nv 7. The basic equation determining the pitch of the organ pipe is either f, n odd integer, for a

Transcription

1 CHAPER : Sound Answers to Questions. Sound exhibits seeral phenomena that gie eidence that it is a wae. he phenomenon o intererence is a wae phenomenon, and sound produces intererence (such as beats). he phenomenon o diraction is a wae phenomenon, and sound can be diracted (such as sound being heard around corners). Reraction is a wae phenomenon, and sound exhibits reraction when passing obliquely rom one medium to another.. Eidence that sound is a orm o energy is ound in the act that sound can do work. A sound wae created in one location can cause the mechanical ibration o an ect at a dierent location. or example, sound can set eardrums in motion, make windows rattle, or shatter a glass.. he child speaking into a cup creates sound waes which cause the bottom o the cup to ibrate. Since the string is tightly attached to the bottom o the cup, the ibrations o the cup are transmitted to longitudinal waes in the string. hese longitudinal waes trael down the string, and cause the bottom o the receier cup to ibrate. his relatiely large ibrating surace moes the adjacent air, and generates sound waes rom the bottom o the cup, traeling up into the cup. hese waes are incident on the receier s ear, and they hear the sound rom the speaker. 4. the requency were to change, the two media could not stay in contact with each other. one medium ibrates with a certain requency, and the other medium ibrates with a dierent requency, 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 wae to be transmitted rom one medium to the other, and so the requency does not change. Since the wae speed changes in passing rom air into water, and the requency does not change, we expect the waelength to change. he wae traels about our times aster in water, so we expect the waelength in water to be about our times longer than it is in air. 5. Listening to music while seated ar away rom the o sound gies eidence that the speed o sound in air does not depend on requency. the speed were highly requency dependent, then high and low sounds created at the same time at the would arrie at your location at dierent times, and the music would sound ery disjointed. he act that the music stays together is eidence that the speed is independent o requency. 6. he sound-production anatomy o a person includes arious resonating caities, such as the throat. he relatiely ixed geometry o these caities will determine the relatiely ixed waelengths o sound that a person can produce. hose waelengths will hae associated requencies gien by. he speed o sound is determined by the gas that is illing the resonant caities. the person has inhaled helium, then the speed o sound will be much higher than normal, since the speed o sound waes in helium is about times that in air. hus the person s requencies will go up about a actor o. his is about a.5 octae shit, and so the person sounds ery high pitched. n 7. he basic equation determining the pitch o the organ pipe is either, n odd integer, or a closed 4L n closed pipe, or, n integer, or an open pipe. n each case, the requency is proportional open L to the speed o sound in air. Since the speed is a unction o temperature, and the length o any 5 Pearson Education, nc., Upper Saddle Rier, NJ. All rights resered. his material is protected under all copyright laws as they currently exist. No portion o this material may be reproduced, in any orm or by any means, without permission in writing rom the 9

2 Giancoli Physics: Principles with Applications, 6 th Edition particular pipe is ixed, the requency is also a unction o temperature. hus when the temperature changes, the resonant requencies o the organ pipes change as well. 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 gien length will resonate (permit standing waes) 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 waes are created or those requencies. requencies ar rom resonant requencies will not persist ery long at all they will die out quickly., or example, two adjacent resonances o a tube are at Hz and Hz, then sound input near one o those requencies will persist and sound relatiely loud. A sound input near 5 Hz would ade out quickly, and so hae a reduced amplitude as compared to the resonant requencies. he length o the tube can be chosen to thus ilter certain requencies, i those iltered requencies are not close to resonant requencies. 9. or a string with ixed ends, the undamental requency is gien by and so the length o L string or a gien requency is L. or a string, i the tension is not changed while retting, the speed o sound waes will be constant. hus or two requencies, the spacing between the rets corresponding to those requencies is gien as ollows. L L Now see table -. Each note there would correspond to one ret on the guitar neck. Notice that as the adjacent requencies get higher, the inter-requency spacing also increases. he change rom C to C# is 5 Hz, while the change rom G to G# is Hz. hus their reciprocals get closer together, and so rom the aboe ormula, the length spacing gets closer together. Consider a numeric example. 4 4 L L.7 L L.4 C C# G G# L L G G#.68 L L C C# he G to G# spacing is only about 68% o the C to C# spacing.. When you irst hear the truck, you cannot see it. here is no straight line path rom the truck to you. he sound waes that you are hearing are thereore arriing at your location due to diraction. Long waelengths are diracted more than short waelengths, and so you are initially only hearing sound with long waelengths, which are low-requency sounds. Ater you can see the truck, you are able to receie all requencies being emitted by the truck, not just the lower requencies. hus the sound brightens due to your hearing more high requency components.. he wae pattern created by standing waes does not trael rom one place to another. he node locations are ixed in space. Any one point in the medium has the same amplitude at all times. hus the intererence can be described as intererence in space moing the obseration point rom one location to another changes the intererence rom constructie (anti-node) to destructie (node). o experience the ull range rom node to anti-node, the position o obseration must change, but all obserations could be made at the same time by a group o obserers. he wae pattern created by beats does trael rom one place to another. Any one point in the medium will at one time hae a amplitude (node) and hal a beat period later, hae a maximum 5 Pearson Education, nc., Upper Saddle Rier, NJ. All rights resered. his material is protected under all copyright laws as they currently exist. No portion o this material may be reproduced, in any orm or by any means, without permission in writing rom the 9

