Chapter15Solutions

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Chapter 15: Sound B Practice Problems page 309 1, ) Sound with a frequency of 261.6 Hz travels through water at a speed of 1435 m/s. Find its wavelength in water. y=Afsods?o 435 m/s F-2%1.6 7z =548 m Find the frequency of a sound wave moving in air at room temperature with a wavelength of 0.667 m, The human ear can detect sounds with frequencies between 20 Hz and 16 kHz. Find the largest and smallest wavelengths the ear can detect, assuming the sound travels through air with a speed of 343 m/s at 20°C. From v = Af the largest wavelength is 343m/s _ ] . 0Hz = 17 m; the smallest is ¥ f 4 343 m/s S 16000 Hz A= i} A = 0.021 m. ] What is the frequency of sound in air at 20°C having a wavelength equal to the diameter of a 15-inch (38 cm) "woofer" loudspeaker? Woofer diameter 38 cm, oY _ (8 m/y ~ 27038 m) = 900 Hz What is the frequency of sound in air at 20°C having a wavelength equal to the diameter of a 3-inch (7.6 cm) diameter "tweeter"? Tweeter diameter 7.6 c¢m =Yy _ (343 m/s) _ Practice Problems 5. A 440-Hz tuning fork is held above a closed pipe. Find the spacings between the resonances when the air temperature is 20°C., Resonance spacing is % S0 using v = Af the resonance spacing is 6. The 440-Hz tuning fork is used with a resonating column to determine the velocity of sound in helium gas. If the spacings between resonances are 110 cm, what is the velocity of sound in He? : Resonance spacing = % = 1.10 m so A=220m and v = fA = (440 Hz)(2.20 m) = 968 m/s. 7. The frequency of a tuning fork is unknown. A student uses an air column at 27°C and finds resonances spaced by 39.2 cm. What is the frequency of the tuning fork? From the previous example problem v = 347 m/s at 27°C and the resonance spacing gives % =0392mor A=078 m. Using v = A, f = %=(o 784 my = 443 Hz, 8. The auditory canal, leading to the ear drum, is a closed pipe 3.0 cm long. Find the approximate value (ignoring end correction) of the lowest resonant frequency. f=77= (343 m)(4 X 003 m) = 3.0 ki Chapter 15 Sound 175

Practice Problems 9. 10. 176 A bugle can be thought of as an open pipe. If a bugle were straightened out, it would be 2.65 m long. The lowest resonant frequency of an open pipe corresponds to the wavelength 2,, where M = L = length of pipe. Further resonances 2 are spaced % apart, giving the series of resonance wavelengths — 2'1 2’2 2'3 L-8afB]5fE) a. If the speed of sound is 343 m/s, find the lowest frequency that is resonant in a bugle, 2L = 2(2.65 m) = 530 m so that the Al = lowest frequency is v_343m/s _ fl—l—f 530 m = 64.7 Hz b. Find the next two higher resonant frequencies in the bugle. v_v _343m/s _ ol e e N R = v_3v_3(343 m/s) _ A= =X Tgesm = 194 He A soprano saxophone is an open pipe. If all keys are closed, it is approximately 65 cm long. Using 343 m/s as the speed of sound, find the lowest frequency that can be played on this instrument (ignoring end corrections). The lowest resonant frequency corresporids to the wavelength given by %— = L, the length of the pipe. A = 2L = 2(0.65 m) = 1.30 m so f=% = 3‘;—33—‘6‘/—; = 20 Hz. Since the saxophone is an open pipe, Amix = 2 X (pipe length) = 2(0.65 m) =130 m v 343 m/s _ Join = 7= 35y = 260 Hz Chapter 15 Sound Practice Problems page 321 11. A 330-Hz and a 333-Hz tuning fork are struck simultaneously. What will the beat frequency be? Beat frequency = ]fz - fll = |333 Hz - 330 Hz| =3 Hz 12. A student has two tuning forks, one with a frequency of 349 Hz and the other with an unknown frequency. When struck together, the tuning forks produce three beats a second. What is the frequency of the unknown tuning fork? : The frequency of the second fork could be either So2=fi + fiur =349 Hz + 3 Hz = 352 Hz or fo = fi = fruw =349 Hz - 3 Hz = 346 Hz. Chapter Review Problems pages 326-327 1. Andrew hears the sound of the firing of a distant cannon 6.00 s after sceing the flash, How far from the cannon is Andrew? d = vt = (343 m/s)(6.00 s) = 2.06 X 10° m 2. Arifle is fired in a valley with parallel vertical walls. The echo from one wall is heard 2.0 s after the rifle was fired. The echo from the other wall is heard 2.0 s after the first echo. How wide is the valley? The time it takes sound to go to wall 1 and back is 2.0 s. The time it takes to go to the wall is half the total time or 1.0 s, ' dy = vt = (343 m/s)(1.0 s) = 3.4 X 102 m. The total time for the sound to go to wall 2 is half of 40 sor 2.0s. dy = vt = (343 m/s)(2.0 s) = 6.8 X 102 m. The total distance is d, + d, = 1.02 X 10° m.

