. v = 5.96 x 103 m/s

    b. steel (from value in Table 17.1)

    49. v = 363 m s

    51. \(\Delta\)x = 924 m

    53. V = 0.05 m3; m = 392.5 kg; \(\rho\) = 7850 kg/m3; v = 5047.54 m/s

    55. TC = 35 °C, v = 351.58 m/s; \(\Delta\)x1 = 35.16 m, \(\Delta\)x2 = 52.74 m \(\Delta\)x = 63.39 m

    57. a. t5.00 °C = 0.0180 s, t35.0 °C = 0.0171 s

    b. % uncertainty = 5.00%

    c. This uncertainty could definitely cause difficulties for the bat, if it didn’t continue to use sound as it closed in on its prey. A 5% uncertainty could be the difference between catching the prey around the neck or around the chest, which means that it could miss grabbing its prey.

    59. 1.26 x 10−3 W/m2

    61. 85 dB

    63. a. 93 dB

    b. 83 dB

    65. 1.58 x 10−13 W/m2

    67. A decrease of a factor of 10 in intensity corresponds to a reduction of 10 dB in sound level: 120 dB − 10 dB = 110 dB.

    69. We know that 60 dB corresponds to a factor of 106 increase in intensity. Therefore, I \(\propto\) X2 \(\Rightarrow \frac{I_{2}}{I_{1}} = \left(\dfrac{X_{2}}{X_{1}}\right)^{2}\), so that X2 = 10−6 atm. 120 dB corresponds to a factor of 1012 increase \(\Rightarrow\) 10−9 atm(1012)1/2 = 10−3 atm.

    71. 28.2 dB

    73. 1 x 106 km

    75. 73 dB − 70 dB = 3 dB; Such a change in sound level is easily noticed.

    77. 2.5; The 100-Hz tone must be 2.5 times more intense than the 4000-Hz sound to be audible by this person.

    79. 0.974 m

    81. 11.0 kHz; The ear is not particularly sensitive to this frequency, so we don’t hear overtones due to the ear canal.

    83. a. v = 344.08 m/s, \(\lambda_{1}\) = 16.00 m, f1 = 21.51 Hz

    b. \(\lambda_{3}\) = 5.33 m, f3 = 64.56 Hz

    85. vstring = 149.07 m/s, \(\lambda_{3}\) = 1.33 m, f3 = 112.08 Hz; \(\lambda_{1}\) = \(\frac{v}{f_{1}}\), L = 1.53 m

    87. a. 22.0 °C

    b. 1.01 m

    89. First overtone = 180 Hz; second overtone = 270 Hz; third overtone = 360 Hz

    91. 1.56 m

    93. The pipe has symmetrical boundary conditions; $$\begin{split} \lambda_{n} & = \frac{2}{n} L, \quad f_{n} = \frac{nv}{2L}, \quad n = 1, 2, 3 \\ \lambda_{1} & = 6.00\; m, \quad \lambda_{2} = 3.00\; m, \quad \lambda_{3} = 2.00\; m \\ f_{1} & = 57.17\; Hz, \quad f_{2} = 114.33\; Hz, \quad f_{3} = 171.50\; Hz \end{split}$$

    95. \(\lambda_{6}\) = 0.5 m; v = 1000 m/s; FT = 6500 N

    97. f = 6.40 kHz

    99. 1.03 or 3%

    101. $$\begin{split} f_{B} & = |f_{1} − f_{2}| \\ |128.3\; Hz − 128.1\; Hz| & = 0.2\; Hz; \\ |128.3\; Hz − 127.8\; Hz| & = 0.5\; Hz; |128.1\; Hz − 127.8\; Hz| & = 0.3\; Hz \end{split}$$

    103. vA = 135.87 m/s, vB = 141.42 m/s, \(\lambda_{A}\) = \(\lambda_{B}\) = 0.40 m, \(\Delta\)f = 15.00 Hz

    105. v = 155.54 m/s, fstring = 971.17 Hz, n = 16.23; fstring = 1076.83 Hz, n = 18.00

    The frequency is 1076.83 Hz and the wavelength is 0.14 m.

