SOUND. Responses to Questions

Transcription

1 SOUND Responses to Questions. Sound exhibits several phenomena that give evidence that it is a wave. ntererence is a wave phenomenon, and sound produces intererence (such as beats). Diraction is a wave phenomenon, and sound can be diracted (such as sound being heard around corners). Reraction is a wave phenomenon, and sound exhibits reraction when passing obliquely rom one medium to another. Sound also requires a medium, a characteristic o mechanical waves.. Evidence that sound is a orm o energy is ound in the act that sound can do work. A sound wave created in one location can cause the mechanical vibration o an object at a dierent location. For example, sound can set eardrums in motion, make windows rattle, or even shatter a glass. See Fig. 9 or a photograph o a goblet shattering rom the sound o a trumpet. 3. The child speaking into a cup creates sound waves that cause the bottom o the cup to vibrate. Since the string is tightly attached to the bottom o the cup, the vibrations o the cup are transmitted to longitudinal waves in the string. These waves travel down the string and cause the bottom o the receiver s cup to vibrate back and orth. This relatively large vibrating surace moves the adjacent air and generates sound waves rom the bottom o the cup that travel up into the cup. These waves are incident on the receiver s ear, and the receiver hears the sound rom the speaker. 4. the requency were to change, the two media could not stay in contact with each other. the two media vibrated with dierent requencies, then particles rom the two media initially in contact could not stay in contact with each other. But particles must be in contact in order or the wave to be transmitted rom one medium to the other, so the requency does not change. Since the wave speed changes in passing rom air into water and the requency does not change, we expect the wavelength to change. Sound waves travel about our times aster in water than in air, so we expect the wavelength in water to be about our times longer than it is in air. 5. the speed o sound in air depended signiicantly on requency, then the sounds that we hear would be separated in time according to requency. For example, i a chord were played by an orchestra, then we would hear the high notes at one time, the middle notes at another, and the lower notes at still another. This eect is not heard or a large range o distances, indicating that the speed o sound in air does not depend signiicantly on requency. Copyright 04 Pearson Education, nc. All rights reserved. This material is protected under all copyright laws as they currently exist. -

2 - Chapter 6. The sound-producing anatomy o a person includes various resonating cavities, such as the throat. The relatively ixed geometry o these cavities determines the relatively ixed wavelengths o sound that a person can produce. Those wavelengths have associated requencies given by λ /. The speed o sound is determined by the gas that is illing the resonant cavities. the person has inhaled helium, then the speed o sound will be much higher than normal, since the speed o sound waves in helium is about 3 times that in air. Thus, the person s requencies will go up by about a actor o 3. This is about a.5-octave shit, so the person s voice sounds very high pitched. n 7. The basic equation determining the pitch o the organ pipe is either closed, n odd integer, or a 4 n closed pipe, or open, n integer, or an open pipe. n each case, the requency is proportional to the speed o sound in air. Since the speed is a unction o temperature, and the length o any particular pipe is very nearly constant over the relatively small range o temperatures in a room, the requency is also a unction o temperature. Thus, when the temperature changes, the resonant requencies o the organ pipes change. Since the speed o sound increases with temperature, as the temperature increases, the pitch o the pipes increases as well. 8. A tube o a given length will resonate (permit standing waves) at certain requencies. When a mix o requencies is input to the tube, only those requencies close to resonant requencies will produce sound that persists, because standing waves are created or those requencies. Frequencies ar rom resonant requencies will not persist very long at all they will die out quickly., or example, two adjacent resonances o a tube are at 00 Hz and 00 Hz, then sound input near one o those requencies will persist and sound relatively loud. A sound input near 50 Hz would ade out quickly and thus have a reduced amplitude as compared to the resonant requencies. The length o the tube can thereore be chosen to ilter certain requencies, i those iltered requencies are not close to resonant requencies. 9. For a string with ixed ends, the undamental requency is given by, so the length o string or a given requency is. For a string, i the tension is not changed while retting, the speed o sound waves will be constant. Thus, or two requencies <, the spacing between the rets corresponding to those requencies is given as ollows: Now see Table 3. Each note there corresponds to one ret on the guitar neck. Notice that as the # adjacent requencies increase, the interrequency spacing also increases. The change rom C to C is # 5 Hz, while the change rom G to G is 3 Hz. Thus, their reciprocals get closer together, so rom the above ormula, the length spacing gets closer together. Consider a numerical example. 4 4 C (.07 0 ) C # G (.4 0 ) 6 77 G # G G # 0.68 C The G to C # # G spacing is only about 68% o the C to # C spacing. Copyright 04 Pearson Education, nc. All rights reserved. This material is protected under all copyright laws as they currently exist.

3 Sound When you irst hear the truck, you cannot see it. There is no straight-line path rom the truck to you. The sound waves that you are hearing are thereore arriving at your location due to diraction. Long wavelengths are diracted more than short wavelengths, so you are initially only hearing sound with long wavelengths, which are low-requency sounds. Ater you can see the truck, you are able to receive all requencies being emitted by the truck, not just the lower requencies. Thus, the sound brightens due to your hearing more high-requency components.. The wave pattern created by standing waves does not travel rom one place to another. The node locations are ixed in space. Any one point in the medium has the same amplitude at all times. Thus, the intererence can be described as intererence in space moving the observation point rom one location to another changes the intererence rom constructive (antinode) to destructive (node). To experience the ull range rom node to antinode, the position o observation must change, but all observations could be made at the same time by a group o observers. The wave pattern created by beats does travel rom one place to another. Any one point in the medium will at one time have a 0 amplitude (node) and hal a beat period later, have a maximum amplitude (antinode). Thus, the intererence can be described as intererence in time. To experience the ull range rom constructive intererence to destructive intererence, the time o observation must change, but all observations could be made at the same position.. the requency o the speakers is lowered, then the wavelength will be increased. Each circle in the diagram will be larger, so the points C and D will move arther apart. 3. Active noise reduction devices work on the principle o destructive intererence. the electronics are ast enough to detect the noise, invert it, and create the opposite wave (80 out o phase with the original) in signiicantly less time than one period o the components o the noise, then the original noise and the created noise will be approximately in a destructive intererence relationship. The person wearing the headphones will then hear a net sound signal that is very low in intensity. 4. For the two waves shown, the requency o beating is higher in wave (a) the beats occur more requently. The beat requency is the dierence between the two component requencies, so since (a) has a higher beat requency, the component requencies are arther apart in (a). 5. There is no Doppler shit i the source and observer move in the same direction, with the same velocity. Doppler shit is caused by relative motion between source and observer, and i both source and observer move in the same direction with the same velocity, there is no relative motion. 6. the wind is blowing but the listener is at rest with respect to the source, the listener will not hear a Doppler eect. We analyze the case o the wind blowing rom the source toward the listener. The moving air (wind) has the same eect as i the speed o sound had been increased by an amount equal to the wind speed. The wavelength o the sound waves (distance that a wave travels during one period o time) will be increased by the same percentage that the wind speed is relative to the still-air speed o sound. Since the requency is the speed divided by the wavelength, the requency does not change, so there is no Doppler eect to hear. Alternatively, the wind has the same eect as i the air were not moving but the source and listener were moving at the same speed in the same direction. See Question 5 or a discussion o that situation. Finally, since there is no relative motion between the source and the listener, there is no Doppler shit. 7. The highest requency o sound will be heard at position C, while the child is swinging orward. Assuming the child is moving with SHM, the highest speed is at the equilibrium point, point C. And to have an increased pitch, the relative motion o the source and detector must be toward each other. The child would also hear the lowest requency o sound at point C, while swinging backward. Copyright 04 Pearson Education, nc. All rights reserved. This material is protected under all copyright laws as they currently exist.