Chapter Review Problems If Karen claps her hands and hears the echo from a distant wall 0.20 seconds later, how far away is the wall? The total distance = vyt = (343 m/5)(0.20 s) = 68.6 m, so the distance to the wall is half this, or 343 m, . If Karen shouts across a canyon and hears an echo 4,00 seconds later, how wide is the canyon? d = vt = (343 m/s)(4.00 §) = 1372 m is the total distance travelled. The distance to the wall is %(1372) = 686 m. A certain instant camera determines the distance to the subject by sending out a sound wave and measuring the time needed for the echo to return to the camera. How long would it take the sound wave to return to the camera if the subject were 3.00 m away? The total distance the sound must travel is 6.00 m. d=vt, so _d_(6.00 m) = v - m= 0.0175 s. t Carol drops a stone into a mine shaft 1225 m deep. How soon after she drops the stone does she hear it hit the bottom of the shaft? First find the time it takes the stone to fall down the shaft by d = lgt’, S0 2 d -[-1225 t = 1 = 1 = ), N % -2-(—9.80) 500 s The time it takes the sound to come back up is found with d = vy, so The total time is 5.00 s + 0.36 § = 5.36 s. If the wavelength of a 4.40 X 102 Hz sound in fresh water is 3.30 m, what is the speed of sound in water? v =fA = (440 X 10 Hz)(3.30 m) = 145 X 10° m/s Chapter Review Problems 8. 10. 11. Sound with a frequency of 442 Hgz travels through steel. A wavelength of 11.66 m is measured. Find the speed of the sound in steel, v =fA = (442 Hz)(11.66 m) = 1.97 X 10 m/s The sound emitted by bats has a wavelength of 3.5 mm. What is its frequency in air? =y _0G43 m/s) _ f= %= 0.0035m) = 9.8 X 10 Hz Ultrasound with a frequency of 4.25 MHz can be used to produce images of the human body. If the speed of sound in the body is the same as in salt water, 1.50 km/s, what is the wavelength in the body? v = Af, s0 2 = ¥ - (1.50 km/s)(1000 m/km) T 7T (825X 10° Hz) =353 X10¢m The equation for the Doppler shift of a sound wave of speed v, reaching a moving detector, is V+ v, 7 = g detector and v, is the speed of the source. If the detector moves toward the source, vy is positive and if the source moves toward the detector v, is positive. A train moving toward a detector at 31 m/s blows a 305-Hz hom. What is detected by a » where v, is the speed of the a. stationary train? _ v+ 7 -2y _ (305 Hz)(343 m/s + 0) T (343 m/s - 31 m/s) = 340 Hz b. train moving toward the first frequency at 21 m/s? : A _ (305 Hz)(343 m/s + (21 m/s)) - (343 m/s - 31 m/s) = 360 Hz Chapter 15 ‘Sound 177

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Chapter Review Problems 12, 13. 14, 178 34 A _ A . The train in the previdus problem is moving - away from the detector, is detected by a Now what frequency a. stationary train? V+y, f =f[v-vj] ) 343 m/s+ 0 = (305 Hz) [343 m/s - (=31 m/s)] = 280 Hz b. train moving away from the first train at 21 m/s? f=ffiiﬂ v—v, _ 343 m/s + (=21 m/s) = (305 Hz) {343 m/s - (=31 m/s)] = 263 Hz A slide whistle has a length of 27 cm. If you want to play a note one octave higher, how long should the whistle be? _ AL _ 4Q7cm) A= 7 = 35— = 36 cm. octave higher is the first overtone of the A note one fundamental. Resonances are spaced by %— wavelength. Since the original whistle length of 27 cm = 3 the wavelength of the first 4 overtone (octave), then the shortest whistle length for the first overtone equals 23" 3 F="g =9cm Adam, an airport worker near a jet plane taking off experiences a sound level of 150 dB. a. If Adam wore ear protectors that reduce the sound level to that of a chain saw, what decrease in dB would be required? Chain saw is 110 dB, so 40 dB reduction is needed. b. If Adam now heard something that sounded like a whisper, what would a person not wearing the protectors hear? A soft whisper is 10 dB, so the actual level would be 50 dB, or that of an average classroom. Chapter 15 Sound ~ Chapter Review Problems 15. A rock band plays at a 80 dB sound level, 16. 17. How many times greater is the sound pressure from another rock band playing at a. 100 dB? Each 20 dB increases pressure by a factor of 10, so 10 dB. b. 120 dB? 100 dB An open vertical tube is filled with water and a tuning fork vibrates over its mouth. As the water level is lowered in the tube, resonance is heard when the water level has dropped 17 cm, and again after 49 cm of distance exists from the water to the top of the tube. What is the frequency of the tuning fork? 49 cm - 17 ¢cm = 32 ¢cm, or 0.32 m. Since the tube is closed at one end, 13, exists between points of resonance. P =032m, s0A=064m v _ 343 m/s f o =71 06am=S40H If you hold a 1.0 m metal rod in the center and hit one end with a hammer, it will oscillate like an open pipe. Antinodes of air pressure correspond to nodes of molecular motion, so there is a pressure antinode in the center of the bar. The speed of sound in aluminum is 5150 m/s. What would be the lowest frequency of oscillation? The rod length is %2., s0A=20m, f=7="Gom =%k

