13-Sound Waves

SECTION - I : STRAIGHT OBJECTIVE TYPE [k.M-I: lh/ks oLrqfu"B izdkj 13.1 A closed organ pipe has length ‘𝑙 ‘. The air in it is vibrating in 3rd overtone with maximum amplitude ‘ a‘. Find the amplitude at a distance of 𝑙 /7 from closed end of the pipe. (A) a (B) a/2 (C) Solution : a/2 (D) a/ The figure shows variation of displacement of particles in a closed organ pipe for 3rd overtone. 7 For third overtone 𝑙 = 4 or  = 4𝑙 7  or 4 = 𝑙 7 Hence the amplitude at P at a distance point [ Made 2003 ] 𝑙 from closed end is ‘a’ because there is an antinode at that 7 Alternate Because there is node at x = 0 the displacement amplitude as function of x can be written as A = a sin 2 kx = a sin  x 7 4𝑙 For third overtone 𝑙 = 4 or  = 7  A = a sin 7 2𝑙 𝑙 = a sin 7  2 = a at x = 𝑙  A = a 7 ,dcanvkxZuikbidhyEckbZ‘𝑙‘ blesaok;qvf/kdrevk;ke'a'dslkFkrhljsvf/kLojdesadfEirgksrhgSaikbidscan fljsls𝑙/7nwjhijvk;keKkrdjks\ Solution : fp=kesa,dcanvkWxZuikbiesarhljsvf/kLojddsfy;sd.kdsfoLFk iuesaifjorZudksfn[k ;kx;kgSA rhljs vf/kLojd gsrq 𝑙 = 7 or  = 4𝑙 or  = 𝑙 4 7 4 7 fodYi vr%P 𝑙 ijvk;ke]canfljsls 7 nwjhij ‘a’gS]D;ksafd;gk¡ij,dizLiancurkgSA D;ksafd ;gk¡x = 0 ij fuLiUn gS] foLFkkiu vk;ke dksx ds Qyu :i esa fuEu izdkj ls fy[k ldrs gSA 2 A = a sin kx = a sin  x rhljs vf/kLojd ds fy,𝑙 = 7 ;k = 4𝑙  A = a sin 7 2𝑙 4 𝑙 = a sin 7 7  2 = a at x = 𝑙  A = a 7 13.2 When a sound wave is reflected from a wall, the phase difference between the reflected and incident pressure wave is: (A*) 0 (B)  (C) /2 (D) /4 Sol. When a sound wave gets reflected from a rigid boundary, the particles at the boundary are unable to vibrate. Thus, a reflected wave is generated which interferes with the oncoming wave to produce zero displacement at the rigid boundary. At these points (zero displacement), the pressure variation is maximum. Thus, a reflected pressure wave has the same phase as the incident wave. 13.2 tc/ofurjax]nhokjlsijkofrZrgksrhgS]ijkofrZrovkifr nkcrjaxdse/;dykUrjgSA (A*) 0 (B)  (C) /2 (D) /4 Sol. tc/ofurjaxdksn`<+ifjlheklsifjofrZrdjizkIrdjrsgS]ifjlhekdkd.kdEiUudjusesavleFkZgSAvr%mRiUu ijkofrZrrjax]tksvkusokyhrjaxdslkFkO;frdj.kdjdsn`<+ifjlhekij'kwU;foLFk iumRiUudjrhgSAblfcUnq ij¼'kwU;foLFk iu½]nkcifjorZuvf/kdregSAvr%ijkofrZrnkcrjaxvkifrrrjax]dslekughdykj[krhgSA 13.3 A source of frequency 'f' is stationary and an observer starts moving towards it at t = 0 with constant small acceleration. Then the variation of observed frequency f ' registered by the observer with time is best represented as : (A*) (B) (C) (D) Sol. After a time t, velocity of observer V0 = at  V  V0   V  at   f =   f . =   f , which is a straight line graph of positive slope. 