3 Chapter Sound amplitude (anti-node). hus the intererence can be described as intererence in time. o experience the ull range rom constructie intererence to destructie intererence, the time o obseration must change, but all obserations could be made at the same position.. the requency o the speakers is lowered, then the waelength will be increased. Each circle in the diagram will be larger, and so the points C and D will moe arther apart.. So-called actie noise reduction deices work on the principle o intererence. the electronics are ast enough to detect the noise, inert it, and create the opposite wae (8 o 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 destructie intererence relationship. he person wearing the headphones will hear a net sound signal that is ery low in intensity. 4. rom the two waes shown, it is seen that the requency o beating is higher in igure (a) the beats occur more requently. he beat requency is the dierence between the two component requencies, and so since (a) has a higher beat requency, the component requencies are urther apart in (a). 5. here is no Doppler shit i the and obserer moe in the same direction, with the same elocity. Doppler shit is caused by relatie motion between and obserer, and i both and obserer moe in the same direction with the same elocity, there is no relatie motion. 6. the wind is blowing but the listener is at rest with respect to the, the listener will not hear a Doppler eect. We analyze the case o the wind blowing rom the towards the listener. he moing air (wind) has the same eect as i the speed o sound had been increased by an amount equal to the wind speed. he waelength o the sound waes (distance that a wae traels during one period o time) will be increased by the same percentage that the wind speed is relatie to the still-air speed o sound. Since the requency is the speed diided by the waelength, the requency does not change, and so there is no Doppler eect to hear. Alternatiely, the wind has the same eect as i the air were not moing but the and listener were moing at the same speed in the same direction. See question 5 or a discussion o that situation. 7. he highest requency o sound will be heard at position C, while the child is swinging orward. Assuming the child is moing with SHM, then the highest speed is at the equilibrium point, point C. And to hae an increased pitch, the relatie motion o the and detector must be towards each other. he child would also hear the lowest requency o sound at point C, while swinging backwards. Solutions to Problems n soling 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 alues might indicate that. or example, in problem 4, alues o 5 Hz and 55 Hz are used. We took both o those alues to hae signiicant igures. We treated the decibel alues similarly. or example, in problem, we treated the alue o db as haing three signiicant igures.. he round trip time or sound is. seconds, so the time or sound to trael the length o the lake is. seconds. Use the time and the speed o sound to determine the length o the lake. d t. s 4 m.4 m 5 Pearson Education, nc., Upper Saddle Rier, NJ. All rights resered. his material is protected under all copyright laws as they currently exist. No portion o this material may be reproduced, in any orm or by any means, without permission in writing rom the 94

4 Giancoli Physics: Principles with Applications, 6 th Edition. he round trip time or sound is.5 seconds, so the time or sound to trael the length o the lake is.5 seconds. Use the time and the speed o sound in water to determine the depth o the lake. d t 56 m s.5 s 95 m. m. (a) (b) 7 m.7 m Hz khz 4 Hz. Hz So the range is rom 7 cm to 7 m. 5.4 m 6 Hz 4. (a) or the ish, the speed o sound in seawater must be used. d. m d t t.64 s 56m s (b) or the ishermen, the speed o sound in air must be used. d t t d. m.9 s 5. he total time 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 t : up t t. Use constant acceleration relationships or an ect up down dropped rom rest that alls a distance h in order to ind t down, with down as the positie direction. Use the constant speed o sound to ind t up or the sound to trael a distance h. down: y y t at h gt up: h t t down down down up up h down up h gt g t g h h g his is a quadratic equation or the height. his can be soled with the quadratic ormula, but be sure to keep seeral signiicant digits in the calculations. h.5 s h.5 s 9.8 m s 6 h h h 64 m.44 m 656 m, 55 m he larger root is impossible since it takes more than.5 sec or the rock to all that distance, so the correct result is h 55 m. 6. he two sound waes trael the same distance. he sound will trael aster in the concrete, and thus take a shorter time. concrete d t t t. s t. s air air concrete concrete concrete air air d t air air air concrete concrete air. s concrete air h 5 Pearson Education, nc., Upper Saddle Rier, NJ. All rights resered. his material is protected under all copyright laws as they currently exist. No portion o this material may be reproduced, in any orm or by any means, without permission in writing rom the 95

