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• 1. Chapter 22. Sound Physics, 6th Edition Chapter 22. Sound Speed of Sound Waves 22-1. Young's modulus for steel is 2.07 x 1011 Pa and its density is 7800 kg/m3. Compute the speed of sound in a steel rod. Y 2.07 x 1011Pa v  ; v = 5150 m/s  7800 kg/m3 22-2. A 3-m length of copper rod has a density 8800 kg/m3, and Young's modulus for copper is 1.17 x 1011 Pa. How much time will it take for sound to travel from one end of the rod to the other? Y 1.17 x 1011Pa v  ; v = 3646 m/s  8800 kg/m3 s s 2(3 m) v ; t  ; t = 1.65 ms t v 3646 m/s 22-3. What is the speed of sound in air (M = 29 g/mol and  = 1.4) on a day when the temperature is 300C? Use the approximation formula to check this result.  RT (1.4)(8.314 J/kg K)(2730  300 )K v  ; v = 349 m/s M 0.029 kg/mol  m/s  v  331 m/s   0.6 0  (300 C); v = 349 m/s  C  22-4. The speed of longitudinal waves in a certain metal rod of density 7850 kg/m3 is measured to be 3380 m/s. What is the Young's modulus for the metal? Y Y v ; v2  ; Y = v2 = (7850 kg/m3)(3380 m/s)2; Y = 8.97 x 1010 Pa   300
• 2. Chapter 22. Sound Physics, 6th Edition 22-5. If the frequency of the waves in Problem 22-4 is 312 Hz, what is the wavelength? v 3380 m/s v  f ;   ;  = 10.8 m f 312 Hz 22-6. Compare the theoretical speeds of sound in hydrogen (M = 2.0 g/mol,  = 1.4) with helium (M = 4.0 g/mol,  = 1.66) at 00C.  RT 1.4)(8.314 J/mol K)(273 K) vH   ; vH = 1260 m/s M 0.002 kg/mol  RT 1.66)(8.314 J/mol K)(273 K) vHe   ; vHe = 971 m/s M 0.004 kg/mol vHe 971 m/s  ; vHe = 0.771 vH vH 1260 m/s *22-7. A sound wave is sent from a ship to the ocean floor, where it is reflected and returned. If the round trip takes 0.6 s, how deep is the ocean floor? Consider the bulk modulus for sea water to be 2.1 x 109 Pa and its density to be 1030 kg/m3. B 2.1 x 109 Pa v  ; v = 1428 m/s  1030 kg/m3 h = vt = (1328 m/s)(0.3 s); h = 428 m Vibrating Air Columns 22-8. Find the fundamental frequency and the first three overtones for a 20-cm pipe at 200C if the pipe is open at both ends. v = 331 m/s + (0.6 )(300) = 343 m/s. nv (1)(343 m/s) fn  ; f1   858 Hz f1 = 858 Hz 2l (2)(0.20 m) (First overtone, n = 2) fn = nf1; f2 = 2(857.5 Hz) = 1715 Hz 301
• 3. Chapter 22. Sound Physics, 6th Edition 22-8 (Cont.) (2nd overtone, n = 3) fn = nf1; f2 = 3(857.5 Hz) = 2573 Hz (3rd overtone, n = 4) fn = nf1; f2 = 4(857.5 Hz) = 3430 Hz 22-9. Find the fundamental frequency and the first three overtones for a 20-cm pipe at 200C if the pipe is closed at one end. nv (1)(343 m/s) fn  ; f1   429 Hz f1 = 429 Hz 4l (4)(0.20 m) (First overtone, n = 3) fn = nf1; f2 = 3(429 Hz) = 1290 Hz (2nd overtone, n = 5) fn = nf1; f2 = 5(429 Hz) = 2140 Hz (3rd overtone, n = 7) fn = nf1; f2 = 7(429 Hz) = 3000 Hz 22-10. What length of closed pipe will produce a fundamental frequency of 256 Hz at 200C? nv (1)(343 m/s) fn  ; l ; l = 0.335 m; l = 33.5 cm 4l 4(256 Hz) 22-11. What length of open pipe will produce a fundamental frequency of 356 Hz at 200C? nv (1)(343 m/s) fn  ; l ; l = 0.482 m; l = 48.2 cm 2l 2(356 Hz) 22-12 What length of open pipe will produce a frequency of 1200 Hz as it first overtone on a day when the speed of sound is 340 m/s? [ For open pipe, first overtone is for n = 2 ] nv 2(340 m/s) fn  ; l ; l = 28.3 cm 2l 2(1200 Hz) 22-13. The second overtone of a closed pipe is 1200 Hz at 200C. What is the length of the pipe. The second overtone for a closed pipe occurs when n = 5, and v = 343 m/s. nv 5(343 m/s) fn  ; l ; l = 35.7 cm 4l 4(1200 Hz) 302
