SOLUTIONS TO CONCEPTS CHAPTER 16

Transcription

1 . air = 30 m/. = 500 m/. Here S = 7 m So, t = t t = SOLUIONS O CONCEPS CHPER 6 = ec =.75 m.. Here gien S = 80 m = 60 m. = 30 m/ So the maximum time interal will be t = 5/ = 60/30 = 0.5 econd. 3. He ha to clap 0 time in 3 econd. So time interal between two clap = (3/0 econd). So the time taken go the wall = (3/ 0) = 3/0 econd. = 333 m/. 4. a) for maximum waelength n = 0 Hz. a b) for minimum waelength, n = 0 khz = 360/ (0 0 3 ) = m = 8 mm x = (/n) = 360/0 = 8 m. 5. a) for minimum waelength n = 0 KHz 450 = n = = 7.5 cm b) for maximum waelength n hould be minium = n = /n 450 / 0 = 7.5 m. 6. ccording to the quetion, a) = 0 cm 0 = 00 cm = m = m/ o, n = / = / = 70 Hz. N = / = Hz = 7 KH (becaue = cm = 0 m) 0 7. a) Gien air = m/, n = Hz air = ( / 4.5) 0 6 = m. b) tiue = 500 m/ t = (500 / 4.5) 0 6 = m. 8. Here gien r y = m a) Gien / =.8 = (/.8) r 5 y 6.0 (.8) 0 m / So, = m b) Let, elocity amplitude = y = dy/dt = 3600 co (600 t.8) 0 5 m/ Here y = m/ gain, = /.8 and = /600 wae peed = = / = 600/.8 = 000 / 3 m/ So the ratio of ( y /) = a) Here gien n = 00, = 350 m/ 350 = = 3.5 m. n 00 In.5 m, the ditance traelled by the particle i gien by x =

2 Chapter 6 So, phae difference = x ( / ). (350 /00) b) In the econd cae, Gien = 0 cm = 0 m 0 So, = x / 35. x (350 /00) 0. a) Gien x = 0 cm, = 5.0 cm = = So phae difference i zero. b) Zero, a the particle i in ame phae becaue of haing ame path.. Gien that p = N/m, = 73 K, M = 3 g = kg =.4 litre = m 3 C/C = r = 3.5 R /.5 R =.4 0cm 0 cm x cm = rp f /.4 5 = 30 m/ (becaue = m/). = 330 m/, =? = = 90 K, = = 305 K We know = 305 = 349 m/ = 73 = = =? We know that = 73 = 4 73 K So temperature will be (4 73) 73 = 89 c. 4. he ariation of temperature i gien by ( ) = + x () d We know that dt = dx du d = dx t = [ ( ) / d)x = = = 0 / ] 73 d d d d [ x] 0 = d 73 Putting the gien alue we get = = 96 m x d 6.

3 Chapter 6 5. We know that = K / Where K = bulk modulu of elaticity K = = (330) 800 N/m F / We know K = / = Pr eure K So, = 0.5 cm 3 6. We know that, ulk modulu = p P0 ( / ) S Where P 0 = preure amplitude P 0 = S 0 = diplacement amplitude S 0 = m = (5.5) 0 6 m m 5 0 = N/m. 7. a) Here gien air = m/., Power = E/t = 0 W f =,000 Hz, =. kg/m 3 So, intenity I = E/t. 0 0 = 44 mw/m (becaue r = 6m) 4r 4 6 b) We know that I = = 0 P air P 3 0 air = 6.0 N/m. 0 c) We know that I = S S 0 = I air Putting the alue we get S g =. 0 6 m. 8. Here I = W /m ; I =? r = 5.0 m, r = 5 m. We know that I r I r = I r I = = 65 5 Ir r = W/m. I 9. We know that = 0 log 0 I 0 I I = 0 log, = 0log I I I / I 0 = I I r r o ( /0) 0 ( /0) I /I o = o ( ) = 40 0 = 0 d. 0 where S 0 = diplacement amplitude 6.3