4 -4 Chapter Responses to MisConceptual Questions. (a) Students may answer that the speed o sound is the same, i they do not understand that the speed o sound is not constant, but depends upon the temperature o the air. When it is hotter, the speed o sound is greater, so it takes less time or the echo to return.. (d) Sound waves are longitudinal waves, so (a) is incorrect. The sound waves can be characterized either by the longitudinal displacement o the air molecules or by the pressure dierences that cause the displacements. 3. A common misconception is to treat the sound intensity level as a linear scale instead o a logarithmic scale. the sound intensity doubles, the intensity level increases by about 3 db, so the correct answer is 73 db. 4. (e) Students oten think that the sound intensity is the same as loudness and thereore mistakenly answer that doubling the intensity will double the loudness. However, the ear interprets loudness on a logarithmic scale. For something to sound twice as loud, it must have an intensity that is 0 times as great. 5. The octave is a measure o musical requency, not loudness. Raising a note by one octave requires doubling the requency. Thereore, raising a note by two octaves is doubling the requency twice, which is the same as quadrupling the requency. 6. (e) n a string or open tube the lowest vibration mode is equal to hal o a wavelength. n a tube closed at one end the lowest vibration mode is equal to a quarter o a wavelength. Thereore, none o the listed objects have a lowest vibration mode equal to a wavelength. 7. (e) A common misconception is that the requency o a sound changes as it passes rom air to water. The requency is the number o wave crests that pass a certain point per unit time. this value were to change as it entered the water, then wave crests would build up or be depleted over time. This would make the interace an energy source or sink, which it is not. The speed o sound in water is greater than in air, so the speed o the wave changes. Since the requency cannot change, the increase in speed results in an increase in wavelength. 8. (e) As the string oscillates, it causes the air to vibrate at the same requency. Thereore, the sound wave will have the same requency as the guitar string, so answers and (c) are incorrect. The speed o sound in air at 0 C is 343 m/s. The speed o sound in the string is the product o the wavelength and requency, 46 m/s, so the sound waves in air have a shorter wavelength than the waves on the string. 9. (c) Pushing the string straight down onto a ret does not aect the tension a signiicant amount, due to the ret being so close to the string. The amplitude o the wave is determined by how hard the string is plucked, not by pushing the string onto the rets. When the string is pushed down, its eective length is shortened, which shortens the wavelength and thus increases the oscillation requency. (The wave speed on the string doesn t change due to retting the string.) 0. (a) The undamental wavelength o an open-ended organ pipe is twice the length o the pipe. one end is closed, then the undamental wavelength is our times the length o the pipe. Since the wavelength doubles when one end o the pipe is closed o, and the speed o sound remains constant, the undamental requency is cut in hal. Copyright 04 Pearson Education, nc. All rights reserved. This material is protected under all copyright laws as they currently exist.

5 Sound -5. (c) The two speakers create sound waves that interact as described by the principle o superposition. When the waves overlap, the requency remains the same; it does not double. the speakers occupied the same location so that each point in the room were equidistant rom the speakers, then the intensity would double everywhere. However, the speakers are separated by a distance o 0 m. Since the path lengths rom each speaker to dierent locations around the room are not the same, at some points in the room the path dierence will be an odd integer number o hal wavelengths, so the sounds will destructively interere. At other locations in the room the speakers will be equidistant, or the path dierence will be an integral number o whole wavelengths, and the sounds will constructively interere. This results in dead spots and loud spots in the room.. (c) A common misconception is that since the cars are moving there must be a Doppler shit. n this situation, however, there is no relative motion between the two vehicles. The two vehicles travel in the same direction at the same speed. Since the distance between them does not change, you and your sister will hear the horn sound at the same requency. 3. (e) As the string vibrates, each part o the string (other than the nodes) oscillates at the same requency, so answer (a) is true. This oscillation excites the air to vibrate at that requency, so answer (c) is true. The wave relationships in answers and (d) are true or any wave, so both are true in this case as well. However, the speed o the wave on the string is determined by the tension and mass o the string, and the speed o sound in the air is determined by the temperature, pressure, and density o the air. The two speeds are not necessarily the same. Since the sound wave and wave on the string have the same requencies, but not necessarily the same wave speeds, they do not necessarily have the same wavelengths. Thus, (e) is not true. Solutions to Problems n solving these problems, the authors did not always ollow the rules o signiicant igures rigidly. We tended to take quoted requencies as correct to the number o digits shown, especially where other values might indicate that. For example, in Problem 49, values o 350 Hz and 355 Hz are used. We took both o those values to have 3 signiicant igures.. The round-trip time or sound is.5 seconds, so the time or sound to travel the length o the lake is.5 seconds. Use the time and the speed o sound to determine the length o the lake. d t (343 m/s)(.5 s) 49 m 430 m. The round-trip time or sound is.0 seconds, so the time or sound to travel the length o the lake is.0 seconds. Use the time and the speed o sound in water to determine the depth o the lake. d t (560 m/s)(.0 s) 560 m 600 m 3. (a) λ 343 m/s 343 m/s 7 m λ.7 0 m 0 Hz.0 0 Hz 0 Hz 0 khz 4 The range is rom.7 cm to 7 m. 343 m/s 5 λ.9 0 m Hz Copyright 04 Pearson Education, nc. All rights reserved. This material is protected under all copyright laws as they currently exist.

