Chapter 14 PROBLEM SOLUTIONS Since vlight v sound, the time required for the flash of light to reach the observer is negligible in

Transcription

1 Chapter 4 PRBLEM LUTN 4. ince light sound, the time required or the lash o light to reach the obserer is negligible in comparison to the time required or the sound to arrie. Thus, we can ignore the time required or the lightning lash to arrie, and knowledge o the actual speed o light is not needed. Then, d 343 m s 6. s m 5.56 km 4. The speed o longitudinal waes in a luid is B. Considering the Earth s crust to consist o a ery iscous luid, our estimate o the aerage bulk modulus o the material in Earth s crust is B kg m 7 0 m s 0 Pa 4.3 The Celsius temperature was T 5 5 C T F C, and the 9 9 absolute temperature was TK 89. C K and the speed o sound in the air would hae been 84 K 33 m s 7 m s 73 K 4.4 The speed o sound in seawater at 5 C is 530 m s. Thereore, the time or the sound to reach the sea loor and return is t d (50 m) 530 m/s 0.96 s Page 4.

2 ound 4.5 ince the sound had to trael the distance between the hikers and the mountain twice, the time required or a one-way trip was.50 s. The speed o sound in air at T =.0 C = 95 K is T 95 K 33 m s 33 m s 3.44 m s 73 K 73 K and the distance the sound traeled to the mountain was d 344 m s.50 s 56 m 4.6 At T = 7 C = 300 K, the speed o sound in air is T 300 K 33 m s 33 m s 347 m s 73 K 73 K The waelength o the 0 Hz sound is 347 m s 0 Hz 7 m and that o the Hz is 347 m s Hz.7 0 m =.7 cm. Thus, range o waelengths o audible sounds at 7 C is.7 cm to 7 m. 4.7 From Table 4., the speed o sound in the saltwater is w = 530 m/s. At T = 0 C = 93 K, the speed o the sound in air is a T 93 K 33 m s 33 m s 343 m s 73 K 73 K d is the width o the inlet, the transit time or the sound in the water is tw d w, and that or the sound in the air is ta tw 4.50 s d a. Thus, d a d w 4.50 s, or d (4.50 s) w w a a Page 4.

3 ound 3 d 530 m s 343 m s 4.50 s m.99 km m s 4.8 At a temperature o T = 0.0 C = 83 K, the speed o sound in air is T 83 K 33 m s 33 m s 337 m s 73 K 73 K The elapsed time between when the stone was released and when the sound is heard is the sum o the time t required or the stone to all distance h and the time t required or sound to trael distance h in air on the return up the well. That is, t + t =.00 s. The distance the stone alls, starting rom rest, in time t is h gt Also, the time or the sound to trael back up the well is t h.00 s t Combining these two equations yields g t.00 s t With 337 m s and g 9.80 m s, this becomes.45 0 s- t t.00 s 0. Applying the quadratic ormula yields one positie solution o t =.95 s, so the depth o the well is h gt 9.80 m s.95 s 8.6 m 4.9 (a) The speed o sound in air at an absolute temperature o T = T C + 73 = = 338 K is Page 4.3

4 ound 4 T 338 K 33 m s 33 m s 368 m s 73 K 73 K (b) When T C = 65 C and the speed o sound is 368 m s, the waelength o sound haing a requency = 845 Hz is 368 m s 845 Hz m 43.6 cm 4.0 (a) The decibel leel,, o a sound is gien 0 log 0, where is the intensity o the sound, and W m is the reerence intensity. Thereore, i = 50 db, the intensity is W m W m (b) The threshold o pain is W m and the answer to part (a) is 000 times greater than this, explaining why some airport employees must wear hearing protection equipment. 4. the intensity o this sound aried inersely with the square o the distance rom the source constant r, the ratio o the intensities at distances r = 6 km and r = km rom the source is gien by constant r r constant r r 6 km km The dierence in the decibel leels at distances r and r rom this source was then 0 log 0 log 0 log log 0 log 6 km km or Page 4.4

5 ound db giing 9.5 db 80 db 9.5 db 50 db 4. The decibel leel due to the irst siren is 00.0 W m 0 log 40 db..0 0 W m - Thus, the decibel leel o the sound rom the ambulance is 0 db 40 db 0 db 50 db 4.3 n terms o their intensities, the dierence in the decibel leel o sounds is 0 log 0 log 0 log log Thus, 0 0 or db and W m, then W m W m 4.4 The sound power incident on the eardrum is P = A where is the intensity o the sound and A = m is the area o the eardrum. (a) At the threshold o hearing,.0 0 W m, and P.0 0 W m m W Page 4.5