5 Chapter Sound he speed o sound in concrete is obtained rom Equation (-4a), able (9-), and able (-). d E N m 9 concrete t air air. kg m 949 m s 949m s. s 47m 4. m 949 m s 7. he 5 second rule says that or eery 5 seconds between seeing a lightning strike and hearing the associated sound, the lightning is mile distant. We assume that there are 5 seconds between seeing the lightning and hearing the sound. (a) At o C, the speed o sound is.6 m s 49m s. he actual distance to the lightning is thereore d t 49 m s 5s 745 m. A mile is 6 m % error 8% 745 (b) At o C, the speed o sound is.6 m s 7 m s. he actual distance to the lightning is thereore d t 7 m s 5s 685 m. A mile is 6 m % error 4% db log. W m. W m - db log. W m. W m he pain leel is times more intense than the whisper. 6. W m log log 6 db. W m. rom Example -4, we see that a sound leel decrease o db corresponds to a haling o intensity. hus the sound leel or one irecracker will be 95 db db 9 db.. rom Example -4, we see that a sound leel decrease o db corresponds to a haling o intensity. hus, i two engines are shut down, the intensity will be cut in hal, and the sound leel will be 7 db. hen, i one more engine is shut down, the intensity will be cut in hal again, and the sound leel will drop by more db, to a inal alue o 4 db.. 58 db log 6. Signal Noise tape Signal Noise tape db log. Signal Noise tape Signal Noise tape Pearson Education, nc., Upper Saddle Rier, NJ. All rights resered. his material is protected under all copyright laws as they currently exist. No portion o this material may be reproduced, in any orm or by any means, without permission in writing rom the 96

6 Giancoli Physics: Principles with Applications, 6 th Edition. (a) According to able -, the intensity in normal conersation, when about 5 cm rom the 6 speaker, is about W m. he intensity is the power output per unit area, and so the power output can be ound. he area is that o a sphere. P P A 4 r W m 4.5 m 9.45 W 9 W A (b) W person people W 4. (a) he energy absorbed per second is the power o the wae, which is the intensity times the area db log. W m. W m (b) P A 7 5. W m 5. m 5. W s yr J 6. yr 7 5. J.6 s 5. he intensity o the sound is deined to be the power per unit area. We assume that the sound spreads out spherically rom the loudspeaker. 5 W.6 W m 5 (a).6 W m log log db m. W m 4 W.6 W m.6 W m log log 4dB m. W m (b) According to the textbook, or a sound to be perceied as twice as loud as another means that the intensities need to dier by a actor o. hat is not the case here they dier only by a actor o.6 6. he expensie amp will not sound twice as loud as the cheaper one (a) ind the intensity rom the db alue, and then ind the power output corresponding to that intensity at that distance rom the speaker..8m db log. W m W m.8m P A 4 r 4.8 m W m 985W 9.9 W (b) ind the intensity rom the 9 db alue, and then rom the power output, ind the distance corresponding to that intensity db log. W m. W m P r r P 4.8 m 985 W 4 4. W m 5 Pearson Education, nc., Upper Saddle Rier, NJ. All rights resered. his material is protected under all copyright laws as they currently exist. No portion o this material may be reproduced, in any orm or by any means, without permission in writing rom the 97

7 Chapter Sound 7. he intensity is proportional to the square o the amplitude. A A A db log log log.59. A A A 8. (a) he intensity is proportional to the square o the amplitude, so i the amplitude is tripled, the intensity will increase by a actor o 9. (b) log log 9 9.5dB 9. he intensity is gien by A. the only dierence in two sound waes is their requencies, then the ratio o the intensities is the ratio o the square o the requencies.. he intensity is gien by 4 A, using the density o air and the speed o sound in air. 4 A.9 kg m 4 m s Hz. m.8 W m.8 W m log log. W m. db db Note that this is aboe the threshold o pain.. rom igure -6, a -Hz tone at 5 db has a loudness o about phons. At 6 Hz, phons corresponds to about 5 db. Answers may ary due to estimation in the reading o the graph.. rom igure -6, at db the low requency threshold o hearing is about 5 Hz. here is no intersection o the threshold o hearing with the db leel on the high requency side o the chart, and so a db signal can be heard all the way up to the highest requency that a human can hear,, Hz.. (a) rom igure -6, at Hz, the threshold o hearing (the lowest detectable intensity by the 9 ear) is approximately 5 W m. he threshold o pain is about 5 W m. he ratio o highest to lowest intensity is thus 5 W m 5 W m 9 (b) At 5 Hz, the threshold o hearing is about W m W m. he ratio o highest to lowest intensity is W m Answers may ary due to estimation in the reading o the graph. 9. W m, and the threshold o pain is about. 5 Pearson Education, nc., Upper Saddle Rier, NJ. All rights resered. his material is protected under all copyright laws as they currently exist. No portion o this material may be reproduced, in any orm or by any means, without permission in writing rom the 98