0  V  s  V  s 13.3 'f' vko`frdkL=k srfLFkjgSrFk t=0ij,disz{kdmldhvksjNksVsvpjRoj.klsxfrizkjEHkdjrkgSArcizs{kd }kjkle;dslkFkekihxbZizsf{krdEiUudhvko`fÙkdkslclslghiznf'kZrdjrkgS (A*) (B) (C) (D) Sol. t le; ckn] izs{kd osx V0 = at  V  V0   V  at   f =  0   f . =  V  s  V  fs , tks /kukRed 3  > 6  f = V 330  < 6 = 55.  If f < 55 Hz, no minimum will occur. 13.12 /ofudsnksdyklEcU/kL=k srS1oS2 izkjEHkesadksbZdykUrjughaj[krsgSaA/ofudkosx330m/sgSA;fnS2ls xqtjusokyhjs[k rFk S1oS2dkstksM+usokyhjs[k dsyEcor~js[k ijdksbZfufEu"BughcurkgS]nksuksaL=k srdh vko`fÙk gS & (A*) 50 Hz (B) 60 Hz (C) 70 Hz (D) 80 Hz Sol. fufEu"Bdsfy, x = (2n – 1)  2 vf/kdrelEHkodykUrj=L=kksrksadse/;nwjh=3m dksbZfufEu"Bdsfy,  2 > 3  > 6  f = V 330  < 6 = 55.  ;fnf < 55Hz, dksbZfufEu"B ughacurk gS 13.13 Under similar conditions of temperature and pressure, In which of the following gases the velocity of sound will be largest. rki o nkc dh leku ifjfLFkfr;ksa esa buesa ls dkSulh xSl /ofu dk mPpre osx j[krh gSA (A*) H2 (B) N2 (C) He (D) CO2 Sol. The speed of sound in air is v =  M of H2 is greatest in the given gases, hence speed of sound in H2 shall be maximum. gy. ok;q esa /ofu dh pky v =  H2 dk M  nh x;h xSlksa esa mPpre gS] vr% H2 esa /ofu dh pky vf/kdre gS 13.14 When beats are produced by two progressive waves of nearly the same frequency, which one of the following is correct? (A) The particles vibrate simple harmonically, with the frequency equal to the difference in the component frequencies. (B*) The amplitude of vibration at any point changes simple harmonically with a frequency equal to the difference in the frequencies of the two waves. (C) The frequency of beats depends upon the position, where the observer is (D) The frequency of beats changes as the time progresses Sol. As y = Absin(2navt) where Ab = 2Acos(2nAt) where nA = n1  n2 2 13.14 tcyxHkxlekuvko`frdhnksizxkehrajxksa}kjkfoLiUnmRiUugksrksgS]fuEuesalsdkSulk,dlghgS? (A) d.k]?kVdvko`fr;ksadsvUrjdslekuvko`fÙk]lsljyvko`rZdEiUudjrkgSA (B*)fdlhfcUnqijdEiUudkvk;kenksrjaxkasdhvko`fÙk;ksadsvUrjdhvko`frls]ljyvko`Ùkxfrds:iesaifjofrZr gksrkgS (C) foLiUnvko`fÙk]isz{kddhfLFkfrijfuHkZjdjrhgS (D) le;c<+usdslkFkfoLiUnvko`fÙkifjofrZrgksrhgS Sol. tSlk fdy = Absin(2navt) tgkaAb = 2Acos(2nAt) tgkanA = n1  n2 2 13.15 S1 and S2 are two coherent sources of sound separated by distance 100.25 , where  is the wave length of sound. S1 leads S2 in phase by /2. A and B are two points on the line joining S and S2 as shown in figure. The ratio of amplitudes of source S1 and S2 are in ratio 1:2. The ratio of intensity at A  A  to that of B    is B  1 (A)  (B*) 9 (C) 0 (D) 9   Sol. For interference at A : S2 is behind of S1 by a distance of 100 + 4 .(equal to phase difference 2 ).  Further S2 lags S1 by 2 . Hence the waves from S1 and S2 interfere at A with a phase difference of 200.5  + 0.5 = 201   Hence the net amplitude at A is 2a – a = a For interference at B : S2  Further S2 lags S1 by 2 . is ahead of S1 by a distance of 100 +  4 .