6 ound 6 (b) At the threshold o pain,.0 W m, and P.0 W m m W 4.5 The decibel leel 0 log 0, where W m. (a) = 00 db, then log 0 0 giing W m 0 (b) all three toadish sound at the same time, the total intensity o the sound produced is W m, and the decibel leel is W m 0 log.00 0 W m 0 log log (a) From the deining equation o the decibel leel, 0 log 0. We sole or the intensity as and ind that.0 0 W m W m W m 0.36 W m (b) 5 trumpets are sounded together, the total intensity o the sound is W m.58 W m 5 (c) the sound propagates uniormly in all directions, the intensity aries inersely as the Page 4.6

7 ound 7 square o the distance rom the source, constant r, and we ind that r r or.0 m r r.58 W m.47 0 W m 8.0 m (d).47 0 W m 0 log 0 log 04 db row row W m (e) The intensity o sound at the threshold o hearing is W m, and rom the discussion and result o part (c), we hae 0 0 r r and with the intensity being.47 0 W m at distance r = 8.0 m, the distance at which the intensity would be W m is.47 0 W m 8.0 m.6 0 m 6 0 r W m r () The sound intensity leel alls as the sound wae traels arther rom the source until it is much lower than the ambient noise leel and is drowned out. 4.7 The intensity o a spherical sound wae at distance r rom a point source is P a 4 r, where P a is the aerage power radiated by the source. Thus, at distances r = 5.0 m and r = 0 km = 0 4, the intensities o the sound wae radiating out rom the elephant are P a a and 4 r 4 r P giing r r Page 4.7

8 ound 8 From the deining equation, 0 log o, the intensity leel o the sound at distance r rom the elephant is seen to be 0 log 0 log r 0 log r 0 log 0 log r 0 log r r r or 5.0 m 0 log 66 db 03 db 37 db 0 m or (a) The intensity o the sound generated by the orchestra ( = 80 db) is 0 0, and that produced by the crying baby ( = 75 db) is rch b 0 0. Thus, the total intensity o the sound enguling you is rch b W m W m 8 4 (b) The combined sound leel is 0 log log db 4.9 (a) The intensity o sound at 0 km rom the horn (where = 50 db) is Page 4.8

9 ound 9 0 / W m W m Thus, rom P 4 r, the power emitted by the source is 3 7 P 4 r m.0 0 W m.3 0 W (b) At r = 50 m, the intensity o the sound will be P.3 0 W W m 4 r 4 50 m and the sound leel is W m 3 0 log 0 log 0 log(4.0 09) 96 db W m P 00 W 4.0 (a) W m 4 r 4 (0.0 m) (b) 0 log 0 log W m.00 0 W m 0 0 log( ) 09 db (c) At the threshold o pain ( = 0 db), the intensity is.00 W m. Thus, rom P 4 r, the distance rom the speaker is Page 4.9

10 ound 0 r P 00 W W m.8 m 4. The sound leel or intensity is 0log 0. Thereore, 0 log 0 log 0 log 0 log ince P P 4 4 r r the ratio o intensities is P 4 r r r P 4 r Thus, 0 log r 0 log r 0 log r r r r 4. The intensity at distance r rom the source is à à 4 4 r r Page 4.0

11 ound (a) A B r B A r 00 m 00 m 00 m (b) A C r C A r 00 m 00 m 00 m When a stationary obserer 0 = 0 hears a moing source, the obsered requency is (a) When the train is approaching, 40.0 m s and approach 345 m s 30 Hz 36 Hz 345 m s 40.0 m s Ater the train passes and is receding, 40.0 m s and 345 m s recede (30 Hz) 87 Hz and 345 m s 40.0 m s Thus, the requency shit that occurs as the train passes is Page 4.

12 ound 75. Hz, or it is a 75. Hz drop recede approach (b) As the train approaches, the obsered waelength is approach 345 m s 36 Hz m 4.4 The general expression or the obsered requency o a sound when the source and/or the obserer are in motion is Here, is the elocity o sound in air, is the elocity o the obserer, is the elocity o the source, and is the requency that would be detected i both the source and obserer were stationary. (a) = 5.00 khz and the obserer is stationary ( = 0), the requency detected when the source moes toward the obserer at hal the speed o sound is 0 (5.00 khz) (5.00 khz) 0.0 khz (b) When = 5.00 khz and the source moes away rom a stationary obserer at hal the speed o sound 343 m s, the obsered requency is Page 4.