Chapter Review Problems 18. The lowest note on an organ is 16.4 Hz. a. What is the shortest open organ pipe that will resonate at this frequency? A =;,V- | R - (205)"‘) =105 m b. What would be the pitch if the same organ pipe were closed? _f fo=3 £ (16"2‘ H2) . 820 12 19. During normal conversation the amplitude of the pressure wave is 0.020 N/m2. a. If the area of the eardrum is 0.52 cm?, what is the force on the eardrum? F = pA = (0.020 N/m»(0.52 X 10~ m?) =104 X 106 N b. The mechanical advantage of the bones in the inner ear is 1.5. What force is exerted on the ova! window? (1.5)(1.04 X 10 N) = 1.56 X 106 N c. The area of the oval window is 0.026 cm2. What is the pressure increase transmitted to the liquid in the cochlea? _ (1.56 X 10 N) _F _ 2 P = 7" 0.06X109my = 0-600 N/m 20. One tuning fork has a 445 Hz pitch. When a second fork is struck, beat notes occur with a frequency of 3 Hz, What are the two possible frequencies of the second fork? 445 Hz - 3 Hz = 442 Hz and 445 Hz + 3 Hz = 448 Hz Chapter Review Problems 21. A flute acts like an open pipe and sounds a note with a 370 Hz pitch. What are the frequencies of the second, third and fourth harmonics of this pitch? 2f = (2)(370 Hz) = 740 Hz 3f = (3)(370 Hz) = 1100 Hz 4f = (4)(370 Hz) = 1500 Hz 22. A clarinet also sounds the same note as the flute in the previous problern, 370 Hz. However, it only produces harmonics that are “odd multiples of the fundamental frequency. What are the frequencies of the lowest three harmonics produced by the clarinet? ' 3f = (3)(370 Hz) = 1100 Hz 5f = (5)(370 Hz) = 1900 Hz 7 = (7)(370 Hz) = 2600 Hz 23. One closed organ pipe has a length of 2.40 m. a. What is the frequency of the note played by this pipe? A=4l= (4)(2.40 m) = 9.60 m v =fA _V _ 343 m/s f=7=Tgom=35TH b. When a second pipe is played at the same time, a 1.40 Hz beat note is heard. By how much is the second pipe too long? f=357Hz ~ 140 Hz = 343 Hz v=fA v_343m/s _ Z.—]:-— 3~—4.3 HZ— 100 m A=4] A _100m I-Z-—4 =250m The difference in lengths is 250 m - 240 m = 0.10 m. Chapter 15 Sound 179

Chapter Review Problems 24. 25. 180 One organ pipe has a length of 836 mm. A second pipe should have a pitch one major third higher. How long should this pipe be? L ='2}:,sozk=2L;andv=ﬂ,S°f= 7 K _ (343m/s) T (2)(0.836 m) The ratio of a frequency one major third higher = 410 Hz, is 5:4, 50 (410 Hz) [g-] = 512 Hz. The length of the second pipe is The Doppler shift was first tested in 1845 by the French scientist B. Ballot. He had a trumpet player sound an A, 440 Hz, while riding on a flatcar pulled by a locomotive. At the same time, a stationary trumpeter played the same note. Ballot heard 3.0 beats per second. How fast was the train moving toward him? 440 Hz + 3.0 Hz = 443 Hz f‘=f[u s0 (v — v)f = (v + v)f and (V_—' vl) v, = - (v_*}v_ B (343 m/s — 0)(440 Hz) = (343 m/s) - YRR = 2.3 m/s A student wants to repeat Ballot's experiment, She plans to have a trumpet played in a rapidly moving car. Rather than listening for beat notes, she wants to have the car move fast enough so the moving trumpet sounds a major third above a stationary trumpet. Chertar 15 Fannd Chapter Review Problems a. How fast would the car have to move? major third ratio = 5:4 V-V =+ and v,=v———(v;,v")f=v—(v+v¢);7 (343 m/s) - (343 m/s - 0) [2] = 69 m/s = 250 km/h b. Should she try the experiment? No, do not try the experiment. Supplemental Problems (Appendix B) 1. The echo of a ship's fog horn, reflected from an iceberg, is heard 5.0 s after the horn is sounded. How far away is the iceberg? d = v, where t = %‘E = 2.5 s, since sound must travel to the iceberg and back. d=vt=(343 m/s)(2.5s) = 8.6 X 10* m What is the speed of sound that has a frequency of 250 Hz and a wavelength of 0.600 m? v = Af = (0.600 m)(250 Hz) = 150 m/s A sound wave has a frequency of 2000 Hz and travels along a steel rod. If the distance between successive compressions is 0.400 m, what is the speed of the wave? v = Af = (0.400 m)(2000 Hz) = 800 m/s What is the wavelength of a sound wave that has a frequency of 250 Hz and a speed of 400 m/s? A==t = 160 m