4 0. We know that, = 0 log 0 J/I 0 ccording to the quetion = 0 log 0 (I/I 0 ) = 0 log (I/I) = = 3 d.. If ound leel = 0 d, then I = intenity = W/m Gien that, audio output = W Let the cloet ditance be x. So, intenity = ( / 4x ) = x = (/) x = 0.4 m = 40 cm.. = 50 d, = 60 d I = 0 7 W/m, I = 0 6 W/m (becaue = 0 log 0 (I/I 0 ), where I 0 = 0 W/m ) gain, I /I = (p /p ) =(0 6 /0 7 ) = 0 (where p = preure amplitude). (p / p ) = Let the intenity of each tudent be I. ccording to the quetion 50 I = 0 log ; = 0log 0 I I = 0 log 0log 0 I 0 00 I I I I0 00 I = 0log 0log I So, = = 53 d. 4. Ditance between tow maximum to a minimum i gien by, /4 =.50 cm = 0 cm = 0 m We know, = nx n = = 0 Hz = 3.4 khz a) ccording to the data /4 = 6.5 mm = 66 mm = =3 m 330 n = = 5 khz b) I minimum = K( ) = I = I maximum = K( + ) = 9 + = 3 3 So, / = / 4 So, the ratio amplitude i. 6. he path difference of the two ound wae i gien by L = = 0.4 m 30 he waelength of either wae = = (m/) ( n ) For detructie interference L = where n i an integer. n 30 or 0.4 m = = n = 30 n 800 Hz = (n + ) 400 Hz 0.4 hu the frequency within the pecified range which caue detructie interference are 00 Hz, 000 Hz, 800 Hz, 3600 Hz and 4400 Hz. Chapter 6 6.4

5 7. ccording to the gien data = 336 m/, /4 = ditance between maximum and minimum intenity = (0 cm) = 80 cm 336 n = frequency = = 40 Hz Here gien = d/ Initial path difference i gien by = d d d If it i now hifted a ditance x then path difference will be d d d = ( d x) d d 4 4 d ( d x) 69d d 64 d x.54 d x =.54 d.44 d = 0.3 d. 9. hown in the figure the path difference.4 = x = (3.) (.4) 3. gain, the waelength of the either ound wae = We know, detructie interference will be occur ( n ) If x = (n ) 30 (3.) (.4) (3.) 30 Soling we get (n )400 = 00(n ) where n =,, 3, 49. (audible region) 30. ccording to the data = 0 cm, S S = 0 cm, D = 0 cm Let the detector i hifted to left for a ditance x for hearing the minimum ound. So path difference I = C 3. = ( 0) (0 x) (0) (0 x) So the minimum ditance hearing for minimum (n ) 0 = = 0 cm ( 0) (0 x) (0) (0 x) S O m m D Q R P y X = 0 oling we get x =.0 cm. S O S D P Q X D 0cm S d ( d / ) d 0cm x d Chapter 6 =x/4 x ( 3.) (.4) 0cm C S S S Gien, F = 600 Hz, and = 330 m/ = /f = 330/600 = 0.55 mm 6.5