13 ound 3 0 (5.00 khz) (5.00 khz) 3.33 khz Both source and obserer are in motion, so ince each train moes toward the other, > 0 and > 0. The speed o the source (train ) is km 000 m h m s h km 3600 s and that o the obserer (train ) is 30 km h 36. m s. Thus, the obsered requency is 345 m s 36. m s (500 Hz) 595 Hz 345 m s 5.0 m s 4.6 ince the obserer hears a reduced requency, the source and obserer are getting arther apart. Hence, the bicyclist is behind the car. With the bicyclist (obserer) behind the car (source) and both moing in the same direction, the obserer moes toward the source ( > 0) while the source moes away rom the obserer ( > 0). Thus, bicyclist car 3 and car where car is the speed o the car. Page 4.3

14 ound 4 The obsered requency is car 3 car 3 car car, giing 45 Hz 440 Hz 345 m s m s car car and car 3. m s 4.7 With the train approaching the stationary obserer ( = 0) at speed t, the source elocity is and the obsered requency is t 44 Hz 345 m s 345 m s t [] As the train recedes, the source elocity is t and the obsered requency is 44 Hz 345 m s 345 m s t [] Diiding equation [] by [] gies m s m s t t Page 4.4

15 ound 5 and soling or the speed o the train yields t 0.39 m s 4.8 We let the speed o the insect be bug and the speed o the bat be bat 5.00 m s, and break the action into steps. n the irst step, the bat is the sound source lying toward the obserer (the insect), so bat, while the insect (obserer) is lying away rom the source, making. 0 is the actual requency sound emitted by the bat, the requency detected (and bug relected) by the moing insect is relect 0 0 bug bat or relect 0 bug bat n the second step o the action, the insect acts as a sound source, relecting a wae o requency relect back to the bat which acts as a moing obserer. ince the source (insect) is moing away rom the obserer, bug, and the obserer (bat) is moing toward the source (insect) giing bat. The requency o the return sound receied by the bat is then return relect relect bat bug or return relect bat bug Combing the results o the steps gies return 0 bug bat bat bug Page 4.5

16 ound 6 or 40.4 khz 40.0 khz 340 m s bug m s bug This reduces to m s bug 340 m s bug or bug 340 m s and yields 3.3 m s. bug Thus, the bat is gaining on the insect at a rate o 5.00 m s 3.3 m s.69 m s. 4.9 For a source receding rom a stationary obserer,. Page 4.6

17 ound 7 Thus, the speed the alling tuning ork must reach is 5 Hz 340 m s 8.9 m s 485 Hz The distance it has allen rom rest beore reaching this speed is 0 a 9.80 m s y y 8.9 m s m The time required or the 485 Hz sound to reach the obserer is t y 8.3 m 340 m s s During this time the ork alls an additional distance y t ayt 8.9 m s s 9.80 m s s.03 m The total distance allen beore the 485 Hz sound reaches the obserer is y y y 8.3 m+.03 m= 9.3 m 4.30 (a) 5 min.0 rad s 60.0 s min Page 4.7

18 ound 8 and or harmonic motion, 3 max A.0 rad s.80 0 m 0.07 m s (b) The heart wall is a moing obserer ( max ) and the detector a stationary source, so the maximum requency relected by the heart wall is wall max max Hz Hz 500 (c) Now, the heart wall is a moing source ( max ) and the detector a stationary obserer. The obsered requency o the returning echo is echo wall max max Hz Hz The hal-angle o the cone o the shock wae is where sound sin sin 4.5 source As shown in the sketch, the angle between the direction o propagation o the shock wae and the direction o the plane s elocity is Page 4.8

19 ound (a) Equation 4. is where is the requency emitted by the source, is the requency detected by the obserer, is the speed o the wae in the propagating medium, is the elocity o the obserer relatie to the medium, and is the elocity o the source relatie to the propagating medium. (b) The yellow submarine is the source or emitter o the sound waes. (c) The red submarine is the obserer or receier o the sound waes. (d) The motion o the obserer away rom the source tends to increase the time obsered between arrials o successie pressure maxima. This eect tends to cause an increase in the obsered period and a decrease in the obsered requency. (e) n this case, the sign o should be negatie to decrease the numerator in Equation 4., and thereby decrease the calculated obsered requency. () The motion o the source toward the obserer tends to decrease the time between the arrial o successie pressure maxima, decreasing the obsered period, and increasing the obsered requency. Page 4.9