6 -6 Chapter 4. The distance that the sound travels is the same on both days and is equal to the speed o sound times the elapsed time. Use the temperature-dependent relationship or the speed o sound in air. d t t [( (3)) m/s](4.80 s) [( ( T)) m/s](5.0 s) T 4 C 5. (a) For the ish, the speed o sound in sea water must be used. d 550 m d t t s 560 m/s For the ishermen, the speed o sound in air must be used. d 550 m d t t 4.5 s 343 m/s 6. The two sound waves travel the same distance. The sound will travel aster in the concrete and thus take a shorter time. d airtair concretetconcrete concrete( tair 0.80s) concrete tair (0.80 s) concrete air concrete d airtair air (0.80 s) concrete air The speed o sound in concrete is obtained rom Eq. 4a, Table 9, and Table 0. 9 E 0 0 N/m 949 m/s ρ.3 0 kg/m concrete m/s d airtair (343 m/s) (0.80 s) 30.5 m 30 m 949 m/s 343 m/s 7. The total time T is the time or the stone to all ( t down ) plus the time or the sound to come back to the top o the cli ( tup ): T tup + tdown. Use constant-acceleration relationships or an object dropped rom rest that alls a distance h in order to ind t down, with down as the positive direction. Use the constant speed o sound to ind t up or the sound to travel a distance h. down: h y y0 + 0tdown + at down h gt down up: h sndtup tup h snd down up snd snd snd g h gt g( T t ) g T h + T h+ T 0 This is a quadratic equation or the height. This can be solved with the quadratic ormula, but be sure to keep several signiicant digits in the calculations. 343 m/s h (343 m/s) +.7 s h+ (.7 s) (343 m/s) m/s ± 5796 h (586 m) h m 0 h 5,89 m, 33 m The larger root is impossible since it takes more than.7 s or the rock to all that distance, so h 33 m. snd Copyright 04 Pearson Education, nc. All rights reserved. This material is protected under all copyright laws as they currently exist.

7 Sound db 0 log 0 0 (.0 0 W/m ).0 W/m db 0 log 0 0 (.0 0 W/m ).0 0 W/m The pain level is 0 0 times more intense than the whisper W/m β 0 log 0 log 6.76 db 6 db.0 0 W/m 0. Compare the two power output ratings using the deinition o decibels. P50 0 W β 0 log 0 log.0 db P 75 W 00 This would barely be perceptible.. From Example 4, we see that a sound level decrease o 3 db corresponds to a halving o intensity. Thus, the sound level or one irecracker will be 85 db 3 db 8 db.. From Example 4, we see that a sound level decrease o 3 db corresponds to a halving o intensity. Thus, i two engines are shut down, the intensity will be cut in hal, and the sound level will be 37 db. Then, i one more engine is shut down, the intensity will be cut in hal again, and the sound level will drop by 3 more db, to a inal value o 34 db For the 8-dB device: 8 db 0 log ( / ) ( / ) signal noise tape signal noise tape For the 98-dB device: 98 db 0 log ( / ) ( / ) signal noise tape signal noise tape 4. (a) The energy absorbed per second is the power o the wave, which is the intensity times the area. 55 db 0 log 0 0 (.0 0 W/m ) W/m P A (3.6 0 W/m )(5.0 0 m ).58 0 W.6 0 J/s s yr 3.0 J 00 yr.0 0 yr J s 5. (a) Find the intensity rom the 30-dB value, and then ind the power output corresponding to that intensity at that distance rom the speaker. β 30 db 0 log 0 0 (.0 0 W/m ) 0 W/m.8 m m 0 0 P A 4πr 4 π(.5 m) (0 W/m ) W 790 W Copyright 04 Pearson Education, nc. All rights reserved. This material is protected under all copyright laws as they currently exist.

8 -8 Chapter Find the intensity rom the 85-dB value, and then rom the power output, ind the distance corresponding to that intensity. β 85 db 0 log 0 0 (.0 0 W/m ) W/m 0 P W P 4π r r m 440 m 4π 4 4 π (3.6 0 W/m ) The irst person is a distance o r 00 m rom the explosion, while the second person is a distance r 5(00 m) rom the explosion. The intensity detected away rom the explosion is inversely proportional to the square o the distance rom the explosion. 5(00 m) β r 00 m r 5; 0log 0log5 6.99dB 7dB 7. (a) The intensity is proportional to the square o the amplitude, so i the amplitude is 3.5 times greater, the intensity will increase by a actor o β 0 log / 0 0 log db db 8. The intensity is given by Eq. 8, ρπ A. the only dierence in two sound waves is their requencies, then the ratio o the intensities is the ratio o the square o the requencies. (. ) The intensity is given by Eq. 8, the speed o sound in air. π ρ A, using the density o air (rom Table 0 ) and 3 4 ρπ A (.9 kg/m )(343 m/s) π (440 Hz) (.3 0 m) W/m W/m β 0 log 0 log db 30 db.0 0 W/m 0 Note that according to Fig. 6, this is above the threshold o pain at that requency. 0. (a) According to Table, the intensity o normal conversation, at a distance o about 50 cm rom 6 the speaker, is about 3 0 W/m. The intensity is the power output per unit area, so the power output can be ound. The area to use is the surace area o a sphere. P P A (4 πr ) (3 0 W/m )4 π(0.50 m) W W A person W million people W Copyright 04 Pearson Education, nc. All rights reserved. This material is protected under all copyright laws as they currently exist.