6 Chapter 6 Let OP = D, PQ = y = y/r () Now path difference i gien by, x = S Q S Q = yd/d Where d = m [he proof of x = yd/d i dicued in interference of light wae] a) For minimum intenity, x = (n + )(/) yd/d = / [for minimum y, x = /] y/d = = / = 0.55 / 4 = rad = (57.) = 7.9 b) For minimum intenity, x = n(/) yd/d = y/d = = /D = 0.55/ = 0.75 rad = 6 c) For more maxima, yd/d =, 3, 4, y/d = = 3, 64, 8 ut ince, the maximum alue of can be 90, he will hear two more maximum i.e. at 3 and S S S 3 P 0 ecaue the 3 ource hae equal intenity, amplitude are equal 0 P So, = = 3 hown in the figure, amplitude of the reultant = 0 (ector method) So, the reultant, intenity at i zero he two ource of ound S and S ibrate at ame phae and frequency. P Reultant intenity at P = I 0 a) Let the amplitude of the wae at S and S be r. When = 45, path difference = S P S P = 0 (becaue S P = S P) So, when ource i witched off, intenity of ound at P i I 0 /4. b) When = 60, path difference i alo 0. S S Similarly it can be proed that, the intenity at P i I 0 / 4 when one i witched off. 34. If = m/, I = 0 cm = 0 0 m Fundamental frequency = = 850 Hz 0 0 We know firt oer tone = (for open pipe) = 750 Hz 0 0 Second oer tone = 3 (/) = = 500 Hz. 35. ccording to the quetion = m/, n = 500 Hz We know that /4I (for cloed pipe) I = m = 7 cm Here gien ditance between two node i = 4. 0 cm, = 4.0 = 8 cm We know that = n 38 = = 4. Hz = m/ Ditance between two node or antinode /4 = 5 cm = 00 cm = m n = / = Hz. 38. Here gien that = 50 cm, = m/ it i an open organ pipe, the fundamental frequency f = (/) = = Hz

7 So, the harmonie are f 3 = 3 = 00 Hz f 5 = 5 = 700, f 6 = 6 = 040 Hz o, the poible frequencie are between 000 Hz and 000 Hz are 00, 360, Here gien I = 0.67 m, l = 0. m, f = 400 Hz We know that = (l l ) = (6 0) = 84 cm = 0.84 m. So, = n = = 336 m/ We know from aboe that, l + d = /4 d = /4 l = 0 = cm. 40. ccording to the quetion 3 f firt oertone of a cloed organ pipe P = 3/4l = 4 30 f fundamental frequency of a open organ pipe P = l Here gien l l = 0 cm 6.7 Chapter 6 length of the pipe P will be 0 cm. 4. Length of the wire =.0 m For fundamental frequency / = l = l = = m Here gien n = 3.8 km/ = 3800 m/ We know = n n = 3800 / =.9 kh. So tanding frequency between 0 Hz and 0 khz which will be heard are = n.9 khz where n = 0,,, 3, Let the length will be l. Here gien that = m/ and n = 0 Hz Here / = l = l 34 We know = n l = 8. 5 cm (for maximum waelength, the frequency i minimum). n a) Here gien l = 5 cm = 5 0 m, = m/ n = = 3.4 KHz l 5 0 b) If the fundamental frequency = 3.4 KHz then the highet harmonic in the audible range (0 Hz 0 KHz) 0000 = = 5.8 = 5 (integral multiple of 3.4 KHz) he reonance column apparatu i equialent to a cloed organ pipe. Here l = 80 cm = 0 0 m ; = 30 m/ 30 n 0 = /4l = = 00 Hz So the frequency of the other harmonic are odd multiple of n 0 = (n + ) 00 Hz ccording to the quetion, the harmonic hould be between 0 Hz and KHz. 45. Let the length of the reonating column will be = Here = 30 m/ ( n ) n hen the two ucceie reonance frequencie are and 4 l 4 l Here gien ( n ) = 59 ; = 4 l n = l (n ) n = = 548 cm = 5 cm. 4 l 4 l

8 46. Let, the piton reonate at length l and l Here, l = 3 cm; =?, n = 5 Hz Now 5 = / = = 38 m/. 47. Let the length of the longer tube be L and maller will be L ccording to the data 440 = () (firt oer tone) 4 and 440 = L L () (fundamental) oling equation we get L = 56.3 cm and L = 8.8 cm. 48. Let n 0 = frequency of the turning fork, = tenion of the tring L = 40 cm = 0.4 m, m = 4g = kg So, m = Ma/Unit length = 0 kg/m n 0 = l m So, nd harmonic n 0 =. ( / l) / m it i union with fundamental frequency of ibration in the air column n 0 = = 85 Hz 4 85 = = 85 (0.4) 0 =.6 Newton Gien, m = 0 g = kg, l = 30 cm = 0.3 m Let the tenion in the tring will be = = ma / unit length = kg he fundamental frequency n 0 = l () he fundamental frequency of cloed pipe n 0 = (/4l) = 70 Hz () ccording equation () () we get 70 = 30 0 = 347 Newton. 50. We know that f ccording to the quetion f + f f f f + / t f +... (neglecting other term) f f (/ ). f 5. We know that the frequency = f, = temperature f So f f f = 93 f = Chapter 6 I I (I -I )