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Supplemental Problems (Appendix B) Supplemental Problems Sam, a train engineer, blows a whistle that has a frequency of 4.0 X 102 Hz as the train approaches a station. If the speed of the train is 25 m/s, what frequency will be heard by a person at the station? 5. What is the wavelength of sound that has a 9./ frequency of 539.8 Hz? =Y _(G4m/s) l—f—m—0635m 6. What is the wavelength of sound that has a frequency of 320.0 Hz? For the Doppler shift, f = f[:::"], where v, is the speed of the detector and v, is the speed _ (343 m/s) of the source. =Y Lom/s) ;L"f"(320.0 ) 107 m . 7 = o0 B[S — a0 3, 7. A stone is dropped into a mine shaft 2500 m deep. How many seconds pass before the stone 10 is heard to strike the bottom of the shaft? First Jane is on a train that is traveling at 95, kmyh. find the time needed for the stone to strike the bottom of the shaft, ‘ 2_d] i _ [2(250 .0 m)] I = 102 = = | 2\«V.0m) d =280 or t [g .80 m/s7) =714 s The time needed for the sound to travel to the top of the shaft is The train passes a factory whose whistle is blowing at 288 Hz. What frequency does Jane hear as the train approaches the factory? f = f[::::’:,, where va = (-95 km/h) [%—é-(*;—s] [gog_m] = 26 m/s vy is negative since its direction is opposite to that of the sound wave. h =-==—7 = (729 v (343m/s) F = (288 Hp[H3ms - (26 mg The total time is thus - 343 m/s-0m/s b=4+6=7145+0729s=787s = 310 Hz 8. A rifle is shot in a valley formed between two 11. What is the sound level of a sound that has a parallel mountains. The echo from one sound pressure one-tenth of 90 dB? mountain is heard after 2.00 s and from the other mountain 2.00 s later. What is the width When the sound pressure is multiplied by of the valley? one-tenth, the sound level decreases by 20 dB, so the sound level = 90 dB ~ 20 dB = 70 dB. Distance to first mountain: dy = vty = (343 m/s)(2.00 §/2) =343 m 12. What is the sound level of a sound that has a Distance to second mountain: d, = vt, = (343 m/s)(4.00 5/2) = 686 m Width of valley: wW=d+d =686m+343m=1030m sound pressure ten times 90 dB? When the sound pressure is multiplied by ten, the sound level increases by 20 dB, so the . sound level = 90 dB + 20 dB = 110 dB. Chapter 15 Sound 181

Supplemental Problems 13, 14. 15. 16. 17. A tuning fork produces a resonance with a closed tube 19.0 ¢m long. What is the lowest possible frequency of the tuning fork? The longest wavelength occurs when the length - of the closed pipe is one—fourth wavelength, so % = 19.0 cm, or A = 4(19.0 cm) = 76.0 cm The longest wavelength corresponds to the lowest frequency, How do the frequencies of notes that are an octave apart compare? The higher note has a frequency twice that of the lower note. Two tuning forks of 320 Hz and 324 Hz are sounded simultaneously. What frequency of sound will the listener hear? beat frequency = 324 Hz - 320 Hz = 4 Hz How many beats will be heard each second when a string with a frequency of 288 Hz is plucked simultaneously with another string that has a frequency of 296 Hz? beat frequency = 296 Hz — 288 Hz = 8 Hz = 8 beaty/s A tuning fork has a frequency of 440 Hz. If another tuning fork of slightly lower pitch is sounded at the same time, 5.0 beats per second are produced. What is the frequency of the second tuning fork? The second tuning fork has a lower frequency than the first, fi - f2=50Hz, or f2=f,-—5.0Hz=440Hz—5.0Hz=435Hz

Chapter15Solutions

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