9 Sound -9. The intensity o the sound is deined to be the power per unit area. We assume that the sound spreads out spherically rom the loudspeaker. (a) β β 0 W 45 W 0.49 W/m W/m 4 π(3.5m) 4 π(3.5m) W/m W/m W/m W/m 0 log 0 log.55 db db 0 log 0 log 4.66 db 5 db According to the textbook, or a sound to be perceived as twice as loud as another, the intensities need to dier by a actor o 0, or dier by 0 db. They dier by only about 7 db. The expensive amp will not sound twice as loud as the cheaper one.. From Fig. 6, a 00-Hz tone at 50 db has a loudness o about 0 phons. At 5000 Hz, 0 phons corresponds to about 0 db. Answers may vary due to estimation in the reading o the graph. 3. From Fig. 6, at 40 db the low-requency threshold o hearing is about Hz. There is no intersection o the threshold o hearing with the 40-dB level on the high-requency side o the chart, so we assume that a 40-dB signal can be heard all the way up to the highest requency that a human can hear, 0,000 Hz. Answers may vary due to estimation in the reading o the graph. 4. (a) From Fig. 6, at 00 Hz, the threshold o hearing (the lowest detectable intensity by the ear) is 9 approximately 5 0 W/m. The threshold o pain is about 5W/m. The ratio o highest to lowest intensity is thus 5W/m W/m 9 0. At 5000 Hz, the threshold o hearing is about 3 0 W/m, 0 W/m. The ratio o highest to lowest intensity is and the threshold o pain is about 0 W/m 3 0 W/m Answers may vary due to estimation in the reading o the graph Each octave is a doubling o requency. The number o octaves, n, can be ound rom the ollowing: n 0,000 Hz (0 Hz) 000 log 000 n log log000 n octaves log 6. For a closed tube, Fig. indicates that. We assume the bass clarinet is at room 4 temperature. 343 m/s. m 4 4 4(69 Hz) n Copyright 04 Pearson Education, nc. All rights reserved. This material is protected under all copyright laws as they currently exist.

10 -0 Chapter 7. For a vibrating string, the requency o the undamental mode is given by Eq. 9b combined with Eq. 3. FT FT 4 FT 4 m 4(0. 3 m)(440 Hz) (3.5 0 kg) 87 N μ m/ 8. (a) the pipe is closed at one end, only the odd harmonic requencies are present. n 343 m/s n n, n,3,5, 73.9Hz 4 4 4(.6 m) 3 Hz Hz 7 57 Hz the pipe is open at both ends, then all the harmonic requencies are present. n 343 m/s n n, n, 3, 5, 48 Hz (.6 m) 96 Hz Hz 4 59 Hz (a) The length o the tube is one-ourth o a wavelength or this (one end closed) tube, so the wavelength is our times the length o the tube. 343 m/s 360 Hz λ 4(0.4 m) the bottle is one-third ull, then the eective length o the air column is reduced to cm. 343 m/s 540 Hz λ 4(0.6 m) 30. For a pipe open at both ends, the undamental requency is given by, so the length or a given undamental requency is. 343 m/s 343 m/s 3 0 Hz 8.6 m 0 khz m (0 Hz) (0,000 Hz) 3. For a ixed string, the requency o the nth harmonic is given by n n. Thus, the undamental or this string is 3 /3 540 Hz/3 80 Hz. When the string is ingered, it has a new length o 70% o the original length. The undamental requency o the vibrating string is also given by, and is constant or the string, assuming its tension is not changed. 80 Hz 60 Hz (0. 70) ingered ingered 3. We approximate the shell as a closed tube o length 5 cm and calculate the undamental requency. 343 m/s 57 Hz 570 Hz 4 4(0.5m) Copyright 04 Pearson Education, nc. All rights reserved. This material is protected under all copyright laws as they currently exist.

11 Sound (a) We assume that the speed o waves on the guitar string does not change when the string is retted. The undamental requency is given by, so the requency is inversely proportional to the length. constant E 330 Hz E E A A A E (0.68 m) 0.5 m A 440 Hz The string should be retted a distance 0.68 m 0.5 m 0.7 m rom the tuning nut o the guitar: (at the right-hand node in Fig. 8a o the textbook). The string is ixed at both ends and is vibrating in its undamental mode. Thus, the wavelength is twice the length o the string (see Fig. 7). λ (0.5m).0m (c) The requency o the sound will be the same as that o the string, 440 Hz. The wavelength is given by the ollowing: λ 343 m/s 0.78 m 440 Hz 34. (a) At T 8 C, the speed o sound is given by ( (8)) m/s 34.8 m/s. For an open pipe, the undamental requency is given by m/s 0.65 m (6 Hz) (c) The requency o the standing wave in the tube is 6 Hz. The wavelength is twice the length o the pipe,.30 m. The wavelength and requency are the same in the air, because it is air that is resonating in the organ pipe. The requency is 6 Hz and the wavelength is.30 m. 35. The speed o sound will change as the temperature changes, and that will change the requency o the organ. Assume that the length o the pipe (and thus the resonant wavelength) does not change Δ.0 λ λ λ. 0 Δ λ ().9 0.9% (.0) λ 36. A lute is a tube that is open at both ends, so the undamental requency is given by, where is the distance rom the mouthpiece (antinode) to the irst open side hole in the lute tube (antinode). 343 m/s 0.49 m (349 Hz) Copyright 04 Pearson Education, nc. All rights reserved. This material is protected under all copyright laws as they currently exist.

12 - Chapter 37. (a) At T C, the speed o sound is ( ()) m/s 344. m/s. For an open pipe, the undamental requency is given by m/s m m (94 Hz) The speed o sound in helium is given in Table as 005 m/s. Use this and the pipe s length to ind the pipe s undamental requency. 005 m/s Hz 858 Hz ( m) 38. (a) The dierence between successive overtones or this pipe is 76 Hz. The dierence between successive overtones or an open pipe is the undamental requency, and each overtone is an integer multiple o the undamental. Since 64 Hz is not a multiple o 76 Hz, 76 Hz cannot be the undamental, so the pipe cannot be open. Thus, it must be a closed pipe. For a closed pipe, the successive overtones dier by twice the undamental requency. Thus, 76 Hz must be twice the undamental, so the undamental is 88 Hz. This is veriied since 64 Hz is three times the undamental, 440 Hz is ive times the undamental, and 66 Hz is seven times the undamental. 39. (a) The dierence between successive overtones or an open pipe is the undamental requency. 330 Hz 75 Hz 55 Hz The undamental requency is given by. Solve this or the speed o sound. (.70 m)(55 Hz) 87 m/s 90 m/s 40. The dierence in requency or two successive harmonics is 40 Hz. For an open pipe, two successive harmonics dier by the undamental, so the undamental could be 40 Hz, with 80 Hz being the 7th harmonic and 30 Hz being the 8th harmonic. For a closed pipe, two successive harmonics dier by twice the undamental, so the undamental could be 0 Hz. But the overtones o a closed pipe are odd multiples o the undamental, and both overtones are even multiples o 30 Hz. So the pipe must be an open pipe. [ (3.0)]m/s m (40 Hz) n 4. (a) The harmonics or the open pipe are n. To be audible, they must be below 0 khz. 4 n 4 (.8 m)( 0 Hz) < 0 Hz n < m/s Since there are 54 harmonics, there are 53 overtones. n The harmonics or the closed pipe are n, n odd. Again, they must be below 0 khz. 4 n 4 4(.8 m)( 0 Hz) < 0 Hz n < m/s The values o n must be odd, so n, 3, 5,, 507. There are 54 harmonics, so there are 53 overtones. 4 Copyright 04 Pearson Education, nc. All rights reserved. This material is protected under all copyright laws as they currently exist.