• 4. Chapter 22. Sound Physics, 6th Edition *22-14. In a resonance experiment, the air in a closed tube of variable length is found to resonate with a tuning fork when the air column is first 6 cm and then 18 cm long. What is the frequency of the tuning fork if the temperature is 200C? [ v = 343 m/s at 200C. ] The distance between adjacent nodes of resonance is one-half of a wavelength.  v 343 m/s  18 cm - 6 cm;   24 cm; f   ; f = 1430 Hz 2  0.24 m *22-15. A closed pipe and an open pipe are each 3 m long. Compare the wavelength of the fourth overtone for each pipe at 200C. ? (Only odd harmonics allowed for closed pipe.) For an open pipe, the fourth overtone is the fifth harmonic, n = 5. 2l 2(3 m) Open (fourth overtone): n  ; 5  ; 5 = 1.20 m n 5 For closed pipe, the fourth overtone is the ninth harmonic, n = 9. 4l 4(3 m) Closed (fourth overtone): n  ; 9  ; 9 = 1.33 m n 9 Sound Intensity and Intensity Level 22-16. What is the intensity level in decibels of a sound whose intensity is 4 x 10-5 W/m2? I 4 x 10-5 W/m2   10log  10log ;  = 76.0 dB I0 1 x 10-12 W/m2 22-17. The intensity of a sound is 6 x 10-8 W/m2. What is the intensity level? I 6 x 10-8 W/m2   10log  10log ;  = 47.8 dB I0 1 x 10-12 W/m2 303
• 5. Chapter 22. Sound Physics, 6th Edition 22-18. A 60 dB sound is measured at a particular distance from a whistle. What is the intensity of this sound in W/m2? I I   10 log  60 dB; 106  ; I = (106)(1 x 10-12 W/m2); I = 1 x 10-6 W/m2 I0 I0 *22-19. What is the intensity of a 40 dB sound? I I   10 log  40 dB; 10 4  ; I = (104)(1 x 10-12 W/m2); I = 1 x 10-8 W/m2 I0 I0 *22-20. Compute the intensities for sounds of 10 dB, 20 dB, and 30 dB. I I   10 log  10 dB; 101  ; I = (101)(1 x 10-12 W/m2); I = 1 x 10-11 W/m2 I0 I0 I I   10 log  20 dB; 10 2  ; I = (102)(1 x 10-12 W/m2); I = 1 x 10-10 W/m2 I0 I0 I I   10 log  30 dB; 103  ; I = (103)(1 x 10-12 W/m2); I = 1 x 10-9 W/m2 I0 I0 22-21. Compute the intensity levels for sounds of 1 x 10-6 W/m2, 2 x 10-6 W/m2, and 3 x 10-6 W/m2. I 1 x 10-6 W/m2   10log  10log ;  = 60.0 dB I0 1 x 10-12 W/m2 I 2 x 10-6 W/m2   10log  10log ;  = 63.0 dB I0 1 x 10-12 W/m2 I 3 x 10-6 W/m2   10log  10log ;  = 64.8 dB I0 1 x 10-12 W/m2 304
• 6. Chapter 22. Sound Physics, 6th Edition 22-22. An isometric source of sound broadcasts a power of 60 W. What are the intensity and the intensity level of a sound heard at a distance of 4 m from this source? P 60 W I   0.2984 W/m 2 ; I = 0.298 W/m2 4 r 2 4 (4 m) 2 I 0.2984 W/m2   10log  10log ;  = 115 dB I0 1 x 10-12 W/m2 22-23. A 3.0-W sound source is located 6.5 m from an observer. What are the intensity and the intensity level of the sound heard at that distance? P 3.0 W I   5.65 x 10-3 W/m 2 ; I = 5.65 x 10-3 W/m2 4 r 2 4 (6.5 m) 2 I 0.2984 W/m2   10log  10log ;  = 97.5 dB I0 1 x 10-12 W/m2 22-24. A person located 6 m from a sound source hears an