8 Giancoli Physics: Principles with Applications, 6 th Edition 4. or a ibrating string, the requency o the undamental mode is gien by 4 L m L m L L L m L. =4 4. m 44 Hz.5 kg 87 N 5. (a) the pipe is closed at one end, only the odd harmonic requencies are present, and are gien by n n, n,,5. n 4L 76.6 Hz 4L 4. m Hz 5 8 Hz 7 56 Hz 5 7 (b) the pipe is open at both ends, all the harmonic requencies are present, and are gien by n n. n L 5 Hz L. m 6 Hz 459 Hz 4 6 Hz 4 6. (a) he length o the tube is one-ourth o a waelength or this (one end closed) tube, and so the waelength is our times the length o the tube. 48 Hz 4.8 m (b) the bottle is one-third ull, then the eectie length o the air column is reduced to cm. 7 Hz 4. m 7. or a pipe open at both ends, the undamental requency is gien by gien undamental requency is L Hz khz L. 8.6 m L 8.6 m Hz, Hz, and so the length or a L 8. or a ixed string, the requency o the n th harmonic is gien by n. hus the undamental or n this string is 54 Hz 8 Hz. When the string is ingered, it has a new length o 6% o the original length. he undamental requency o the ibrating string is also gien by, and is a constant or the string, assuming its tension is not changed. L 8 Hz Hz ingered L.6 L.6.6 ingered 5 Pearson Education, nc., Upper Saddle Rier, NJ. All rights resered. his material is protected under all copyright laws as they currently exist. No portion o this material may be reproduced, in any orm or by any means, without permission in writing rom the 99

9 Chapter Sound 9. (a) We assume that the speed o waes on the guitar string does not change when the string is retted. he undamental requency is gien by, and so the requency is inersely L proportional to the length. L constant L Hz E L L L L.7 m.5475 m E E A A A E 44 Hz A he string should be retted a distance.7 m.5475 m.85 m.8 m rom the nut o the guitar. (b) he string is ixed at both ends and is ibrating in its undamental mode. hus the waelength is twice the length o the string (see ig. -7). L.5475 m.95 m. m (c) he requency o the sound will be the same as that o the string, 44 Hz. he waelength is gien by the ollowing..78 m 44 Hz o. (a) At C, the speed o sound is gien by.6 m s 4.6 m s. or an open pipe, the undamental requency is gien by 4.6 m s L L 6 Hz.656 m L. (b) he requency o the standing wae in the tube is 6 Hz. he waelength is twice the length o the pipe,. m (c) he waelength and requency are the same in the air, because it is air that is resonating in the organ pipe. he requency is 6 Hz and the waelength is. m. he 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 waelength) does not change %.6. A lute is a tube that is open at both ends, and so the undamental requency is gien by L, where L is the distance rom the mouthpiece (antinode) to the irst open side hole in the lute tube (antinode). 5 Pearson Education, nc., Upper Saddle Rier, NJ. All rights resered. his material is protected under all copyright laws as they currently exist. No portion o this material may be reproduced, in any orm or by any means, without permission in writing rom the

10 Giancoli Physics: Principles with Applications, 6 th Edition L L 94 Hz.58 m. (a) At gien by o C, the speed o sound is. or an open pipe, the undamental requency is L. L L 94 Hz.58 m (b) he speed o sound in helium is 5 m s, rom able -. Use this and the pipe s length to to ind the pipe s undamental requency. 5m s 86 Hz L.58 m 4. (a) he dierence between successie oertones or this pipe is 76 Hz. he dierence between successie oertones or an open pipe is the undamental requency, and each oertone is an integer multiple o the undamental. Since 64 Hz is not a multiple o 76 Hz, 76 Hz cannot be the undamental, and so the pipe cannot be open. hus it must be a closed pipe. (b) or a closed pipe, the successie oertones dier by twice the undamental requency. hus 76 Hz must be twice the undamental, so the undamental is 88. Hz. his is eriied since 64 Hz is times the undamental, 44 Hz is 5 times the undamental, and 66 Hz is 7 times the undamental. 5. (a) he dierence between successie oertones or an open pipe is the undamental requency. Hz 75 Hz 55 Hz (b) he undamental requency is gien by L.8 m 55 Hz 98m s. m s. Sole this or the speed o sound. L 6. he dierence in requency or two successie harmonics is 4 Hz. or an open pipe, two successie harmonics dier by the undamental, so the undamental could be 4 Hz, with 4 Hz being the 6 th harmonic and 8 Hz being the 7 th harmonic. or a closed pipe, two successie harmonics dier by twice the undamental, so the undamental could be Hz. But the oertones o a closed pipe are odd multiples o the undamental, and both oertones are een multiples o Hz. So the pipe must be an open pipe. L 4. m L 4 Hz 7. (a) he harmonics or the open pipe are 4.4 m Hz 4 n Hz n 49.6 L Since there are 49 harmonics, there are 48 oertones n n. o be audible, they must be below khz. L 5 Pearson Education, nc., Upper Saddle Rier, NJ. All rights resered. his material is protected under all copyright laws as they currently exist. No portion o this material may be reproduced, in any orm or by any means, without permission in writing rom the