20 ound 0 (g) n this case, the sign o should be positie to decrease the denominator in Equation 4., and thereby increase the calculated obsered requency. 53 a m s 3.00 m s (h) Hz Hz 53 m s.0 m s 4.33 The waelength o the waes being generated by the speakers is 343 m s 343 Hz.00 m so m ince the two speakers are drien by the same sound source or oscillator, they must ibrate in phase with each other. This means that the point at x =.00 m, being equidistant rom the two speakers must be a point o constructie intererence. ther points o constructie intererence along the line connecting the speakers will be at integral multiples o a hal waelength rom this antinode. This means that constructie intererence occurs at x.00 m,.00 m,.00 m, and.00 m 3, or at x m,.00 m,.50 m,.00 m,.50 m, 3.00 m, and 3.50 m 4.34 The waelength o the sound emitted by the speaker is 345 m s 756 Hz m and a hal waelength is 0.8 m. (a) a condition o constructie intererence currently exists, this can be changed to a case o Page 4.0

21 ound destructie intererence by adding a distance o 0.8 m to the path length through the upper arm. (b) To moe rom a case o constructie intererence to the next occurrence o constructie intererence, one should increase the path length through the upper arm by a ull waelength, or by m At point D, the distance o the ship rom point A is 800 m 600 m 800 m 000 m d d ince destructie intererence occurs or the irst time when the ship reaches D, it is necessary that d d, or d d 000 m 600 m 800 m 4.36 The speakers emit sound o waelength 345 m s 450 Hz m, so m.at point, r r.50 m 8.00 m 8.4 m. To create destructie intererence at point, we moe the top speaker upward a distance y rom its original location until we hae r r. ince this did not change r, we must now hae r r 8.4 m m 8.5 m But, ater moing the speaker, this gies Page 4.

22 ound r.50 m y 8.00 m 8.5 m or.50 m y 8.5 m 8.00 m 8.59 m and y 8.59 m.50 m.43 m 4.37 The waelength o the sound is 345 m s 690 Hz m (a) At the irst relatie maximum (constructie intererence), d d d m Using the Pythagorean theorem, m m d d, giing d 0.40 m (b) At the irst relatie minimum (destructie intererence), Page 4.

23 ound 3 d d d 0.50 m Thereore, the Pythagorean theorem yields 0.50 m m d d, or d m 4.38 n the undamental mode o ibration, the waelength o waes in the wire is L m.400 m the wire is to ibrate at 6.6 Hz, the speed o the waes must be.400 m 6.6 Hz 366. m s With m L kg m kg m the required tension is gien by F as F m s kg m 84.0 N Page 4.3

24 ound n the third harmonic, the string orms a standing wae o three loops, each o length 8.00 m 3.67 m. The waelength o the wae is then 5.33 m. (a) The nodes in this string ixed at each end will occur at distances o 0,.67 m, 5.33 m, and 8.00 m rom the end. Antinodes occur halway between each pair o adjacent nodes, or at.33 m, 4.00 m, and 6.67 m rom the end. (b) The linear density is m L kg 8.00 m kg m and the wae speed is F 49.0 N kg m 99.0 m s Thus, the requency is 99.0 m s 5.33 m 8.6 Hz 4.40 With antinodes at each end and a single node located at the center o the rod, the length o the rod is one-hal waelength, or L.00 m.00 m Page 4.4

25 ound 5 The speed o sound in aluminum is requency o the resonance in the rod is 5 00 m s (see Table 4. in the textbook), so the 5 00 m s.00 m Hz.55 khz 4.4 The acing speakers produce a standing wae in the space between them, with the spacing between nodes being d NN 343 m s 800 Hz 0.4 m the speakers ibrate in phase, the point halway between them is an antinode o pressure, at.5 m 0.65 m rom either speaker. Then there is a node at 0.65 m 0.4 m 0.58 m, and nodes at 0.58 m 0.4 m m m 0.4 m m 0.58 m 0.4 m 0.73 m, 0.73 m 0.4 m m and m 0.4 m.6 m Page 4.5

26 ound 6 rom one speaker. 4.4 n a wire o length is ixed at both ends, the waelength o the undamental mode o ibration is. The speed o transerse waes in the wire is F, where F is the tension in the wire and is the mass per unit length o the wire. The undamental requency or the wire is then F we hae two wires with the same mass per unit length, one o length L and under tension F while the second has length L and tension 4F, the ratio o the undamental requencies o the two wires is, long L 4F 4 L F, short or the two wires hae the same undamental requency o ibration. this requency is = 60 Hz, then the requency o the second harmonic or both wires is 60 Hz 0 Hz 4.43 (a) The linear density is m L kg.35 m.85 0 kg m (b) n a string ixed at both ends, the undamental mode has a node at each end and a single Page 4.6