9 5. rod =?, air = m/, L r = 5 0, d = 5 0 metre r L r r = a D a = 0 m/. 53. a) Here gien, L r =.0/ = 0.5 m, d a = 6.5 cm = m Kundt tube apparatu i a cloed organ pipe, it fundamental frequency r n = r = = 500 m/. 4L b) r L r a = a d a r = 338 m/. Chapter the tunning fork produce beat with the adjutable frequency the frequency of the tunning fork will be n = ( ) / = tuning fork produce 4 beat with a known tuning fork whoe frequency = 56 Hz So the frequency of unknown tuning fork = either 56 4 = 5 or = 60 Hz Now a the firt one i load it ma/unit length increae. So, it frequency decreae. it produce 6 beat now original frequency mut be 5 Hz. 60 Hz i not poible a on decreaing the frequency the beat decreae which i not allowed here. 56. Group I Group II Gien = 350 = 350 = 3 cm = 3. cm = 3 0 m = 3. 0 m So = frequency = 093 Hz = 350 / 3. 0 = 086 Hz So beat frequency = = 7 Hz. 57. Gien length of the cloed organ pipe, l = 40 cm = 40 0 m air = So, it frequency = = = 00 Hertz. 4 l the tuning fork produce 5 beat with the cloed pipe, it frequency mut be 95 Hz or 05 Hz. Gien that, a it i loaded it frequency decreae. So, the frequency of tuning fork = 05 Hz. 58. Here gien n = 600 = l 4 the tenion increae frequency increae It i gien that 6 beat are produce when tenion in i increae. So, n 606 = n n l (/ l) (/ l) M ( /M) ( /M) 606 = Gien that, l = 5 cm = 5 0 m =.0. y hortening the wire the frequency increae, [f = ( / l) ( /M) ] the ibrating wire produce 4 beat with 56 Hz, it frequency mut be 5 Hz or 60 Hz. It frequency mut be 5 Hz, becaue beat frequency decreae by hortening the wire. So, 5 = () 5 0 M Let length of the wire will be l, after it i lightly hortened, 6.9

10 Chapter 6 56 = l M Diiding () by () we get () 5 l l = 0.43 m So, it hould be horten by (5 4.6) = 0.39 cm. 60. Let u = elocity of ound; m = elocity of the medium; o = elocity of the oberer; a = elocity of the ource. u m o f = F m uing ign conention in Doppler effect, m = 0, u = m/, = 0 and o = 0 m (36 km/h = 0 m/) 0 ( 0) = KHz = 350/ KHz =.06 KHz f u m o = f u m [8 km/h = 5 m/] uing ign conention, 0 0 app. Frequency = 400 = 436 Hz I II 00 m/ (36km/h = 0m/) 8km/h = 5m/ a) Gien = 7 km/hour = 0 m/, = apparent frequency = 50 = 38 H b) For econd cae apparent frequency will be = 50 = 8 Hz. 0 ( 0) 63. Here gien, apparent frequency = 60 Hz So original frequency of the train i gien by = f f = Hz So, apparent frequency of the train obered by the oberer in f = f = 60 = 480 Hz Let, the bat be flying between the wall W and W. So it will liten two frequency reflecting from wall W and W So, apparent frequency, a receied by wall W = fw = f = 330/ herefore, apparent frequency receied by the bat from wall W i gien by ( 6) F of wall W = fw f Similarly the apparent frequency receied by the bat from wall W i f (34/336)f So the beat frequency heard by the bat will be = = = 370 Hz. 65. Let the frequency of the bullet will be f Gien, u = 330 m/, = 0 m/ w bat w 6.0