11 Chapter Sound (b) he harmonics or the closed pipe are m Hz 4 n n, n odd. Again, they must be below khz. 4L n Hz n L he alues o n must be odd, so n =,, 5,, 499. here are 5 harmonics, and so there are 49 oertones 8. he ear canal can be modeled as a closed pipe o length.5 cm. he resonant requencies are gien n by, n odd. he irst seeral requencies are calculated here. n 4L n n n 4 Hz, n odd n 4L 4.5 m 4 Hz Hz 7 Hz 5 n the graph, the most sensitie requency is between and 4 Hz. his corresponds to the undamental resonant requency o the ear canal. he sensitiity decrease aboe 4 Hz, but is seen to latten out around, Hz again, indicating higher sensitiity near, Hz than at surrounding requencies. his, Hz relatiely sensitie region corresponds to the irst oertone resonant requency o the ear canal. 9. he beat period is. seconds, so the beat requency is the reciprocal o that,.5 Hz. hus the other string is o in requency by.5 Hz. he beating does not tell the tuner whether the second string is too high or too low. 4. he beat requency is the dierence in the two requencies, or 77 Hz 6 Hz 5 Hz. the requencies are both reduced by a actor o 4, then the dierence between the two requencies will also be reduced by a actor o 4, and so the beat requency will be 4 5 Hz.75Hz.8 Hz. 4. he 5 Hz shrill whine is the beat requency generated by the combination o the two sounds. his means that the brand X whistle is either 5 Hz higher or 5 Hz lower than the knownrequency whistle. it were 5 Hz lower, then it would be in the audible range or humans. Since it cannot be heard by humans, the brand X whistle must be 5 Hz higher than the known requency whistle. hus the brand X requency is.5 khz 5 khz 8.5 khz 4. Since there are 4 beats/s when sounded with the 5 Hz tuning ork, the guitar string must hae a requency o either 46 Hz or 54 Hz. Since there are 9 beats/s when sounded with the 55 Hz tuning ork, the guitar string must hae a requency o either 46 Hz or 64 Hz. he common alue is 46 Hz. 4. he undamental requency o the iolin string is gien by L L m L 94 Hz. Change the tension to ind the new requency, and then subtract the two requencies to ind the beat requency. 5 Pearson Education, nc., Upper Saddle Rier, NJ. All rights resered. his material is protected under all copyright laws as they currently exist. No portion o this material may be reproduced, in any orm or by any means, without permission in writing rom the

12 Giancoli Physics: Principles with Applications, 6 th Edition L m L L m L Hz.98. Hz 44. Beats will be heard because the dierence in the speed o sound or the two lutes will result in two dierent requencies. We assume that the lute at 5. o C will accurately play the middle C m s L.66 m L 6 Hz.6 5. m s L.66 m 5 Hz 6 Hz 5 Hz 9 beats sec 45. uning ork A must hae a requency o Hz either higher or lower than the 44 Hz ork B. uning ork C must hae a requency o 4 Hz either higher or lower than the 44 Hz ork B. A 48 Hz or 444 Hz 47 Hz or 445 Hz C he possible beat requencies are ound by subtracting all possible requencies o A and C. Hz or 7 Hz A C 46. (a) or destructie intererence, the smallest path dierence must be one-hal waelength. hus the waelength in this situation must be twice the path dierence, or. m.. m 4 Hz (b) here will also be destructie intererence i the path dierence is.5 waelengths,.5 waelengths, etc. L.5.5 m. m.5. m 9 Hz Hz L.5.5 m. m.5. m 75 Hz 7 Hz 47. he beat requency is beats per seconds, or.5 Hz. (a) he other string must be either Hz.5 Hz.5 Hz or Hz.5 Hz.5 Hz. (b) Since m L L L, we hae o change.5 Hz to Hz: o change.5 Hz to Hz:.5.5.,.% increase.978,.% decrease. 5 Pearson Education, nc., Upper Saddle Rier, NJ. All rights resered. his material is protected under all copyright laws as they currently exist. No portion o this material may be reproduced, in any orm or by any means, without permission in writing rom the

13 Chapter Sound 48. o ind the beat requency, calculate the requency o each sound, and then subtract the two requencies Hz beat.64 m.76 m 49. (a) Obserer moing towards stationary.. m s (b) Obserer moing away rom stationary.. m s obs 55 Hz 4 Hz obs 55 Hz 69 Hz 5. (a) Source moing towards stationary obserer. 55 Hz 7 Hz m s src (b) Source moing away rom stationary obserer. 55 Hz 4 Hz m s src 5. (a) or the 5 m/s relatie elocity: Hz 9 Hz moing 5 m s src obserer moing src Hz 87 Hz 5 m s he requency shits are slightly dierent, with obserer. he two requencies are moing moing close, but they are not identical. o signiicant igures they are the same. (b) or the 5 m/s relatie elocity: Hz.55 Hz moing 5 m s src obserer moing src Hz.87 Hz 5 m s he dierence in the requency shits is much larger this time, still with obserer. moing moing 5 Pearson Education, nc., Upper Saddle Rier, NJ. All rights resered. his material is protected under all copyright laws as they currently exist. No portion o this material may be reproduced, in any orm or by any means, without permission in writing rom the 4