13 Sound A tube closed at both ends will have standing waves with displacement nodes at each end, so it has the same harmonic structure as a string that is astened at both ends. Thus, the wavelength o the undamental requency is twice the length o the hallway, λ 8m. 343 m/s Hz 9 Hz ; 38 Hz λ 8 m 43. The tension and mass density o the string do not change, so the wave speed is constant. The requency / ratio or two adjacent notes is to be. The requency is given by. st st ret ret / uningered 75.0 cm st cm / / uningered ret uningered st ret uningered uningered n/ nd / / nth ; x n/ nth uningerd nth uningered ( ) ret ret ret ret / / x (75.0 cm)( ) 4. cm ; x (75.0 cm)( ) 8. cm 3 3/ x4 4/ 5 5/ x6 6/ x x (75.0 cm)( ).9 cm ; (75.0 cm)( ) 5.5 cm (75.0 cm)( ) 8.8 cm ; (75.0 cm)( ).0 cm 44. The ear canal can be modeled as a closed pipe o length.5 cm. The resonant requencies are given by n n, n odd. The irst several requencies are calculated here. 4 n n(343 m/s) n n(3430 Hz), n odd 4 4(.5 0 m) 3430Hz 0,300Hz 7,00Hz 3 5 n the graph, the most sensitive requency is between 3000 and 4000 Hz. This corresponds to the undamental resonant requency o the ear canal. The sensitivity decrease above 4000 Hz, but is seen to latten out around 0,000 Hz again, indicating higher sensitivity near 0,000 Hz than at surrounding requencies. This 0,000-Hz relatively sensitive region corresponds to the irst overtone resonant requency o the ear canal. 45. From Eq. 8, the intensity is proportional to the square o the amplitude and the square o the A requency. From Fig. 5, the relative amplitudes are 0.4 A and A A π ρ A A A π ρ π ρ A A A (0.5) 0.0 A (0.4) 0.64 A A β 0 log 0 log 0.64 db ; 0 log 0 log db 3 β3 Answers may vary due to the reading o the igure. Copyright 04 Pearson Education, nc. All rights reserved. This material is protected under all copyright laws as they currently exist.

14 -4 Chapter 46. The beat period is.0 seconds, so the beat requency is the reciprocal o that, 0.50 Hz. Thus, the other string is o in requency by ± 0.50 Hz. The beating does not tell the tuner whether the second string is too high or too low. 47. The 5000-Hz shrill whine is the beat requency generated by the combination o the two sounds. This means that the brand X whistle is either 5000 Hz higher or 5000 Hz lower than the known-requency whistle. it were 5000 Hz lower, then it would just barely be in the audible range or humans. Since humans cannot hear it, the brand X whistle must be 5000 Hz higher than the known requency whistle. Thus, the brand X requency is 3.5 khz + 5 khz 8.5 khz. Since the original requencies are good to 0. KHz, we assume that the 5000-Hz value is 5.0 khz. 48. The beat requency is the dierence in the two requencies, or 77 Hz 6 Hz 5 Hz. both requencies are reduced by a actor o 4, then the dierence between the two requencies will also be reduced by a actor o 4, so the beat requency will be (5 Hz) 3.75 Hz 3.8 Hz Since there are 3 beats/s when sounded with the 350-Hz tuning ork, the guitar string must have a requency o either 347 Hz or 353 Hz. Since there are 8 beats/s when sounded with the 355-Hz tuning ork, the guitar string must have a requency o either 347 Hz or 363 Hz. The common value is 347 Hz. FT 50. The undamental requency o the violin string is given by 94 Hz. Change the μ tension to ind the new requency and then subtract the two requencies to ind the beat requency. (0.975) FT FT μ μ Δ ( 0.975) (94 Hz)( 0.975) 3.7 Hz 5. (a) Since the sounds are initially 80 out o phase, another 80 o phase must be added by a path x d x A length dierence. Thus, the dierence o the distances rom the speakers to the point o constructive intererence must be hal o a wavelength. See the diagram. 343 m/s ( d x) x λ d x+ λ d min λ 0.56m (305Hz) This minimum distance occurs when the observer is right at one o the speakers. the speakers are separated by more than m, the location o constructive intererence will be moved away rom the speakers, along the line between the speakers. Since the sounds are already 80 out o phase, as long as the listener is equidistant rom the speakers, there will be completely destructive intererence. So even i the speakers have a tiny separation, the point midway between them will be a point o completely destructive intererence. The minimum separation between the speakers is The beat requency is 3 beats per.5 seconds, or. Hz. We assume the strings are the same length and the same mass density. (a) The other string is either 0.0 Hz. Hz 8.8 Hz or 0.0 Hz +. Hz. Hz. B Copyright 04 Pearson Education, nc. All rights reserved. This material is protected under all copyright laws as they currently exist.