27 ound 7 antinode in the center, so that L, or = L = (.0 m) =.0 m. Then, the wae speed in the wire is.0 m 4. Hz 90.6 m s (c) The speed o transerse waes in a string is F, so the required tension is F.85 0 kg m 90.6 m s 5 N (d) L.0 m.0 m [ee part (b) aboe.] (e) The waelength o the longitudinal sound waes produced in air by the ibrating string is air air 343 m s 4. Hz 8.33 m 4.44 (a) A string ixed at each end orms standing wae patterns with a node at each end and an integer number o loops, each o length two ends. Thus, L n or L n. and with an antinode at its center, between the the string has tension F and mass per unit length, the speed o transerse waes is T. Thus, when the string orms a standing wae o n loops (and hence n antinodes), the requency o ibration is T n T na A L n L L A T A Page 4.7

28 ound 8 (b) Assume the length is doubled, L B = L A, and a new standing wae is ormed haing n B = n A and T B = T A. Then, B nb TB na TA na TA A L L L B A A (c) oling the general result obtained in part (a) or the tension in the string gies T 4 L n. Thus, i B = A, L B = L A, and n B = n A +, we ind 4 B LB 4 A LA na 4 A LA na na B A A nb n A A A n n n A na T T T (d) now we hae B = 3 A, L B = L A, L L, and n n, then B A B A T B 4 L 4 9 L T n 6 6 B B A A A LA B 4n n A A A or T T B A (a) From the sketch at the right, notice that when d.0 m, 5.0 m d L.5 m, and / sin d 4 L Page 4.8

29 ound 9 Then ealuating the net ertical orce on the lowest bit o string, F F cos mg 0 gies the tension in the string as y F mg kg 9.80 m s cos cos 4 79 N (b) The speed o transerse waes in the string is F 79 N kg m.8 0 m s For the pattern shown, 3 d, so d 4.0 m 3 3 Thus, the requency is m s 4.0 m. 0 Hz 4.46 (a) For a standing wae o 6 loops, 6 ( ) L, or L.0 m 3 3 The speed o the waes in the string is then Page 4.9

30 ound 30.0 m 50 Hz.0 0 m s 3 ince the tension in the string is F mg 5.0 kg 9.80 m s 49 N, F gies F 49 N.0 0 m kg m (b) m 45 kg, then F 45 kg 9.80 m s N, and N kg m m s Thus, m s 50 Hz.0 m and the number o loops is n L.0 m.0 m (c) m = 0 kg, the tension is F 0 kg 9.80 m s 98 N, and Page 4.30

31 ound 3 98 N kg m.4 0 m s Then,.4 0 m s 50 Hz 0.94 m and n L.0 m 0.47 m is not an integer, so no standing wae will orm The speed o transerse waes in the string is F N kg m 70.7 m s The undamental waelength is = L =.00 0 m and its requency is 70.7 m s.00 0 m Hz The harmonic requencies are then n n n Hz, with n being an integer. The largest one under Hz is Hz khz 4.48 The distance between adjacent nodes is one-quarter o the circumerence. Page 4.3

32 ound 3 d d NN AA 0.0 cm 5.00 cm 4 so = 0.0 cm = 0.00 m, and 900 m s 0.00 m Hz 9.00 khz The singer must match this requency quite precisely or some interal o time to eed enough energy into the glass to crack it Assuming an air temperature o T = 37 C = 30 K, the speed o sound inside the pipe is T 30 K 33 m s 33 m s 353 m s 73 K 73 K n the undamental resonant mode, the waelength o sound waes in a pipe closed at one end is = 4 L. Thus, or the whooping crane t.0 0 t and 353 m s 3.8 t.0 0 t m 58 Hz 4.50 (a) n the undamental resonant mode o a pipe open at both ends, the distance between antinodes is daa L. Page 4.3

33 ound 33 Thus, = L = (0.30 m) = m, and 340 m s m 53 Hz (b) d AA 340 m s Hz m 4.5 cm 4.5 Hearing would be best at the undamental resonance, so = 4 L = 4 (.8 cm) and 340 m s 00 cm 4.8 cm m Hz 3.0 khz 4.5 (a) To orm a standing wae in the tunnel, open at both ends, one must hae an antinode at each end, a node at the middle o the tunnel, and the length o the tunnel must be equal to an integral number o hal-waelengths [ L n or L n]. The resonance requencies o the tunnel are then sound in air 345 m s 345 m s n n Hz n,, 3, n n Ln m (b) t would be good to make such a rule. Any car horn would produce seeral closely spaced resonance requencies o the air in the tunnel, so the sound would be greatly ampliied. ther driers might be rightened directly into dangerous behaior or might blow their horns also (a) The speed o sound is 33 m s at 0 C, so the undamental waelength o the pipe open at both ends is Page 4.33