14 Giancoli Physics: Principles with Applications, 6 th Edition (c) or the m/s relatie elocity: Hz 6. Hz moing m s src obserer moing src Hz.75 Hz m s he dierence in the requency shits is quite large, still with obserer.. he Doppler ormulas are asymmetric, with a larger shit or the moing than or the moing obserer, when the two are getting closer to each other. As the moes toward the obserer with speeds approaching the speed o sound, the obsered requency tends towards ininity. As the obserer moes toward the with speeds approaching the speed o sound, the obsered requency tends towards twice the emitted requency. 5. he requency receied by the stationary car is higher than the requency emitted by the stationary car, by 5.5 Hz. obs 5.5 Hz Hz 5 m s 5. he moing ect can be treated as a moing obserer or calculating the requency it receies and relects. he bat (the ) is stationary. ect ect bat moing hen the ect can be treated as a moing emitting the requency ect stationary obserer. ect ect ect bat bat bat ect ect ect 5. m s Hz 4. Hz 5. m s moing, and the bat as a 54. he wall can be treated as a stationary obserer or calculating the requency it receies. he bat is lying toward the wall. wall bat bat 5 Pearson Education, nc., Upper Saddle Rier, NJ. All rights resered. his material is protected under all copyright laws as they currently exist. No portion o this material may be reproduced, in any orm or by any means, without permission in writing rom the 5

15 Chapter Sound hen the wall can be treated as a stationary emitting the requency, and the bat as a wall moing obserer, lying toward the wall. bat bat bat bat wall bat bat bat bat 4 4. Hz.9 Hz 5. m s 5. m s 55. We assume that the comparison is to be made rom the rame o reerence o the stationary tuba. he stationary obserers would obsere a requency rom the moing tuba o 75 Hz 77 Hz 77 Hz 75 Hz Hz. obs beat. m s 56. he beats arise rom the combining o the original.5 MHz requency with the relected signal which has been Doppler shited. here are two Doppler shits one or the blood cells receiing the original signal (obserer moing away rom stationary ) and one or the detector receiing the relected signal ( moing away rom stationary obserer). blood blood blood blood original detector original original blood blood blood original detector original original. blood blood original blood blood 6.5 Hz 9Hz.54 m s. 57. he maximum Doppler shit occurs when the heart has its maximum elocity. Assume that the heart is moing away rom the original o sound. he beats arise rom the combining o the original.5 MHz requency with the relected signal which has been Doppler shited. here are two Doppler shits one or the heart receiing the original signal (obserer moing away rom stationary ) and one or the detector receiing the relected signal ( moing away rom stationary obserer). heart heart heart heart original detector original original heart heart heart original detector original original blood blood original blood blood heart blood 5 Pearson Education, nc., Upper Saddle Rier, NJ. All rights resered. his material is protected under all copyright laws as they currently exist. No portion o this material may be reproduced, in any orm or by any means, without permission in writing rom the 6

16 Giancoli Physics: Principles with Applications, 6 th Edition.54 m s.7m s blood 6 original 5 Hz.5 Hz 5 Hz instead we had assumed that the heart was moing towards the original o sound, we would get. Since the beat requency is much smaller than the original requency, blood original the term in the denominator does not signiicantly aect the answer. 58. he Doppler eect occurs only when there is relatie motion o the and the obserer along the line connecting them. n the irst our parts o this problem, the whistle and the obserer are not moing relatie to each other and so there is no Doppler shit. he wind speed increases (or decreases) the elocity o the waes in the direction o the wind, and the waelength o the waes by the same actor, while the requency is unchanged. (a), (b), (c), (d) 57 Hz (e) he wind makes an eectie speed o sound in air o 4 +. = 55 m/s, and the obserer is moing towards a stationary with a speed o 5. m/s. 5. m s obs 57 Hz 594 Hz 55 m s sns () Since the wind is not changing the speed o the sound waes moing towards the cyclist, the speed o sound is 4 m/s. he obserer is moing towards a stationary with a speed o 5. m/s. 5. m s obs 57 Hz 595 Hz sns 59. (a) We represent the Mach number by the symbol M. M M. m s (b) M M km h m s 97.5km h..6 km h 6 m s 6. (a) he angle o the shock wae ront relatie to the direction o motion is gien by Eq. -7. o o sin sin (b) he displacement o the plane t rom the time it passes oerhead to the time the shock wae reaches the obserer is shown, along with the shock wae ront. rom the displacement and height o the plane, the time is ound. h h tan t t tan 7 m. m s tan 5.77 o.6s s h t 5 Pearson Education, nc., Upper Saddle Rier, NJ. All rights resered. his material is protected under all copyright laws as they currently exist. No portion o this material may be reproduced, in any orm or by any means, without permission in writing rom the 7