15 Sound -5 FT Since, we have μ To change 8.8 Hz to 0.0 Hz: To change. Hz to 0.0 Hz: F F F T T T FT FT 0.0 F FT.0 FT,. % increase F FT FT,.% decrease (a) For destructive intererence, the smallest path dierence must be one-hal wavelength. Thus, the wavelength in this situation must be twice the path dierence, or.00 m. 343 m/s 343 Hz λ.00 m There will also be destructive intererence i the path dierence is.5 wavelengths,.5 wavelengths, etc. Δ.5λ 0.50 m 343 m/s λ m 09 Hz 000 Hz.5 λ 0.33 m Δ.5λ 0.50 m 343 m/s λ 0.0 m 75 Hz 700 Hz.5 λ 0.0 m 54. (a) The microphone must be moved to the right until the dierence in distances rom the two sources is hal a wavelength. See the diagram. We square the expression, collect terms, isolate the remaining square root, and square again. d d λ ( d x) ( d x ) ( d x ) λ ( d x ) λ ( d x ) λ ( λ ) ( d x) ( d x) dx λ λ 4 ( d x ) + d x dx λ + λ λ 4 6 ( d x ) + 4 ( ) ( d + λ 4 6 ) 4 λ + λ λ λ + λ + λ λ 6 4 4d x dx d dx x x (4 d λ ) The values are d 3.00 m, 3.0 m, and λ / (343 m/s)/(474 Hz) m. d d l d x. (3.00 m) + (3. 0 m) (0.736 m) 4 6 x (0.736 m) 0.49 m 4(3.00 m) (0.736 m) When the speakers are exactly out o phase, the maxima and minima will be interchanged. The intensity maxima are 0.49 m to the let or right o the midpoint, and the intensity minimum is at the midpoint. Copyright 04 Pearson Education, nc. All rights reserved. This material is protected under all copyright laws as they currently exist.

16 -6 Chapter 55. (a) To ind the beat requency, calculate the requency o each sound and then subtract the two requencies. beat (343 m/s) Hz 8.9 Hz λ λ.54 m.7 m The speed o sound is 343 m/s, and the beat requency is Hz. The regions o maximum intensity are one beat wavelength apart. λ 343 m/s m 38 m Hz 56. (a) Observer moving toward stationary source: obs 30.0 m/s + + (650 Hz) 790 Hz snd 343 m/s Observer moving away rom stationary source: obs 30.0 m/s (650Hz) 50Hz snd 343 m/s 57. The moving object can be treated as a moving observer or calculating the requency it receives and relects. The bat (the source) is stationary. object object bat snd Then the object can be treated as a moving source emitting the requency object and the bat as a stationary observer. object ( ) ( ) object snd snd object bat bat bat object object snd + object + + snd snd m/s 7.0 m/s 4 ( Hz) Hz 343 m/s m/s 58. The requency received by the stationary car is higher than the requency emitted by the stationary car, by Δ 4.5 Hz. source obs source +Δ source snd snd 343 m/s source Δ (4.5 Hz) 8.5 Hz 8 Hz source 8 m/s Copyright 04 Pearson Education, nc. All rights reserved. This material is protected under all copyright laws as they currently exist.

17 Sound The wall can be treated as a stationary observer or calculating the requency it receives. The bat is lying toward the wall. wall bat bat snd Then the wall can be treated as a stationary source emitting the requency wall and the bat as a moving observer, lying toward the wall. bat bat ( snd bat ) bat wall bat bat snd bat snd ( snd bat ) snd m/s m/s 4 ( Hz) 3. 0 Hz 343 m/s 6.0 m/s 60. We assume that the comparison is to be made rom the rame o reerence o the stationary tuba. The stationary observers would observe a requency rom the moving tuba o obs source 75 Hz 78.9 Hz beat 78.9 Hz 75 Hz 3.9 Hz 3 Hz source 4.0 m/s snd 343 m/s 6. The ocean wave has λ 44 m and 8 m/s relative to the ocean loor. The requency o the ocean 8 m/s wave is then Hz. λ 44 m (a) For the boat traveling west, the boat will encounter a Doppler shited requency, or an observer moving toward a stationary source. The speed 8 m/s represents the speed o the waves in the stationary medium, so it corresponds to the speed o sound in the Doppler ormula. The time between encountering waves is the period o the Doppler shited requency. obs 4 m/s observer + + (0.409 Hz) Hz moving snd 8 m/s T.375 s.4 s Hz For the boat traveling east, the boat will encounter a Doppler shited requency, or an observer moving away rom a stationary source. obs 4 m/s observer (0.409 Hz) Hz moving snd 8 m/s T s Hz Copyright 04 Pearson Education, nc. All rights reserved. This material is protected under all copyright laws as they currently exist.

18 -8 Chapter 6. (a) The observer is stationary, and the source is moving. First the source is approaching, and then the source is receding. m/s 0.0 km/h m/s 3.6 km/h source (580 Hz) 750 Hz moving src m/s toward snd 343 m/s source (580 Hz) 440 Hz moving src m/s away + + snd 343 m/s Both the observer and the source are moving, so use Eq. 4. m/s 90.0 km/h 5 m/s 3.6 km/h ( snd + obs ) (343 m/s + 5 m/s) approaching (580 Hz) 880 Hz ( snd src ) (343 m/s m/s) ( snd obs ) (343 m/s 5 m/s) receding (580 Hz) 340 Hz ( + ) (343 m/s m/s) snd src (c) Both the observer and the source are moving, so again use Eq. 4. m/s 80.0 km/h. m/s 3.6 km/h ( ) ( 343 m/s. m/s) (580 Hz) 640 Hz ( ) ( 343 m/s m/s) snd obs police car snd src approaching ( ) (343 m/s +. m/s) ( 580 Hz) ( ) (343 m/s m/s) snd obs police car snd src receding 530 Hz 63. The maximum Doppler shit occurs when the heart has its maximum velocity. Assume that the heart is moving away rom the original source o sound. The beats arise rom the combining o the original.5-mhz requency with the relected signal which has been Doppler shited. There are two Doppler shits one or the heart receiving the original signal (observer moving away rom stationary source) and one or the detector receiving the relected signal (source moving away rom stationary observer). heart heart heart snd ( snd heart ) heart original ; detector original original snd heart heart ( snd + heart ) + + snd snd ( original detector snd blood ) blood Δ original original original ( ) + ( + ) 3 blood snd 6 original snd blood snd blood Δ 40 Hz (.54 0 m/s) 0.08 m/s Δ (.5 0 Hz) 40 Hz Copyright 04 Pearson Education, nc. All rights reserved. This material is protected under all copyright laws as they currently exist.