34 ound 34 L giing L 33 m s 300 Hz 0.55 m (b) At T = 30 C = 303 K, T 303 K 33 m s 33 m s 349 m s 73 K 73 K and 349 m s L 0.55 m 36 Hz 4.54 bsere rom Equations 4.8 and 4.9 in the textbook that the dierence between successie resonance requencies is constant, regardless o whether the pipe is open at both ends or is closed at one end. Thus, the resonance requencies o 650 Hz or less or this pipe must be 650 Hz, 550 Hz, 450 Hz, 350 Hz, 50 Hz, 50 Hz, and 50.0 Hz, with the lowest or undamental requency being 50.0 Hz. Note, rom the list gien aboe, the resonance requencies are only the odd multiples o the undamental requency. This is a characteristic o a pipe that is closed at one end and open at the other. The length o such a pipe (with an antinode at the open end and a node at the closed end) is one-quarter o the waelength o the undamental requency, so the length o this pipe must be Page 4.34

35 ound 35 L sound 340 m s Hz.70 m 4.55 n a string ixed at both ends, the length o the string is equal to a hal-waelength o the undamental resonance requency, so = L. The undamental requency may then be written as T T L 4L a second identical string with tension T < T is struck, the undamental requency o ibration would be T T T T 4L 4L T T When the two strings are sounded together, the beat requency heard will be beat T N.0 0 Hz 5.64 beats s T N 4.56 By shortening her string, the second iolinist increases its undamental requency. Thus, beat Hz 98Hz. ince the tension and the linear density are both identical or the two strings, the speed o transerse waes, F, has the same alue or both strings. Thereore,, or. The undamental waelength o a string ixed at both ends is L, and this yields L L cm 9.7 cm 98 Page 4.35

36 ound The commuter, stationary relatie to the station and the irst train, hears the actual source requency ( 0, = = 80 Hz) rom the irst train. The requency the commuter hears rom the second train, moing relatie to the station and commuter, is gien by, beat 80 Hz Hz 78 Hz or 8 Hz This stationary obserer ( 0 = 0) hears the lower requency ( 0, = 78 Hz) i the second train is moing away rom the station, so gies the speed o the receding second train as 345 m s m s 0 78 Hz 80 Hz 80 Hz 345 m s 345 m s or 345 m s 345 m s 80 Hz 78 Hz and 80 Hz 345 m s 3.88 m s 78 Hz so one possibility or the second train is 3.88 m s away rom the station The other possibility is that the second train is moing toward the station and the Page 4.36

37 ound 37 commuter is detecting the higher o the possible requencies ( 0, = 8 Hz). n this coase, [( ) /( )] yields 8 Hz 80 Hz 345 m s m s and 345 m s 345 m s 80 Hz 8 Hz or 80 Hz 345 m s 3.79 m s 8 Hz n this case, the elocity o the second train is 3.79 m s toward the station The temperatures o the air in the two pipes are T = 7 C = 300 K and T = 3 C = 305 K. The speed o sound in the two pipes is T T 33 m s and 33 m s 73 K 73 K ince the pipes hae the same length, the undamental waelength, 4L, is the same or them. Thus, rom, the ratio o their undamental requencies is seen to be, which gies. Page 4.37

38 ound 38 The beat requency produced is then T beat T or beat 305 K 480 Hz = 3.98 Hz 300 K 4.59 (a) First consider the wall a stationary obserer receiing sound rom an approaching source haing elocity a. The requency receied and relected by the wall is relect a. Now consider the wall as a stationary source emitting sound o requency relect to an obserer approaching at elocity a. The requency o the wae heard by the obserer is a a a relect a a Thus, the beat requency between the tuning ork and its echo is a a.33 beat (56 Hz).98 Hz a a Page 4.38