17 Chapter Sound 6. (a) he Mach number is the ratio o the ect s speed to the speed o sound. 4.5 km hr m s.6 km hr obs M m s sound (b) Use Eq..5 to ind the angle. o sin sin sin.48 M rom Eq. -7, sin (a) (b) sin sin. 85 m s o 56 m s sin sin 85 m s o. 6. Consider one particular wae as shown in the diagram, created at the location o the black dot. Ater a time t has elapsed rom the creation o that wae, the supersonic has moed a distance t, and the wae ront has moed a distance t. he line rom the position o the at time t is tangent to all o the wae ronts, showing the location o the shock wae. A tangent to a circle at a point is perpendicular to the radius connecting that point to the center, and so a right angle is ormed. rom the right triangle, the angle deined. sin t t t t can be 64. (a) he displacement o the plane rom the time it passes oerhead to the time the shock wae reaches the listener is shown, along with the shock wae ront. rom the displacement and height o the plane, the angle o the shock wae ront relatie to the direction o motion can be ound, using Eq km.5 o tan tan 7. km. (b) M.7 o sin sin 7.5 km. km 65. he minimum time between pulses would be the time or a pulse to trael rom the boat to the maximum distance and back again. he total distance traeled by the pulse will be 4 m, at the speed o sound in resh water, 44 m/s. d 4 m d t t.8s 44 m s 5 Pearson Education, nc., Upper Saddle Rier, NJ. All rights resered. his material is protected under all copyright laws as they currently exist. No portion o this material may be reproduced, in any orm or by any means, without permission in writing rom the 8

18 Giancoli Physics: Principles with Applications, 6 th Edition 66. Each octae is a doubling o requency. he number o octaes, n, can be ound rom the ollowing. n n, Hz Hz log nlog n log log 9.97 octaes 67. Assume that only the undamental requency is heard. he undamental requency o an open pipe is gien by L. (a) 57 Hz 69 Hz..5 L. m L.5 m Hz 4. Hz Hz L. m L.5 m L. m 7.5 Hz 7 Hz (b) On a noisy day, there are a large number o component requencies to the sounds that are being made more people walking, more people talking, etc. hus it is more likely that the requencies listed aboe will be a component o the oerall sound, and then the resonance will be more prominent to the hearer. the day is quiet, there might be ery little sound at the desired requencies, and then the tubes will not hae any standing waes in them to detect. 68. he single mosquito creates a sound intensity o W m create a sound intensity o times that o a single mosquito. log log db.. hus mosquitoes will 69. he two sound leel alues must be conerted to intensities, then the intensities added, and then conerted back to sound leel : 8 db log : 87 db log total 8 87 total log log db 7. he power output is ound rom the intensity, which is the power radiated per unit area db log. W m.6 W m P P P 4 r 4. m.6 W m 57. W A 4 r 5 Pearson Education, nc., Upper Saddle Rier, NJ. All rights resered. his material is protected under all copyright laws as they currently exist. No portion o this material may be reproduced, in any orm or by any means, without permission in writing rom the 9

19 Chapter Sound 7. Relatie to the Hz output, the 5 khz output is db. P P P 5 khz 5 khz 5 khz db log log. P 5 W 5 khz 5 W 5 W 5 W 7. he 4 db leel 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 db log. W m. W m P A r. W m.. W 7. he gain is gien by P W out log log 5dB P W in. 74. Call the requencies o our strings o the iolin,,, with the lowest pitch. he mass A B C D A per unit length will be named. All strings are the same length and hae the same tension. or a string with both ends ixed, the undamental requency is gien by A B A B A L L B A.5 L L A C B A C 4 A L L C A A D C A D 8 A L L D A (a) he wae speed on the string can be ound orm the length and the undamental requency. L. m 44 Hz m s L he tension is ound rom the wae speed and the mass per unit length kg m 8.6 m s 48 N (b) he length o the pipe can be ound rom the undamental requency and the speed o sound. L.95 m 4L Hz (c) he irst oertone or the string is twice the undamental. 88 Hz he irst oertone or the open pipe is times the undamental. Hz 76. he apparatus is a closed tube. he water leel is the closed end, and so is a node o air displacement. As the water leel lowers, the distance rom one resonance leel to the next corresponds to the distance between adjacent nodes, which is one-hal waelength. 5 Pearson Education, nc., Upper Saddle Rier, NJ. All rights resered. his material is protected under all copyright laws as they currently exist. No portion o this material may be reproduced, in any orm or by any means, without permission in writing rom the

20 Giancoli Physics: Principles with Applications, 6 th Edition L L.95 m.5 m.54 m 65 Hz.54 m 77. he requency o the guitar string is to be the same as the third harmonic (n = ) o the closed tube. n he resonance requencies o a closed tube are gien by, n,,5, and the requency o n 4L a stretched string is gien by. Equate the two requencies and sole or the tension. L m L m 9 string string tube string 9 4 m s. kg L m L 4L 4L 4.75 m 7.4 N 78. By anchoring the oerpass to the ground in the middle, the center o the oerpass is now a node point. his orces the lowest requency or the bridge to be twice the undamental requency, and so now the resonant requency is 8. Hz. Since the earthquakes don t do signiicant shaking aboe 6 Hz, this modiication should be eectie. 79. Since the sound is loudest at points equidistant rom the two s, the two s must be in phase. he dierence in distance rom the two s must be an odd number o hal-waelengths or destructie intererence..4 m.68 m.68 m 54 Hz.4 m.7 m 4 m s.7 m 5 Hz out o range 8. he Doppler shit is. Hz, and the emitted requency rom both trains is 44 Hz. hus the requency receied by the conductor on the stationary train is 47 Hz. Use this to ind the moing train s speed. 44 Hz.4m s 47 Hz 8. As the train approaches, the obsered requency is gien by approach recedes, the obsered requency is gien by equate them, and then sole or train. approach train train recede approach approach recede recede train recede train 58 Hz 486 Hz 58 Hz 486 Hz train. As the train. Sole each expression or, 7 m s 5 Pearson Education, nc., Upper Saddle Rier, NJ. All rights resered. his material is protected under all copyright laws as they currently exist. No portion o this material may be reproduced, in any orm or by any means, without permission in writing rom the