(equal to phase difference  2 ). Hence waves from S1 and S2 interfere at B with a phase difference of 200.5  – 0.5 = 200  0. Hence the net amplitude at A is 2a + a = 3a  A   a 2 1 Hence    =  3a  = 9  B    13.15 100.25  nwjh }kjk i`FkDd`r nks dyk lEc) /ofu ds L=kksr S1 o S2 gS] tgka  / o f u d h r j a x n S / ; Z g S A S1,S2 ls/2dykvkxsgSAS1 oS2 dkstksM+usokyhjs[k ijnksfcUnqAoBfp=k uqlkjgSAL=k srS1 oS2 dsvk;keksa  A  dk vuqikr 1:2 gSA A o B dh rhozrkvksa ds vuqikr    gSA B  1 (A)  (B*) 9  (C) 0 (D) 9   Sol. A ij O;frdj.k ds fy, : S2, S1 ls100 + 4 nwjh }kjk ihNs ( 2 dykUrj ds cjkcj) gS ,oa S2, S1 ls 2 vkxs gSA vr%S1 oS2 ls rjaxksa ds A ij O;frdj.k ds fy, dykUrj 200.5 + 0.5 = 201 gSA vr% A ij ifj.kkeh vk;ke 2a– a=a gSA    B ij O;frdj.k ds fy,: S2 , S1 ls 100+ 4 dh nwjh }kjk vkxs( 2 dykUrj ds cjkcj) gSA S2 , S1 ls 2 vkxs gSA vr% S1 o S2 ls rjaxksa ds B ij O;frdj.k ds fy, dykUrj200.5 – 0.5 = 200 0 gSA vr% B ij ifj.kkeh vk;ke2a + a= 3a gSA  A   a 2 1 vr%    =   = 3a 9  B    13.16 There is a set of four tuning forks, one with the lowest frequency vibrating at 550 Hz. By using any two tuning forks at a time, the following beat frequencies are heard: 1, 2, 3, 5, 7, 8. The possible frequencies of the other three forks are: pkjLofj=k sadk,dlewggSftuesa,d550gV~ZtdhU;wurevko`fÙkdslkFkdEiUudjjgkgSA,dlkFk,dle; ijnksLofj=k sadksmi;ksxesaysdj1,2,3,5,7,8 foLianvko`fÙk;k¡lquhtkrhgSArksvU;rhuLofj=k sadhlaHk for vko`fÙk;k¡ gksaxh & (A) 552, 553, 560 (B) 557, 558, 560 (C) 552, 553, 558 (D*) 551, 553, 558 Sol. To get beat frequency 1, 2, 3, 5, 7, 8, it is possible when other three tuning fork have frequencies 551, 553, 558, etc. 1, 2, 3, 5, 7, 8, foLianvko`frizkIrdjusdsfy;s;glEHkogSfdvU;rhuksLofj=kdhvko`fr551,553, 558bR;kfn gksA 13.17 A 100 m long rod of density 10.0 x 104 kg/m3 and having Young’s modulus Y = 1011 Pa, is clamped at one end. It is hammered at the other free end. The longitudinal pulse goes to right end, gets reflected and again returns to the left end. How much time, the pulse take to go back to initial point. ,d fljsij ca/kh gqbZ] ,d100 m yEch NM+ dk?kuRo10.0x 104 kg/m3 o;ax izR;kLFkrk xq.kkadY=1011 PagSA blsvU;eqDrfljsijgFk SM+slsBksdrsgSAvuqns/;ZLiUnnka;safljsaijtkrsgSa]vkSjijkofrZrgksdjokilck;safljs ij vkrs gSaA izkjfEHkd fcUnq ij LiUn dks okil vkus esa fdruk le; yxsxkA (A) 0.1 sec. (B*) 0.2 sec. (C) 0.3 sec. (D) 2 sec. Sol. Vs = = = 103 m/sec. t = 2𝑙 V 2 x 100 = 1000 = 0.2 sec Ans. is (B) 13.18 For a sound wave travelling towards +x direction, sinusoidal longitudinal displacement  at a certain time is given as a function of x. If Bulk modulus of air is B = 5 x 105 N/m2, the variation