11 Chapter a) pparent frequency before croing = f = f = 3f b) pparent frequency after croing = f = f = 0.6 f f 0.6f So, = 0. f 3f herefore, fractional change = 0. = 0.8. he peron will receie, the ound in the direction and C making an angle with the track. Here, = tan (0.5/.4) = So the elocity of the ource will be co when heard by the oberer. So the apparent frequency receied by the man from train. 0 0 f = Hz co co 0.5km nd the apparent frequency heard but the man from train C,.km.km 0 0 f = 500 = 476 Hz. co 67. Let the elocity of the ource i = a) he beat heard by the tanding man = 4 So, frequency = = 444 Hz or 436 Hz = 400 On oling we get = 3.06 m/ = km/hour. b) he itting man will liten le no.of beat than Here gien elocity of the ource = 0 elocity of the oberer 0 = 3 m/ 33 3 So, the apparent frequency heard by the man = 56 = 58.3 Hz. 33 from the approaching tuning form = f f = [(33 3)/33] 56 = 53.7 Hz. So, beat produced by them = = 4.6 Hz. 69. ccording to the data, = 5.5 m/ for each turning fork. So, the apparent frequency heard from the tuning fork on the left, 330 f = 5 = Hz = 57.5 Hz imilarly, apparent frequency from the tunning fork on the right, 330 f = 5 = 50 Hz So, beat produced = 7.5 Hz. 70. ccording to the gien data Radiu of the circle = 00/ 0 m = (/) metre; = 5 re/ec. So the linear peed = r = 5/ =.59 () S S So, elocity of the ource =.59 m/ () hown in the figure at the poition the oberer will liten maximum and at the poition it will liten minimum frequency. 33 So, apparent frequency at = 500 = 55 Hz pparent frequency at = 500 = 485 Hz

12 7. ccording to the gien data = 90 km/hour = 5 m/ec. 0 = 5 m/ec So, apparent frequency heard by the oberer in train or oberer in = 500 = 577 Hz Here gien f = Hz pparent frequency f = Hz (greater than that alue) Let the elocity of the oberer = o Gien = 0 So o = (330 + o ) = o = m/ = 97 km/h 4 4 b) hi peed i not practically attainable ordinary car. 73. ccording to the quetion elocity of car = = 08 km/h = 30 m/ = 7 km/h = 0 m/, f = 800 Hz So, the apparent frequency heard by the car i gien by, 30m/ f = = 87 Hz a) ccording to the quetion, = 500 m/, f = 000 Hz, = 0 m/, o = 5 m/ So, the apparent frequency heard by the ubmarine, = 000 = 034 Hz 0m/ b) pparent frequency receied by ubmarine, = 034 = 068 Hz Chapter Gien that, r = 0.7 m, F = 800 Hz, u = m/ Frequency band = f f = 6 Hz Where f and f correpond to the maximum and minimum apparent frequencie (both will occur at the mean poition becaue the elocity i maximum). Now, f = f f = 8 f and f = f f S S O 5m/ 0.7 m D = Soling for we get, =.695 m/ For SHM, = r = (.695/0.7) = 0 So, = / = /5 = 0.63 ec. 76. u = 334 m/, b = 4 m/, o = 0 o, = b co = 4 (/ ) = 4 m/. u o, the apparent frequency f = f 650 u b co = 670 Hz W N S 4 45 m / co E