39 ound 39 (b) When the student moes away rom the wall, a changes sign so the beat requency heard is beat a a a a giing a beat beat The receding speed needed to obsere a beat requency o 5.00 Hz is a 345 m s 5.00 Hz 56 Hz 5.00 Hz 3.40 m s 4.60 The extra sensitiity o the ear at 3000 Hz appears as downward dimples on the cures in Figure 4.9 o the textbook. At T = 37 C = 30 K, the speed o sound in air is T 30 K 33 m s 33 m s 353 m s 73 K 73 K Thus, the waelength o Hz sound is 353 m s Hz 0.8 m Page 4.39

40 ound 40 For the undamental resonant mode in a pipe closed at one end, the length required is L 0.8 m m.94 cm At normal body temperature o T = 37 C = 30 K, the speed o sound in air is T 30 K 33 m s 33 m s 353 m s 73 K 73 K and the waelength o Hz sound is 353 m s Hz.76 0 m.76 cm Thus, the diameter o the eardrum is.76 cm 4.6 The decibel leel o a sound haing intensity is 0 log 0, where 0 is a constant. oling or the intensity in terms o the decibel leel gies = 0 0 /0. Thus, i the decibel leel o the sound rom a single iolin is = 70 db, and the decibel leel o the sound rom the ull orchestra is = 85 db, the ratio o the intensity o the ull orchestra to that rom the single iolin is Page 4.40

41 ound (a) With a decibel leel o 03 db, the intensity o the sound at.60 m rom the speaker is ound rom 0 log 0 as W m W m the speaker broadcasts equally well in all directions, the intensity (power per unit area) at.60 m rom the speaker is uniormly distributed oer a spherical wae ront o radius r =.60 m centered on the speaker. Thus, the power radiated is P A r W m 4.60 m W (b) eiciency P output P input W or 0.43% 50 W 4.64 (a) At point C, the distance rom speaker A is r A 3.00 m 4.00 m 5.00 m and the intensity o the sound rom this speaker is A P A W 4 r A m W m 6 The sound leel at C due to speaker A alone is then Page 4.4

42 ound 4 A 6 A W m 0 log 0 log 65.0 db.00 0 W m 0 (b) The distance rom point C to speaker B is r.00 m 4.00 m 4.47 m and B the intensity o the sound rom this speaker alone is B P 3 B.50 0 W W m 4 rb m The sound leel at C due to speaker B alone is thereore B 6 B W m 0 log 0 log 67.8 db W m 0 (c) both speakers are sounded together, the total sound intensity at point C is AB A B W m W m W m and the total sound leel in decibels is AB 6 AB W m 0 log 0 log 69.6 db.00 0 W m We assume that the aerage intensity o the sound is directly proportional to the number o cars passing each minute. the sound leel in decibels is 0 log 0, the intensity o the sound is 0 0 0, so the aerage intensity in the aternoon, when 00 cars per minute are passing, is Page 4.4

43 ound W m W m The expected aerage intensity at night, when only 5 cars pass per minute, is gien by the ratio , or W m W m and the expected sound leel in decibels is W m 0 log 0 log 67.0 db W m 4.66 The well will act as a pipe closed at one end (the bottom) and open at the other (top). The resonant requencies are the odd integer multiples o the undamental requency, or n = (n ) where n =,, 3,..... Thus, i n and n+ are two successie resonant requencies, their dierence is n n n n n n n this case, we hae 60.0 Hz 5.0 Hz =, giing the undamental requency or the well as = 4.00 Hz. n the undamental mode, the well (pipe closed at one end) orms a standing wae pattern with a node at the bottom and the irst antinode at the top, making the depth o the well d sound 345 m s Hz.6 m Page 4.43

44 ound The requency heard rom the irst train, moing toward the stationary obserer at 30.0 m s, is 345 m s 300 Hz 38.6 Hz 345 m s 30.0 m s The second train moes toward the obserer at 30.0 m s. The requency heard rom this train must be beat 38.6 Hz 3.0 Hz=33.6 Hz Then, rom the speed o the approaching second train must be 300 Hz 345 m s 3.9 m s 33.6 Hz 4.68 (a) a source emits sound o requency (as detected by an obserer stationary relatie to the source), the requency detected by the obserer when the source and/or the obserer is in motion is where is the elocity o sound in air, is the elocity o the obserer, and is the elocity o the source. n the gien situation, = 30 Hz, = 0, and when the train is approaching the obserer, requency heard by the obserer is 40 m s. Thus, the Page 4.44

