Chapter 14. Answers to Even Numbered Problems s cm to 17 m m K. 10. (a) (b) W db. 14.
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1 Answers to Een Numbered Problems Chpter s 4..7 cm to 7 m m K 0. () W (b) W. 3.0 db 4. () 6. () W m (b) 8. db W m (b) 09 db (c).8 m IA 8. () I (b) IA I 5 B 0. () 0.0 khz (b) 3.33 khz. 3. m s, behind the cr 4. 4 khz 6. () 0.07 m s (b) Hz (c) Hz 8. () 57. s (b) 56.6 km 30. () 0.43 m (b) m 3. () m (b) constructie interference N khz Hz C 40. () kg m (b) (c) no stnding we will form khz 50
2 50 CHAPTER () 53 Hz (b) 4.5 cm 46. n( 06 Hz ) for n to 9 nd n( 84.5 Hz ) for n to m s cm Hz cm 56. () 36 Hz (b) 87 Hz (c) m nd.0 m m m s 6. () m (b) 3.0 C 64. () 0.33 m (b) 4 mm 66. () pth difference of λ produces destructie interference (b) Along hyperbol gien by m s 70. (b) 53 Hz y x () 55.8 m s (b).50 khz.00 m
3 ound 503 Problem olutions 4.6 At temperture of T 0.0 C 83 K, the speed of sound in ir is T 83 K ( 33 m s) ( 33 m s) 337 m s 73 K 73 K The elpsed time between when the stone ws relesed nd when the sound is herd is the sum of the time t required for the stone to fll distnce h nd the time t required for sound to trel distnce h in ir on the return up the well. Tht is, t + t.00 s. The distnce the stone flls, strting from rest, in time t is gt h Also, the time for the sound to trel bck up the well is h t.00 s t Combining these two equtions yields g t.00 s With 337 m s nd g 9.80 m s, this becomes s t + t.00 s 0 Applying the qudrtic eqution yields one positie solution of t.945 s, so the depth of the well is gt ( 9.80 m s )(.945 s) h 8.5 m 4.7 From Tble 4., the speed of sound in the sltwter is 530 m s. At T 0 C 93 K, the speed of the sound in ir is T 93 K ( 33 m s) ( 33 m s) 343 m s 73 K 73 K If d is the width of the inlet, the trnsit time for the sound in the wter is t w d tht for the sound in the ir is t tw s. d d w Thus, s, or d ( 4.50 s) w d ( 530 m s )( 343 m s ) ( ) m s s.99 0 m.99 km w w d, nd w 4. bsere tht I 00 W m I 00 W m.00 Thus, the difference in the intensity leels of these two sounds is I I I I 0 I β β 0log 0log 0log 0log I I I I I or β β log db
4 504 CHAPTER () The intensity of the sound generted by the orchestr (β 80 db) is β I I0 I 0, nd tht produced by the crying bby (β 75 db) is Ib rch I 0 0. Thus, the totl intensity of the sound engulfing you is rch b ( 0 0 ) I I + I I W m W m 8 4 (b) The combined sound leel is 8 β 0log II 0log db ( 0) 4.9 The sound leel for intensity I is β 0log( II 0 ). Therefore, ince Thus, I I I I 0 I β β 0log 0log 0log 0log I 0 I 0 I I 0 I à ( 4π ) I Ã, the rtio of intensities is 4π r r I à 4π r r I r à 4π r β r r r β 0log 0log 0log r r r 4. When sttionry obserer ( 0) + f f f. () When the trin is pproching, m s nd ( f ) pproch hers moing source, the obsered frequency is 345 m s 30 Hz 36 Hz 345 m s 40.0 m s After the trin psses nd is receding, 40.0 m s nd 345 m s 30 H z 87 Hz. 345 m s ( 40.0 m s) ( f ) recede Thus, the frequency shift tht occurs s the trin psses is 75. Hz f f f, or it is 75. Hz drop recede pproch (b) As the trin pproches, the obsered welength is
5 ound m s λ 36 Hz ( f ) pproch m Both source nd obserer re in motion, so f f. ince ech trin moes towrd the other, > 0 nd > 0. The speed of the source (trin ) is km 000 m h m s h km 3600 s nd tht of the obserer (trin ) is 30 km h 36. m s. Thus, the obsered frequency is 345 m s+ 36. m s f ( 500 Hz) 595 Hz 345 m s 5.0 m s 4.7 For source receding from sttionry obserer, Thus, the speed the flling tuning fork must rech is f 5 Hz ( 340 m s) 8.9 m s f 485 Hz The distnce it hs fllen from rest before reching this speed is m s 0 y 8.3 m y 9.80 m s. f f f + ( ) The time required for the 485 Hz sound to rech the obserer is y 8.3 m t s 340 m s During this time the fork flls n dditionl distnce y t + t y 8.9 m s s m s s.03 m The totl distnce fllen before the 485 Hz sound reches the obserer is y y+ y 8.3 m+.03 m 9.3 m 4.33 The welength of the sound is λ 345 m s m f 690 Hz. () At the first reltie mximum (constructie interference), d d+ λ d m peker m peker d d bserer