    107. f2 = f1 ± fB = 260.00 Hz ± 1.50 Hz, so that f2 = 261.50 Hz or f2 = 258.50 Hz

    109. $$\begin{split} f_{ace} & = \frac{f_{1} + f_{2}}{2}; f_{B} = f_{1} − f_{2}\; (assume\; f_{1} \geq f_{2}); \\ f_{ace} & = \frac{(f_{B} + f_{2}) + f_{2}}{2} \Rightarrow f_{2} = 4099.750\; Hz, f_{1} = 4100.250\; Hz \end{split}$$

    111. a. 878 Hz

    b. 735 Hz

    113. 3.79 x 103 Hz

    115. a. 12.9 m/s

    b. 193 Hz

    117. The first eagle hears 4.23 x 103 Hz. The second eagle hears 3.56 x 103 Hz.

    119. vs = 31.29 m/s; fo = 1.12 kHz

    121. An audible shift occurs when \(\frac{f_{obs}}{f_{s}}\) ≥ 1.003; $$\begin{split} f_{obs} & = f_{s} \frac{v}{v − v_{s}} \Rightarrow \frac{f_{obs}}{f_{s}} = \frac{v}{v − v_{s}} \Rightarrow \\ v_{s} & = 0.990\; m/s \end{split}$$

    123. \(\theta\) = 30.02°; vs = 680.00 m/s; tan \(\theta\) = \(\frac{y}{v_{s} t}\), t = 21.65 s

    125. sin \(\theta\) = 1 M, \(\theta\) = 56.47°; y = 9.31 km

    127. s1 = 6.34 nm; s2 = 2.30 nm; kx1 + \(\phi\) = 0 rad; kx2 + \(\phi\) = 1.20 rad; k(x2 − x1) = 1.20 rad; k = 3.00 m−1; \(\omega\) = 1019.62 s−1; s1 = smax cos(kx1 − \(\phi\)); \(\phi\) = 5.66 rad; s(x, t) = (6.30 nm)cos(3.00 m−1 x − 1019.62 s−1 t + 5.66)

    Additional Problems

    129. vs = 346.40 m/s; $$\begin{split} \lambda_{n} & = \frac{2}{n} L \quad f_{n} = \frac{v_{s}}{\lambda_{n}} \\ \lambda_{1} & = 1.60\; m \quad f_{1} = 216.50\; Hz \\ \lambda_{2} & = 0.80\; m \quad f_{2} = 433.00\; Hz \end{split}$$

    131. a. \(\lambda_{6}\) = 0.40 m; v = 57.15 m/s; f6 = 142.89 Hz

    b. \(\lambda_{s}\) = 2.40 m

    133. v = 344.08 m/s; vA = 29.05 m/s, vB = 33.52 m/s; fA = 961.18 Hz, fB = 958.89 Hz; fA, beat = 161.18 Hz, fB, beat = 158.89 Hz

    135. v = 345.24 m/s

    a. I = 31.62 \(\mu\)W/m2

    b. I = 0.16 \(\mu\)W/m2

    c. smax = 104.39 \(\mu\)m

    d. smax = 7.43 \(\mu\)m

    137. \(\frac{f_{A}}{f_{D}} = \frac{v + v_{s}}{v − v_{s}}\), (v − vs)\(\frac{f_{A}}{f_{D}}\) = v + vs, v = 347.39 m/s; TC = 27.70 °C

    Challenge Problems

    139. \(\sqrt{x^{2} + d^{2}} − x = \lambda\), x2 + d2 = \(\lambda\) + x)2; x2 + d2 = \(\lambda^{2}\) + 2x\(\lambda\) + x2, d= \(\lambda^{2}\) + 2x\(\lambda\); x = \(\frac{d^{2} − \left(\dfrac{v}{f}\right)^{2}}{2 \frac{v}{f}}\) 

    141. a. For maxima: \(\Delta\)r = d sin \(\theta\); d sin \(\theta\) = n\(\lambda\) n = 0, ± 1, ± 2 ..., \(\theta\) = sin−1 \(\left(\dfrac{n \lambda}{d}\right)\) n = 0, ± 1, ± 2 ...

    b. For minima: \(\Delta\)r = d sin \(\theta\); d sin \(\theta\) = \(\left(n + \dfrac{1}{2}\right) \lambda\) n = 0, ± 1, ± 2 ... \(\theta\) = sin−1 \(\left(\left(n + \dfrac{1}{2}\right) \dfrac{\lambda}{d}\right)\) n = 0, ± 1, ± 2 ...

    143. a. vstring = 160.73 m/s, fstring = 535.77 Hz

    b. ffork = 512 Hz

    c. ffork = \(\frac{n \sqrt{\frac{F_{T}}{\mu}}}{2L}\), FT = 141.56 N

    145. a. f = 268.62 Hz

    b. \(\Delta\) f ≈ \(\frac{1}{2} \frac{\Delta F_{T}}{F_{T}}\)f = 1.34 Hz

    147. a. v = 466.07 m/s

    b. \(\lambda_{9}\) = 51.11 mm

    c. f9 = 9.12 kHz

    d. fsound = 9.12 kHz

    e. \(\lambda_{air}\) = 37.86 mm

    Contributors

    Samuel J. Ling (Truman State University), Jeff Sanny (Loyola Marymount University), and Bill Moebs with many contributing authors. This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0).