13 77. u = 330 m/, 0 = 6 m/ a) pparent frequency at, y = 336 m = f uin 330 = in3 6.3 Chapter 6 [becaue, = tan (40/336) = 3 ] = 680 Hz. b) t the point y = 0 the ource and litener are on a x-axi o no apparent change L in frequency i een. So, f = 660 Hz. 40 c) hown in the figure = tan (40/336) = 3 Here gien, = 330 m/ ; = in 3 = 0.6 m/ u So, F = 660 = 640 Hz. u in3 78. train or = 08 km/h = 30 m/; u = m/ S 336 a) he frequency by the paenger itting near the open window i 500 Hz, he i inide the train and doe not hair any relatie motion. b) fter the train ha paed the apparent frequency heard by a peron tanding near the track will be, 0 o f = 500 = 459 Hz 30 c) he peron inide the ource will liten the original frequency of the train. Here, gien m = 0 m/ For the peron tanding near the track u m 0 pparent frequency = 500 = 458 Hz. u ( ) 79. o find out the apparent frequency receied by the wall, a) = km/h = 0/3 = m/ o = 0, u = 330 m/ m 330 So, the apparent frequency i gien by = f = 600 = 66 Hz / 3 b) he reflected ound from the wall whitle now act a a ource whoe frequency i 66 Hz. So, u = 330 m/, = 0, o = 0/3 m/ So, the frequency by the man from the wall, / 3 f = 66 = 63 m/ Here gien, u = 330 m/, f = 600 Hz So, apparent frequency receied by the car u o f = f 600 u Hz [ o = 0 m/, = 0] 330 he reflected ound from the car act a the ource for the peron. Here, = 0 m/, o = So f = f 60 = 47 Hz hi i the frequency heard by the peron from the car. 8. a) f = 400 Hz,, u = 335 m/ (/f) = (335/400) = 0.8 m = 80 cm b) he frequency receied and reflected by the wall, u o 335 f = f 400 u [ = 54 m/ and o = 0] 30 f f m 0 6m/ 0m/

14 x = (/f) = = 0.8 m = 80 cm c) he frequency receied by the peron itting inide the car from reflected wae, f = f 400 = 467 [ = 0 and o = 5 m/] Chapter 6 d) ecaue, the difference between the original frequency and the apparent frequency from the wall i ery high ( = 37 Hz), he will not hear any beat.mm) u ( ) f = 400 Hz, u = 34 m/, f = f 400 () u (0) 34 for the reflected wae, u 0 f = 40 = f u = = = = 4 m/ f = khz, = 330 m/, u = m/ t t = 0, the ource croe P a) ime taken to reach at Q i S 330 t = = ec 330 b) he frequency heard by the litner i f = f uco ince, = 90 f = (/u) = KHz. c) fter ec, the ource i at m from P toward right. 84. t = 4000 Hz, u = m/ Let t be the time taken by the ource to reach at O. Since oberer hear the ound at the intant it croe the O, t i alo time taken to the ound to reach at P. OQ = ut and QP = t S Co = u/ elocity of the ound along QP i (u co ). u=m/ P f = 0 f f uco u f u 330 Putting the alue in the aboe equation, f = 4000 = = 408 Hz a) Gien that, f = 00 Hz, u = 70 m/, L = 00 m, = m/ From Doppler equation (a in problem no.84) (Detector) f = D f = 00 = 600 Hz. u 70 L=t b) = elocity of ound, u = elocity of ource let, t be the time taken by the ound to reach at D DO = t = L, and SO = ut ut t = L/ S u O S S O 660m/ Q P 330m 6.4

15 L L SD = SO DO u L u Putting the alue in the aboe equation, we get 0 SD = 70 = 3.6 m. 86. Gien that, r =.6 m, f = 500 Hz, u = 330 m/ a) t, elocity of the particle i gien by = rg m/ and at C, c = 5rg m/ So, maximum frequency at C, u 330 f c = f Hz. u u 330 Similarly, maximum frequency at i gien by f f (500) 494 Hz. u ( ) b) elocity at = 3rg m/ So, frequency at i gien by, u 330 f = f 500 = 490 Hz u and frequency at D i gien by, u 330 f D = f 500 u Let the ditance between the ource and the oberer i x (initially) So, time taken for the firt pule to reach the oberer i t = x/ and the econd pule tart after (where, = /) and it hould trael a ditance x a. x / a So, t = x / a t t = Putting = = /, we get u a t t = x a o, frequency heard = (becaue, f = ) u a t t t=0 S r C Chapter 6 D D C t= S x x ½ at D 6.5