of pressure excess will be : +x fn'kk dh vksj lapfjr /ofurjax ds fy,] T;koØh; vuqnS/;Z foLFkkiu ,d fuf'pr le; ijx dsQyu ds:i esafn;kx;kgSA;fnok;qdkvk;ruizR;kLFkrkxq.k adB=5x 105 N/m2,gS]nkcvkf/kD;esaifjorZugksxk& 10-4m 0 10-4m x(m) Pex +2 x 102 Pa (A) 0 -2 x 102 Pa 0.1 0.2 x(m) Pex +5 x 102 Pa (B) 0 -5 x 102 Pa 0.1 0.2 x(m) Pex +2 x 102 Pa 0 0.2 x(m) 0.1 -2 x 102 Pa Pex +5 x 102 Pa (D*) 0 0.2 x(m) -5 x 102 Pa Sol.  = A sin (kx – t) d Pex = – B dx = – BAk cos (kx – t) amplitude of Pex (Pex dk vk;ke) = BAk  2  = (5 x 105) (10–4)  0.2    = 5 x 102 Pa So correct ans is (D) blfy, lgh mÙkj gS(D) 13.19 Figure shows a stretched string of length L and pipes of length L, 2L, L/2 and L/2 in options (A), (B), (C) and (D) respectively. The string’s tension is adjusted until the speed of waves on the string equals the speed of sound waves in the air. The fundamental mode of oscillation is then set up on the string. In which pipe will the sound produced by the string cause resonance ? fp=kesaLyEckbZdhruhgqbZMksjhrFkkfodYi(A), (B), (C)o(D)esaØe'k%L,2L, L/2oL/2 yEckbZdsikbin'kkZ;s x;sgSAMksjhesarukodksrcrdlek;ksftrdjrsgStcrdMksjhesarjaxdhpky]ok;qesa/ofudhpkydscjkcj ughagkstkrhgSAnksyudhewyfo/k ]rcMksjhesaO;ofLFkrgkstkrhgSAdkSulsikbZiesa]MksjhesamRiUu/ofuvuqukn mRiUu djrh gS ? (A) Sol. (B) (B*) 2L (C) (D) Fundamental frequency of wire (f (A) wire) = v 2𝑙 f = v , 4𝑙 3v , 4𝑙 5v 4𝑙 cannot match with f wire (B) f = v v 2(2𝑙) 2v 2v , 2(2𝑙) , 3v 2(2𝑙) its second harmonic 2v 2(2𝑙) matches with fwire. (C) , f = 2(𝑙/2) , v 2(𝑙 / 2) 3v cannot match with fwire f = 4(𝑙 / 2) , 4(𝑙 / 2) cannot match with fwire Sol. (B) rkj dh ewy vko`fr(f ) = v (A) wire 2𝑙 f = v , 4𝑙 3v , 4𝑙 5v 4𝑙 fwire lsughfeyrhgS (B) f = v v 2(2𝑙) 2v 2v , 2(2𝑙) , 3v 2(2𝑙) 2v bldknwljk lauknh 2(2𝑙) fwire lsfeyrk gS (C) , f = 2(𝑙/2) , v 2(𝑙 / 2) 3v fwire lsughafeyrhgSA f = 4(𝑙 / 2) , 4(𝑙 / 2) ..... fwire ls ugha feyrh gSA 13.20 Earthquakes generate sound waves inside Earth. Unlike a gas, Earth can experience both transverse (S) and longitudinal (P) sound waves. Typically, the speed of S waves is about 4 km/s. A seismograph records P and S waves from an earthquake. The first P waves arrive 3.0 min before the first S wave (figure). Assuming the waves travel in a straight line, how far away does the earthquake occur ? (Yearth = 12.8 × 1010 pa, earth = 2000 kg/m3) 0 1 2 3 4 5 6 Time (min) (A) 1900 km (B*) 1440 km (C) 1800 km (D) 1200 km Sol. (B) vs = 4 km/sec vp = = = 8000 m/sec. = 8 km/sec 𝑙  𝑙 = 3 min = 3 × 60 sec. vs vp 𝑙  𝑙 = 3 × 60 𝑙 = 1440 km 4 8 13.20 HkwdEii`FohdsvUnj/ofurjaxmRiUudjrkgSA,dxSldsfoijhr]i`FohvuqizLFk(S)ovuqnS/;Z(P)nksukasizdkj dh/ofurjaxvuqHkodjldrhgSAfo'ks"kr%]SrjaxksadhpkyyxHkx4km/sgSA,dflLeksxzkQ¼HkwdEiys[kh½ HkwdEilsizkIrPoSrjaxksadksvfHkysf[krdjrkgSAizFkeP