intensity of 2 x 10-4 W/m2. What intensity is heard by a person located 2.5 m from the source? I1r12 (2 x 10-4 W/m2 )(6 m)2 I1r12  I 2 r22 ; I2   ; I = 1.15 x 10-3 W/m2 r22 (2.5 m)2 *22-25. The intensity level 6 m from a source is 80 dB. What is the intensity level at a distance of 15.6 m from the same source? I I   10 log  80 dB; 108  ; I = (108)(1 x 10-12 W/m2); I = 1 x 10-4 W/m2 I0 I0 I1r12 (1 x 10-4 W/m2 )(6 m)2 I1r12  I 2 r22 ; I2   ; I2 = 1.48 x 10-5 W/m2 r22 (15.6 m)2 I 1.48 x 10-5 W/m2   10log  10log ;  = 71.7 dB I0 1 x 10-12 W/m2 305
• 7. Chapter 22. Sound Physics, 6th Edition The Doppler Effect Assume that the speed of sound in 343 m/s for all of these problems. 22-26. A stationary source of sound emits a signal at a frequency of 290 Hz. What are the frequencies heard by an observer (a) moving toward the source at 20 m/s and (b) moving away from the source at 20 m/s? (Approach = +, recede = -) V  v0  343 m/s + 20 m/s  f0  f s  290 Hz  ; f0 = 307 Hz V  vs  343 m/s - 0  V  v0  343 m/s + (-20 m/s)  f0  f s  290 Hz  ; f0 = 273 Hz V  vs  343 m/s - 0  22-27. A car blowing a 560-Hz horn moves at a speed of 15 m/s as it first approaches a stationary listener and then moves away from a stationary listener at the same speed. What are the frequencies heard by the listener? (Approach = +, recede = -) V  v0  343 m/s + 0  f0  f s  560 Hz  ; f0 = 586 Hz V  vs  343 m/s - 15 m/s  V  v0  343 m/s + 0  f0  f s  360 Hz  ; f0 = 537 Hz V  vs  343 m/s - (-15 m/s)  22-28. A person stranded in a car blows a 400-Hz horn. What frequencies are heard by the driver of a car passing at a speed of 60 km/h? (Approach = +, recede = -) km  1000 m   1 h  v0  60     16.7 m/s h  1 km   3600 s  V  v0  343 m/s + 16.7 m/s  Approaching: f 0  f s  400 Hz  ; f0 = 419 Hz V  vs  343 m/s - 0  At same point as car there is no change: f0 = 400 Hz V  v0  343 m/s + (-16.7 m/s)  Leaving: f0  f s  400 Hz  ; f0 = 381 Hz V  vs  343 m/s - 0  306
• 8. Chapter 22. Sound Physics, 6th Edition 22-29. A train moving at 20 m/s blows a 300-Hz whistle as it passes a stationary observer. What are the frequencies heard by the observer as the train passes? V  v0  343 m/s + 0  Approaching: f 0  f s  300 Hz  ; f0 = 319 Hz V  vs  343 m/s - 20 m/s  When at the same position there is no change: f0 = 300 Hz V  v0  343 m/s + 0  Leaving: f0  f s  300 Hz  ; f0 = 283 Hz V  vs  343 m/s - (-20 m/s)  *22-30. A child riding a bicycle north at 6 m/s hears a 600 Hz siren from a police car heading south at 15 m/s. What is the frequency heard by the child? (Approaches are +) V  v0  343 m/s + 6 m/s  Approaching: f 0  f s  600 Hz  ; f0 = 638 Hz V  vs  343 m/s - 15 m/s  *22-31. An ambulance moves northward at 15 m/s. Its siren has a frequency of 600 Hz at rest. A car heads south at 20 m/s toward the ambulance. What frequencies are heard by the car driver before and after they pass? (Approach = +, recede = -) V  v0  343 m/s + 20 m/s  Before passing: f 0  f s  600 Hz  ; f0 = 664 Hz V  vs  343 m/s - 15 m/s  V  v0  343 m/s + (-20 m/s)  After passing: f0  f s  600 Hz  ; f0 = 541 Hz V  vs  343 m/s - (-15 m/s)  *22-32. A truck traveling at 24 m/s overtakes a car traveling at 10 m/s in the same direction. The trucker blows a 600-Hz horn. What frequency is heard by the car driver? The car is moving away, so v0 = -10 m/s; Truck is approaching, so vs = +24 m/s V  v0  343 m/s + (-10 m/s)  f0  f s  600 Hz  343 m/s - (+24 m/s)  ; f0 = 626 Hz V  vs   307