19 Sound -9 instead we had assumed that the heart was moving toward the original source o sound, we would Δ get blood snd. Since the beat requency is much smaller than the original requency, original +Δ the Δ term in the denominator does not signiicantly aect the answer. 64. We represent the Mach number by the symbol M. obj (a) M obj Msnd (0. 33)[(33 m/s) (4)] 3.98 m/s 0 m/s snd obj obj 3000 km/h m/s M snd km/h 70 m/s snd M km/h 65. The speed is ound rom Eq. 5. wave wave. km/h sin θ obj 0.58 km/h km/h sin θ sin obj 66. (a) The angle o the shock wave ront relative to the direction o motion is given by Eq. 5. snd snd sin θ θ sin obj. snd.. The displacement o the plane ( objt) rom the time v t obj it passes overhead to the time the shock wave reaches the observer is shown, along with the shock θ wave ront. From the displacement and height o the plane, the time is ound. h h tan θ t obj h 6500 m t 8 44 s 8 s tan θ (. )(30 m/s) tan 8.44 obj 67. (a) The Mach number is the ratio o the object s speed to the speed o sound. 4 m/s (.5 0 km/h) obs 3.6 km/h M m/s sound Use Eq. 5 to ind the angle. snd θ sin sin sin v M 99. obj snd 68. From Eq. 7, sin θ. The speed o sound in the ocean is taken rom Table. (a) obj obj snd 343 m/s θ sin sin. 900 m/s snd 560 m/s θ sin sin m/s obj Copyright 04 Pearson Education, nc. All rights reserved. This material is protected under all copyright laws as they currently exist.

20 -0 Chapter 69. (a) The displacement o the plane rom the time it passes overhead to the time the shock wave reaches the listener is shown, along with the shock wave ront. From the displacement and height o the plane, the angle o the shock wave ront relative to the direction o motion can be ound..45 km.0 km θ.45 km.45 tan θ θ tan km.0 The speed and Mach number are ound rom Eq. 5. snd 330 m/s obj 56. m/s 560 m/s sin θ sin obj M.7 sin θ sin snd 70. Find the angle o the shock wave, then the distance the plane has traveled when the shock wave reaches the observer. Use Eq. 5. snd snd θ sin sin sin 30 obj.0snd m 9500 m tan θ D 6,454 m 6 km D tan m D θ 7. The minimum time between pulses would be the time or a pulse to travel rom the boat to the maximum distance and back again. The total distance traveled by the pulse will be 70 m at the speed o sound in resh water, 440 m/s. d 70 m d t t 0. s 440 m/s 7. The single mosquito creates a sound intensity o 0 0 W/m. Thus, 00 mosquitoes will create a sound intensity o 00 times that o a single mosquito β 0 log 0 log 00 3 db The two sound level values must be converted to intensities; then the intensities are added and converted back to sound level : 8 db 0 log : 87 db 0 log total (6.7 0 ) total 0 β 0 log 0 log db Copyright 04 Pearson Education, nc. All rights reserved. This material is protected under all copyright laws as they currently exist.

21 Sound The power output is ound rom the intensity, which is the power radiated per unit area. 5 db 0 log 0 0 (.0 0 W/m ) W/m 5 khz P P P 4πr 4 π(8. 5 m) (3.6 0 W/m ) 70.45W 70 W A 4π r The answer has 3 signiicant igures. 75. Relative to the 000-Hz output, the 5-kHz output is db. P5 khz P5 khz P. 5 khz db 0 log. log 0 5 W 5 W 5 W P (5 W)(0 ) 4W 76. The 30-dB level is used to ind the intensity, and the intensity is used to ind the power. t is assumed that the jet airplane engine radiates equally in all directions. β 30 db 0 log 0 0 (.0 0 W/m ).0 0 W/m P A πr (.0 0 W/m ) π(.0 0 ) 0.03 W 77. (a) Pout 35 W The gain is given by β 0 log 0 log 5 db. P 3 in.0 0 W We solve the gain equation or the noise power level. P P 0 W β 0 log 5 0 W P signal signal 9 Pnoise β /0 93/0 noise The strings are both tuned to the same requency, and they have the same length. The mass per unit length is the density times the cross-sectional area. The requency is related to the tension by Eqs. 3 and 9b. FT FT FT ; F T 4 ρ π r μ μ ρπ r T high 4 ρ π high high high 0.74 mm T low 4 ρ π r low low low mm F r r d.07 F r d 79. The apparatus is a closed tube. The water level is the closed end, so it is a node o air displacement. As the water level lowers, the distance rom one resonance level to the next corresponds to the distance between adjacent nodes, which is one-hal wavelength. Δ λ λ Δ (0.395 m 0.5 m) m 343 m/s 635 Hz λ m Copyright 04 Pearson Education, nc. All rights reserved. This material is protected under all copyright laws as they currently exist.

22 - Chapter 80. The undamental requency o a tube closed at one end is given by. The change in air 4 temperature will change the speed o sound, resulting in two dierent requencies. 3.0 C 3.0 C C 3.0 C 3.0 C 5.0 C 5.0 C 3.0 C 5.0 C 5.0 C C (3.0) Δ 3.0 C 5.0 C 5.0 C (349 Hz) 3.63 Hz 4 Hz 5.0 C (5.0) 8. Call the requencies o our strings o the violin A, B, C, and D, with A the lowest pitch. The mass per unit length will be named μ. All strings are the same length and have the same tension. For a string with both ends ixed, the undamental requency is given by F T T A B.5 A.5 μb 0.44μ A μb μa (.5) F F FT. μ T T A C.5 B (. 5) A (.5) μc 0.0μ 4 A μc μa (.5) F 3 T 3 T A D.5 C (.5) A (.5) μd 0.088μ 6 A μd μa (.5) F F μ μ μ 8. We combine the expression or the requency in a closed tube with the Doppler shit or a source moving away rom a stationary observer, Eq. b. snd 4 snd 343 m/s 7 Hz 30 Hz source source 5 m/s (7.0 0 m) + snd snd 343 m/s 83. The eective length o the tube is + D 0.55 m + (0.030 m) 0.56 m. e 3 3 Uncorrected requencies: Corrected requencies: n (n ), n,,3, m/s 4 (n ) 56 Hz, 468 Hz, 770 Hz, 090 Hz 4(0.55 m) n (n ), n,,3, 4 e 343 m/s 4 (n ) 53 Hz, 459 Hz, 766 Hz, 070 Hz 4(0.56 m) 50 Hz, 460 Hz, 770 Hz, 00 Hz Copyright 04 Pearson Education, nc. All rights reserved. This material is protected under all copyright laws as they currently exist.