45 ound 45 a, 345 m s 0 30 Hz 36 Hz 345 m s 40.0 m s (b) When the train is receding rom the stationary obserer, 40.0 m s and the detected requency will be b, 345 m s m s 30 Hz 30 Hz 87 Hz 345 m s 40.0 m s 385 m s c) The waelengths measured by the obserer in each o the cases aboe are a a, 345 m s 36 Hz m and b b, 345 m s 87 Hz.0 m 4.69 This situation is ery similar to the undamental resonance o an organ pipe that is open at both ends. The waelength o the standing waes in the crystal is = t, where t is the thickness o the crystal, and the requency is m s m Hz 6 khz 4.70 The distance rom the balcony to the man's head is y 0.0 m.75 m 8.3 m The time or a warning to trael this distance is Page 4.45

46 ound 46 t 8.3 m 345 m s s so the total time needed to receie the warning and react is t t s s. The time or the pot to all this distance, starting rom rest, is t y 8.3 m a 9.80 m s y.93 s Thus, the latest the warning should be sent is at t t t.93 s s.58 s into the all. n this time interal, the pot has allen 9.80 m s.58 s. m y g t and is h 0.0 m. m= 7.8 m aboe the sidewalk. 4.7 n the weekend, there are one-ourth as many cars passing per minute as on a week day. Thus, the intensity,, o the sound on the weekend is one-ourth that,, on a week day. The dierence in the decibel leels is thereore: Page 4.46

47 ound 47 0 log 0 log 0 log 0 log (4) 6 db o o so, 6 db 70 db 6 db 64 db 4.7 (a) At T = 0 C = 93 K, the speed o sound in air is T 93 K 33 m s 33 m s 343 m s 73 K 73 K The irst harmonic or undamental o the lute (a pipe open at both ends) is gien by 343 m s L.3 m. 6.6 Hz Thereore, the length o the lute is L.3 m m (b) n the colder room, the length o the lute and hence its undamental waelength is essentially unchanged (that is, = =.3 m). Howeer, the speed o sound and thus the requency o the undamental will be lowered. At this lower temperature, the requency must be beat 6.6 Hz 3.00 Hz=58.6 Hz Page 4.47

48 ound 48 The speed o sound in this room is.3 m 58.6 Hz 339 m s From 33 m s T 73 K, the temperature in the colder room is gien by T 339 m s 73 K 73 K 86 K 3.0 C 33 m s 33 m s 4.73 The maximum speed o the oscillating block and speaker is k max A A m 0.0 N m m.00 m s 5.00 kg When the speaker (sound source) moes toward the stationary obserer, then = + max and the maximum requency heard is max max 345 m s 440 Hz 44 Hz 345 m s.00 m s When the speaker moes away rom the stationary obserer, the source elocity is = max and the minimum requency heard is min max 345 m s 440 Hz 439 Hz 345 m s.00 m s Page 4.48

49 ound The speed o transerse waes in the wire is F F L m 400 N m kg 365 m s When the wire ibrates in its third harmonic, wire and the sound produced by the wire is = L/3 = m, so the requency o the ibrating 365 m s m 730 Hz ince both the wire and the wall are stationary, the requency o the wae relected rom the wall matches that o the waes emitted by the wire. Thus, as the student approaches the wall at speed, he approaches one stationary source and recedes rom another stationary source, both emitting requency = 730 Hz. The two requencies that will be obsered are and The beat requency is beat so Page 4.49

50 ound 50 beat 8.30 Hz 730 Hz 340 m s.93 m s 4.75 The speeds o the two types o waes in the rod are long Y Y m V Y A L m and trans F F F L m L m Thus, i long = 8 trans, we hae Y A L F L 64 m m or the required tension is F N m m Y A N 4.76 (a) As the student walks along the line connecting the locations o the two speakers, she is walking away rom the irst speaker with elocity.50 m s, and toward the second speaker with elocity. Thus, she will experience a Doppler shit in the sound rom each o the stationary ( = 0) speakers. The requency detected rom the irst speaker is,, and the requency detected rom the second speaker is,. The beat requency she will hear as she receies these slightly dierent requencies together is Page 4.50

51 ound 5 beat,, s s.50 m s s 456 Hz 3.97 beats s 345 m s (b) The waelength o the sound produced by each o the speakers will be 345 m s 456 Hz m n the standing wae pattern that orms along the line connecting the two speakers, successie antinodes will be separated by a distance x m. As the student walks along this line with speed.50 m s, she will hear an intensity maximum eery time she passes an antinode. The time interal (period) between hearing successie maxima is gien by (requency) will be T x, so the number o intensity maxima heard per second maxima.50 m s T x m 3.97 Hz (c) The answers are identical. The models are equally alid. We may think o the intererence o the two waes as intererence in space or in time, linked to space by the steady motion o the student. Page 4.5