• 9. Chapter 22. Sound Physics, 6th Edition *22-33. A 500-Hz sound source is heard by a stationary observer at a frequency of 475 Hz. What is the speed of the train? Is it moving toward the observer or away from the observer? V  v0 fo V  0 f 0 475 Hz f0  f s ;  ;   0.950 V  vs f s V  vs f s 500 Hz V  0.950; V  0.95V  0.95 vs ; 0.95 vs  0.05V V  vs 0.05(343 m/s) vs  ; vs = -18.1 m/s, away 0.950 The negative sign means the train is moving away from the observer. Challenge Problems 22-34. The speed of sound in a certain metal rod is 4600 m/s and the density of the metal is 5230 kg/m3. What is Young’s modulus for this metal? Y Y v ; v2  ; Y =  v2 = (5230 kg/m3)(4600 m/s)2; Y = 1.11 x 1011 Pa   22-35. A sonar beam travels in a fluid for a distance of 200 m in 0.12 s. The bulk modulus of elasticity for the fluid is 2600 MPa. What is the density of the fluid? 200 m B B B 2.60 x 109 Pa v  1667 m/s; v  ; v2  ;   0.12 s   v 2 (1667 m/s)2  = 936 kg/m3 22-36. What is the frequency of the third overtone for a closed pipe of length 60 cm? 7v (7)(343 m/s) Third overtone is for n = 7: f 7  ; f7  f1 = 1000 Hz 4l (4)(0.60 m) 308
• 10. Chapter 22. Sound Physics, 6th Edition 22-37. A 40-g string 2 m in length vibrates in three loops. The tension in the string is 270 N. What is the wavelength? What is the frequency? 2l 2(2 m) Fl (270 N)(2 m)   ;  = 1.33 m v  n 3 m 0.040 kg v 116 m/s v = 116 ft/s; f   ; f = 87.1 Hz  1.33 m 22-38. How many beats per second are heard when two tuning forks of 256 Hz and 259 Hz are sounded together? f’- f = 259 Hz – 256 Hz = 3 beats/s *22-39. What is the length of a closed pipe if the frequency of its second overtone is 900 Hz on a day when the temperature is 20C? The second overtone for a closed pipe occurs when n = 5, and v = 343 m/s. nv 5(343 m/s) fn  ; l ; l = 47.6 cm 4l 4(900 Hz) *22-40. The fundamental frequency for an open pipe is 360 Hz. If one end of this pipe is closed, what will be the new fundamental frequency? (n = 1 for fundamental ) We will first find the length of an open pipe that has a frequency of 360 Hz nv 1(343 m/s) fn  ; l ; l = 47.6 cm 2l 2(360 Hz) Now, take this length for a closed pipe to find new frequency: (1)v (343 m/s) f1  ; f  ; f = 180 Hz 4l 4(0.476 m) 309
• 11. Chapter 22. Sound Physics, 6th Edition *22-41. A 60-cm steel rod is clamped at one end as shown in Fig. 23-13a. Sketch the fundamental and the first overtone for these boundary conditions. What are the wavelengths in each case? Boundary conditions = node at right, antinode to left: Fundamental 1 = 4l = 4(0.60 m); 1 = 2.40 m First overtone adds one node, 1st ovt. = 3rd harmonic. First overtone 4l 4(0.60 m) 3   ; 3 = 0.800 m 3 3 *22-42. The 60-cm rod in Fig. 22-13b is now clamped at its midpoint. What are the wavelengths for the fundamental and first overtone? Fundamental Boundary conditions: A node must be at the center, and an antinode must be at each end in both cases. First overtone 2l