23 Sound As the train approaches, the observed requency is given by approach. As the train train snd recedes, the observed requency is given by recede. Solve each expression or, train + snd equate them, and then solve or train. train train approach recede snd snd train + ( approach recede ) (565 Hz 486 Hz) snd (343 m/s) 6 m/s ( + ) (565 Hz Hz) approach recede 85. The Doppler shit is 3.5 Hz, and the emitted requency rom both trains is 508 Hz. Thus, the requency received by the conductor on the stationary train is 5.5 Hz. Use this to ind the moving train s speed. snd 508 Hz source snd (343 m/s).35 m/s ( snd source ) 5.5 Hz 86. (a) Since both speakers are moving toward the observer at the same speed, both requencies have the same Doppler shit, and the observer hears no beats. (c) The observer will detect an increased requency rom the speaker moving toward him and a decreased requency rom the speaker moving away. The dierence in those two requencies will be the beat requency that is heard. towards away train train + snd snd snd snd towards away train train ( snd train ) ( snd train ) + + snd snd 343 m/s 343 m/s (348 Hz) 4.38 Hz 4 Hz 343 m/s.0 m/s (343 m/s +.0 m/s) Since both speakers are moving away rom the observer at the same speed, both requencies have the same Doppler shit, and the observer hears no beats. 87. For each pipe, the undamental requency is given by. Find the requency o the shortest pipe. 343 m/s 7.46 Hz (.40m) The longer pipe has a lower requency. Since the beat requency is 6.0 Hz, the requency o the longer pipe must be Hz. Use that requency to ind the length o the longer pipe. 343 m/s.6 m (65.46Hz) Copyright 04 Pearson Education, nc. All rights reserved. This material is protected under all copyright laws as they currently exist.

24 -4 Chapter 88. Use Eq. 4, which applies when both source and observer are in motion. There will be two Doppler shits in this problem irst or the emitted sound with the bat as the source and the moth as the observer and then or the relected sound with the moth as the source and the bat as the observer. ( snd + moth ) ( snd + bat ) ( snd + moth ) ( snd + bat ) moth bat bat moth bat ( snd bat ) ( snd moth ) ( snd bat ) ( snd moth ) ( ) ( ) (5.35 khz) 55.3 khz ( ) ( ) 89. t is 70.0 ms rom the start o one chirp to the start o the next. Since the chirp itsel is 3.0 ms long, it is 67.0 ms rom the end o a chirp to the start o the next. Thus, the time or the pulse to travel to the moth and back again is 67.0 ms. The distance to the moth is hal the distance that the sound can travel in 67.0 ms, since the sound must reach the moth and return during the 67.0 ms. sndt 3 d (343 m/s) ( s).5 m walk 90. The person will hear a requency toward + rom the speaker that they walk toward. The snd walk person will hear a requency away rom the speaker that they walk away rom. The snd beat requency is the dierence in those two requencies. walk walk walk.6 m/s towards away + (8Hz).6Hz snd snd snd 343 m/s 9. The ratio o sound intensity passing through the door to the original sound intensity is a 30-dB decrease β 0 log / 30 log / 3 0 Only /000 o the sound intensity passes through the door. 9. The alpenhorn can be modeled as an open tube, so the undamental requency is, and the n overtones are given by n, n,,3,. 343 m/s 50.44Hz 50Hz (3.4m) 370 n n F# n(50.44 Hz) 370 Hz n Thus, the 7th harmonic, which is the 6th overtone, is close to F #. 93. The walls o the room must be air displacement nodes, so the dimensions o the room between two parallel boundaries correspond to a hal wavelength o sound. Fundamental requencies are then given by. 343 m/s 343 m/s Length: 36 Hz Width: 48 Hz (4.7 m) (3.6m) 343 m/s Height: 6 Hz (.8m) Copyright 04 Pearson Education, nc. All rights reserved. This material is protected under all copyright laws as they currently exist.

25 Sound (a) The singing rod is maniesting standing waves. When holding the rod at its midpoint, it has a node at its midpoint and antinodes at its ends. Thus, the length o the rod is a hal wavelength. The speed o sound in aluminum is ound in Table. 500 m/s Hz 300 Hz λ.6m The wavelength o sound in the rod is twice the length o the rod,.6 m. (c) The wavelength o the sound in air is determined by the requency and the speed o sound in air. λ 343 m/s m 0. m Hz 95. Equation 8 gives the relationship between intensity and the displacement amplitude: π ρ A, where A is the displacement amplitude. Thus, A, or A. Since the intensity increased by a actor o 0, the amplitude would increase by a actor o the square root o 6 the intensity increase, or The beats arise rom the combining o the original 3.5-MHz requency with the relected signal, which has been Doppler shited. There are two Doppler shits one or the blood cells receiving the original signal (observer moving away rom stationary source) and one or the detector receiving the relected signal (source moving away rom stationary observer). blood blood blood snd ( snd blood ) blood original detector original original snd blood blood ( snd + blood ) + + snd snd ( snd blood ) blood Δ original detector original original original ( snd + blood ) ( snd + blood ) 6 (3.0 0 ) (3.5 0 Hz) Hz 40 Hz 3 (.54 0 m/s ) Solutions to Search and Learn Problems. The intensity can be ound rom the sound level in decibels. β β 0 log (0 W/m ).0 W/m 0 /0 Consider a square perpendicular to the direction o travel o the sound wave. The intensity is the energy transported by the wave across a unit area perpendicular to the direction o travel, per unit time. ΔE So, where S is the area o the square. Since the energy is moving with the wave, the S Δ t speed o the energy is, the wave speed. n a time Δ t, a volume equal to Δ V SΔ t would contain all o the energy that had been transported across the area S. Combine these relationships to ind the energy in the volume. 3 ΔE ΔV (.0 W/m )(0.00 m) 9 Δ E SΔ t.9 0 J SΔt 343 m/s Copyright 04 Pearson Education, nc. All rights reserved. This material is protected under all copyright laws as they currently exist.