6 506 CHAPTER 4 Using the Pythgoren theorem, ( d m ) d ( m ) + +, giing d 0.40 m (b) At the first reltie minimum (destructie interference), d d + λ d m Therefore, the Pythgoren theorem yields ( d 0.50 m ) d ( m ) + +, or d m 4.37 The fcing spekers produce stnding we in the spce between them, with the spcing between nodes being λ 343 m s dnn 0.4 m f 800 Hz If the spekers ibrte in phse, the point hlfwy between them is n ntinode of pressure, t.5 m 0.65 m from either speker. 0.4 m Then there is node t 0.65 m 0.58 m, node t 0.58 m 0.4 m m, node t m 0.4 m m, node t 0.58 m m 0.73 m, node t 0.73 m m m, nd node t m m.6 m from one speker () From the sketch t the right, notice tht when 5.0 m d d.0 m, L.5 m, d/ nd θ sin 4 L Then eluting the net erticl force on the lowest bit of string, Σ Fy Fcosθ mg 0 gies the tension in the string s mg ( kg)( 9.80 m s ) F 79 N cos θ cos 4 (b) The speed of trnserse wes in the string is F 79 N.8 0 m s µ kg m For the pttern shown, 3( λ ) d, so d 4.0 m λ 3 3 L d q ur F mg urfur
7 ound 507 Thus, the frequency is m s f λ 4.0 m. 0 Hz λ L.0 m 4.40 () For stnding we of 6 loops, 6 L, or λ 3 3 The speed of the wes in the string is then.0 m λ f ( 50 Hz).0 0 m s 3 F mg 5.0 kg 9.80 m s 49 N (b) If ince the tension in the string is F µ gies F 49 N µ (.0 0 m ) kg m m 45 kg, then F ( 45 kg)( 9.80 m s ) N, nd N m s kg m m s Thus, λ.0 m f 50 Hz L.0 m nd the number of loops is n λ.0 m (c) If m 0 kg, the tension is F ( 0 kg)( 9.80 m s ) 98 N, nd 98 N.4 0 m s kg m.4 0 m s Then, λ 0.94 m f 50 Hz L.0 m nd n is not n integer, λ 0.47 m so no stnding we willform () The speed of sound is 33 m s t 0 C, so the fundmentl welength of the pipe open t both ends is λ L giing f 33 m s L f 300 Hz 0.55 m T 303 K (b) At T 30 C 303 K, ( 33 m s) ( 33 m s) 349 m s 73 K 73 K 349 m s nd f 36 Hz λ L 0.55 m
8 508 CHAPTER For pipe open t both ends, the frequency of the nth hrmonic is, fn n( L) the difference between two successie resonnt frequencies is f fn+ fn ( n+ ) n L L L In this cse, the pipe is. Thus, L.00 m nd f 49 Hz 40 Hz 8 Hz. Thus, the speed of sound in L f.00 m 8 Hz m s 4.5 If the second trin is moing towrd the sttionry obserer with speed, the Doppler effect gies f f > f Therefore, f f + fbet 80 Hz+ H z 8 Hz, nd the speed of the trin is f 80 ( 345 m s) 3.79 m s f 8 nd the elocity of the trin is 3.79 m s tow rd the sttion If the second trin is moing wy from the sttionry obserer ( ) +, then f f < f, giing f f fbet 80 Hz H z 78 Hz Thus, f 80 ( 345 m s) 3.88 m s f 78 nd the elocity of the trin is 3.88 m s w y from the sttion
9 ound () First consider the wll sttionry obserer receiing sound from n pproching source hing elocity. The frequency receied nd reflected by the wll is freflect f. Now consider the wll s sttionry source emitting sound to n obserer pproching t elocity. The frequency of the we herd by the obserer is f f f f reflect Thus, the bet frequency between the tuning fork nd its echo is +.33 fbet f f f f ( 56 Hz).98 Hz (b) When the student moes wy from the wll, chnges sign so the bet frequency herd is f bet ( ) ( ) f f +, giing fbet f f bet The receding speed needed to obsere bet frequency of 5.00 Hz is ( 345 m s)( 5.00 Hz) 56 Hz 5.00 Hz 3.40 m s 4.6 The mximum speed of the oscillting block nd speker is k 0.0 N m mx Aω A ( m ).00 m s m 5.00 kg When the speker (sound source) moes towrd the sttionry obserer, then + nd the mximum frequency herd is 345 m s ( f ) f ( 440 Hz) 44 Hz mx mx 345 m s.00 m s When the speker moes wy from the sttionry obserer, the source elocity is mxnd the minimum frequency herd is 345 m s ( f ) f ( 440 Hz) 439 Hz min + mx 345 m s+.00 m s mx 4.68 The speed of trnserse wes in the wire is
10 50 CHAPTER 4 ( 400 N )( m ) F F L 365 m s 3 µ m.5 0 kg When the wire ibrtes in its third hrmonic, λ L m, so the frequency of the ibrting wire nd the sound produced by the wire is 365 m s f 730 Hz λ m ince both the wire nd the wll re sttionry, the frequency of the we reflected from the wll mtches tht of the wes emitted by the wire. Thus, s the student pproches the wll t speed, he pproches one sttionry source nd recedes from nother sttionry source, both emitting frequency f 730 Hz. The two frequencies tht will be obsered re + ( f ) f nd ( f ) f The bet frequency is so + f fbet f f f f 8.30 Hz bet ( 340 m s).93 m s f ( 730 Hz) 4.7 () When the obserer is sttionry with source emitting sound of frequency f.80 kh z moing towrd it, the frequency detected by the obserer is f f f where is the speed of the source, nd ir. Thus, if the detected frequency is f.80 kh z ( 343 m s) 55.8 m s f.5 kh z 343 m s is the speed of sound in the f.5 kh z, the speed of the source is (b) In this cse, the skydier is moing obserer with elocity m s, moing towrd sttionry source (the ground) tht is emitting (reflecting) sound of frequency f.5 kh z. The frequency detected by the skydier will be m s m s f f (.5 kh z).50 kh z 343 m s 0
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