Fundamental: 1   2(0.60 m); 1 = 1.20 m 1 First overtone is first possibility after the fundamental. Because of clamp at midpoint, 2l 2(0.60 m) First overtone:     = 0.400 m 3 3 *22-43. The velocity of sound in a steel rod is 5060 m/s. What is the length of a steel rod mounted as shown in Fig. 22-13a if the fundamental frequency of vibration for the rod is 3000 Hz? v 5060 m/s Fundamental v  f ;   ;  = 1.69 m f 3000 Hz  1.69 m For fundamental: 1 = 4l; l  ; l = 42.2 cm 4 4 310
• 12. Chapter 22. Sound Physics, 6th Edition *22-44. Find the ratio of the intensities of two sounds if one is 12 dB higher than the other? I1 I2 I2 I 1  10 log ;  2  10 log ;  2  1  10 log  10 log 2 I0 I0 I0 I0 Recall that: log A – log B = log (A/B) and apply to the above relation I2 I I2  2  1  10 log  10 log 2 ;  2  1  10 log I0 I0 I1 I2 I2 I2  2  1  10 log  12 dB; log  1.2; 101.2  I1 I1 I1 (I2/I1) = 15.8 *22-45. A certain loud speaker has a circular opening of area 6 cm2. The power radiated by this speaker is 6 x 10-7 W. What is the intensity of the sound at the opening? What is the intensity level? P 6 x 10-7 W A  6 cm 2  6 x 10-4 m 2 ; I  ; I = 1 x 10-3 W/m2 A 6 x 10-4 m 2 I 1 x 10-3 W/m2   10log  10log ;  = 90 dB I0 1 x 10-12 W/m2 *22-46. The noon whistle at the textile mill has a frequency of 360 Hz What are the frequencies heard by the driver of a car passing the mill at 25 m/s on a day when sound travels at 343 m/s? V  v0  343 m/s + 25 m/s  Approaching: f 0  f s  360 Hz  ; f0 = 386 Hz V  vs  343 m/s - 0  At same point as source: f0 = 360 Hz V  v0  343 m/s - 25 m/s  Leaving: f 0  f s  360 Hz  ; f0 = 334 Hz V  vs  343 m/s - 0  311
• 13. Chapter 22. Sound Physics, 6th Edition *22-47. What is the difference in intensity levels (dB) for two sounds whose intensities are 2 x 10-5 W/m2 and 0.90 W/m2? [ See solution for Prob. 22-44 above. ] I2 I I2  2  1  10 log  10 log 2 ;  2  1  10 log I0 I0 I1 0.90 W/m 2  2  1  10 log ;  = 46.5 dB 2 x 10-5 W/m 2 Critical Thinking Questions *22-48. By inhaling helium gas, one can raise the frequency of the voice considerably. For air M = 29 g/mol and  = 1.4; for helium M = 4.0 g/mol and  = 1.66. At a temperature of 270C, you sing a “C” note at 256 Hz. What is the frequency that will be heard if you inhale helium gas and other parameters are unchanged? Notice that both v and f were increased. How do you explain this in view of the fact that v = f. Discuss.  RT 1.4)(8.314 J/mol K)(300 K) vair   ; vH = 347 m/s M 0.029 kg/mol  RT 1.66)(8.314 J/mol K)(300 K) vHe   ; vHe = 1017 m/s M 0.004 kg/mol The fundamental wavelength is a property of the boundary conditions which don’t change. Therefore, the frequencies are directly proportional to the velocities. f He 1017 m/s   2.93; fHe = 2.93 fair = 2.93(256 Hz); fHe = 750 Hz f air 347 m/s The ratio of v/f is constant and equal to the wavelength in each case. 312