21 Chapter Sound 8. he sound is Doppler shited up as the car approaches, and Doppler shited down as it recedes. he obserer is stationary in both cases. he octae shit down means that. approach recede car car approach engine recede engine approach recede car car 4 m s engine engine car 8. or each pipe, the undamental requency is gien by. ind the requency o the shortest L pipe Hz L.4 m he longer pipe has a lower requency. Since the beat requency is Hz, the requency o the longer pipe must be 6.46 Hz. Use that requency to ind the length o the longer pipe. L.84 m L 6.46 Hz 84. (a) Since both speakers are moing towards the obserer at the same speed, both requencies hae the same Doppler shit, and the obserer hears no beats. (b) he obserer will detect an increased requency rom the speaker moing towards him and a decreased requency rom the speaker moing away. he dierence in those two requencies will be the beat requency that is heard. towards away train train towards away Hz train train. m s train train. m s Hz (c) Since both speakers are moing away rom the obserer at the same speed, both requencies hae the same Doppler shit, and the obserer hears no beats. 85. he beats arise rom the combining o the original 5.5 MHz requency with the relected signal which has been Doppler shited. here are two Doppler shits one or the blood cells receiing the original requency (obserer moing away rom stationary ) and one or the detector receiing the relected requency ( moing away rom stationary obserer). blood blood blood blood original detector original original blood blood blood blood 5 Pearson Education, nc., Upper Saddle Rier, NJ. All rights resered. his material is protected under all copyright laws as they currently exist. No portion o this material may be reproduced, in any orm or by any means, without permission in writing rom the

22 Giancoli Physics: Principles with Applications, 6 th Edition blood blood original detector original original original. m s blood blood Hz.9 Hz.54 m s. m s 86. Use Eq. -4, which applies when both and obserer are in motion. here will be two Doppler shits in this problem irst or the emitted sound with the bat as the and the moth as the obserer, and then the relected sound with the moth as the and the bat as the obserer. moth bat moth bat moth bat bat moth bat bat moth bat moth khz 54.9 khz t is 7. ms rom the start o one chirp to the start o the next. Since the chirp itsel is. ms long, it is 67. ms rom the end o a chirp to the start o the next. hus the time or the pulse to trael to the moth and back again is 67. ms. he distance to the moth is hal the distance that the sound can trael in 67. ms, since the sound must reach the moth and return during the 67. ms. d t 67. s.5 m 88. he Alpenhorn can be modeled as an open tube, and so the undamental requency is n the oertones are gien by, n,,,. n L 5.44 Hz 5 Hz L.4 m 7 n n 5.44 Hz 7 Hz n 7.4 n # 5.44 hus the 7 th harmonic, which is the 6 th oertone, is close to sharp. L, and 89. he walls o the room must be air displacement nodes, and so the dimensions o the room between two parallel boundaries corresponds to a hal-waelength o sound. undamental requencies are then gien by L. Length: 4 Hz Width: 4 Hz L 5. m L 4. m Height: L.8 m 6 Hz 9. (a) he singing rod is maniesting standing waes. By holding the rod at its midpoint, it has a node at its midpoint, and antinodes at its ends. hus the length o the rod is a hal waelength. 5 m s 8 Hz.8 Hz L.8 m 5 Pearson Education, nc., Upper Saddle Rier, NJ. All rights resered. his material is protected under all copyright laws as they currently exist. No portion o this material may be reproduced, in any orm or by any means, without permission in writing rom the

23 Chapter Sound (b) he waelength o sound in the rod is twice the length o the rod,.8 m. (c) he waelength o the sound in air is determined by the requency and the speed o sound in air.. m 8 Hz 9. Eq. -8 gies the relationship between intensity and the displacement amplitude: where A is the displacement amplitude. hus a actor o or 6. A, A, or A. Since the intensity increased by, the amplitude would increase by a actor o the square root o the intensity increase, 9. he angle between the direction o the airplane and the shock wae ront is ound rom Eq. -5. o sin sin sin. he distance that the plane has traeled horizontally rom the obserer is ound rom the time and the speed: x t. he altitude is ound rom the angle and the horizontal distance. o o 4 tan y x y x tan t tan 9 s tan.6 m 9. he apex angle is 5 o, so the shock wae angle is 7.5 o. he angle o the shock wae is also gien by sin. wae ect o sin sin. km h sin km h wae ect ect wae y o 9 x 5 Pearson Education, nc., Upper Saddle Rier, NJ. All rights resered. his material is protected under all copyright laws as they currently exist. No portion o this material may be reproduced, in any orm or by any means, without permission in writing rom the 4