• 14. Chapter 22. Sound Physics, 6th Edition *22-49. A toy whistle is made out of a piece of sugar cane that is 8 cm long. It is essentially an open pipe from the air inlet to the far end. Now suppose a hole is bored at the midpoint so that the finger can alternately close and open the hole. (a) If the velocity of sound is 340m/s, what are the two possible fundamental frequencies obtained by closing and opening the hole at the center of the cane? (b) What is the fundamental frequency if the finger covers the hole and the far end is plugged? An opening forces an antinode to occur at that point. 0.08 m (a) With the finger closing the hole, the fundamental is:  = 2l = 2(0.08 m); 1 = 1.60 m 340 m/s f1  ; f1 = 212 Hz 1.60 m Now, with the finger removed, the fundamental is:  = l = 0.08 m, so that 340 m/s f1  ; f1 = 425 m 0.80 m (b) Now if the far end is plugged and the hole is covered, f1 = 4l: 340 m/s  = 4(0.08 m) = 3.2 m; f  ; f = 106 Hz 3.20 m *22-50. A tuning fork of frequency 512 Hz is moved away from an observer and toward a flat wall with a speed of 3 m/s. The speed of sound in the air is 340 m/s. What is the apparent frequency of the unreflected sound? What is the apparent frequency of the reflected sound? How many beats are heard each second? For unrelfected sound, vs = +3 m/s; for reflected sound, vs = -3 m/s V  v0  340 m/s + 0  Unreflected: f0  f s  (512 Hz)   f0 = 517 Hz V  vs  340 m/s -(+3 m/s)  313
• 15. Chapter 22. Sound Physics, 6th Edition V  v0  340 m/s + 0  *22-50. (Cont.) Reflected: f0  f s  (512 Hz)   f0 = 508 Hz V  vs  340 m/s -(-3 m/s)  Number of beats = 517 Hz – 508 Hz = 9 beats/s *22-51. Using the logarithmic definition of decibels, derive the following expression which relates the ratio of intensities of two sounds to the difference in decibels for the sounds: I2  2   1  10 log I1 Use this relationship to work Problems 22-44 and 22-47. I1 I2 I2 I 1  10 log ;  2  10 log ;  2  1  10 log  10 log 2 I0 I0 I0 I0 Recall that: log A – log B = log (A/B) and apply to the above relation I2 I I2  2  1  10 log  10 log 2 ;  2  1  10 log I0 I0 I1 Refer to Problems 22-44 and 22-47 for applications of this formula. 22-52. The laboratory apparatus shown in Fig. 22-14 is used to measure the speed of sound in air by the resonance method. A vibrating tuning fork of frequency f is held over the open end of a tube, partly filled with water. The length of the air column can be varied by changing the water level. As the water level is gradually lowered from the top of the tube, the sound intensity reaches a maximum at the three levels shown in the figure. The maxima occur whenever the air column resonates with the tuning fork. Thus, the distance between successive resonance positions is the distance between adjacent notes for the standing waves in the air column. The frequency of the fork is 512 Hz, and the 314
• 16. Chapter 22. Sound Physics, 6th Edition resonance positions occur at 17, 51, and 85 cm from the top of the tube. What is the velocity of sound in the air? What is the approximate temperature in the room? The distance between adjacent resonance points must be equal to one-half of a wavelength. Therefore, the wavelength of the sound must be:  = 2(51 cm – 17 cm) = 68.0 cm v = f = (512 Hz)(0.680 m); v = 348 m/s  m/s  v  331 m/s   0.6 2  t  348 m/s t = 28.30C  C  *21-53. What is the difference in intensity levels of two sounds, one being twice the intensity of the other? I2 2I Assume I2 = 2I1:  2  1  10 log  10 log 1 ;  = 3.01 dB I1 I1 315