rjaxsaizFkeSrjaxksals3 feuVigysvkrhgS]¼fp=k½A ekuk fd rjax ,d ljy js[kk es lapfjr gS] fdruh nwjh ij HkwdEi vk;k gksxkA (Yearth = 12.8 × 1010 pa,  = 2000 kg/m3) 0 1 2 3 4 5 6 l e; ¼feuV½ (A) 1900 km (B*) 1440 km (C) 1800 km (D) 1200 km Sol. (B) vs = 4 km/sec vp = = = 8000 m/sec. = 8 km/sec 𝑙  𝑙 = 3 min = 3 × 60 sec. vs vp 𝑙  𝑙 = 3 × 60 𝑙 = 1440 km 4 8 13.21 If the source is moving towards right, wave front of sound waves get modified to – (A) (B*) (C) (D) None of these Sol. (B) Towards right wavelength gets compressed, towards left, wavelength gets expended 13.21 ;fn L=kksr] nk¡;ha vksj xfr'khy gS] /ofu rjax dk izkIr :ikUrfjr rjaxzkx gS – (A) (B*) (C) (D) buesa ls dksbZ ugha Sol. (B) nkW;hvksjrjaxnS/;ZlEihfM+rizkIrgksrkgS]ckW;hvksjrjaxnS/;ZQSykgqvkizkIrgksrkgSA 13.22 Equation of a stationary and a travelling waves are as follows y = a sin kx cos t and y = a sin (t –  kx). The phase difference between two points x1 = 3k 1 3 and x2 = 2k is  in the standing wave (y1) and is  in travelling wave (y ) then ratio 2 is Modified (A) 1 (B) 5/6 (C) 3/4 (D*) 6/7 Sol. (D) x1 and x2 are in successive loops of std. waves. so,  =   3    7  1  6 and  = K(x) = K  2K  = 3K  6 = 2 7 13.22 ,d vizxkeh o izxkeh rajxks dh lehdj.k fuEuy1 = asin kx cost oy2 = a sin(t – kx) gSA nks fcUnqvksa  3 1 x1 = 3k ox2 = 2k ds e/; dykUrj vizxkeh rjax(y1) es1 gS o izxkeh rjax(y2) esa 2 gS rc 2 gS &Modified (A) 1 (B) 5/6 (C) 3/4 (D*) 6/7 Sol. (D) x1 ox2 vizxkeh rjax ds Øekxr ywi gS rkfd, 1 =   3    7  1  6 rFk 2 = K(x) = K  2K  = 3K  6 = 2 7 13.23 In the resonance tube experiment, the first resonance is heard when length of air column is 𝑙1 and second resonance is heard when length of air column is 𝑙2. What should be the minimum length of the tube so that third resonance can also be heard. (A*) 2𝑙2 – 𝑙1 (B) 2 𝑙1 (C) 5 𝑙1 (D) 7 𝑙1 Sol. (A) 𝑙1 +  = v 4f0 𝑙2 +  = 3v 4f0 𝑙3 +  = 5v 4f0 Solving get 𝑙3 = 2𝑙2 – 𝑙1 13.23 vuquknuyhiz;ksxesa]izFkevuquknok;qLrEHkdh𝑙1 yEckbZijlqukbZnsrkgSrFk nwljkvuquknok;qLrEHkdh𝑙2 yEckbZijlqukbZnsrkgSAuyhdhU;wureyEckbZfdruhgksuhpkfg,rkfdrhljkvuquknHkhlquldsA (A*) 2𝑙2 – 𝑙1 (B) 2 𝑙1 (C) 5 𝑙1 (D) 7 𝑙1 Sol. (A) 𝑙1 +  = v 4f0 𝑙2 +  = 3v 4f0 𝑙3 +  = 5v 4f0 gy djus ij izkIr gksrk gS𝑙3 = 2𝑙2 – 𝑙1 13.24 Radio waves coming at   to vertical are received by a radar after reflection from a nearby water surface & directly. What can be height of antenna from water surface so that it records a maximum intensity (a maxima). (wavelength = ) (Assume phase changes by  after reflection) (A)  2 cos  (B)  2 sin  (C)  4 sin  (D*)  4 cos  Sol. (D) radio wave are electromagnetic wave. So it get extra phase after reflection C h sec   h  A (h sec ) cos 2 B total path difference = AB + BC + /2 =  for maxima h sec  cos 2 + h sec  = /2 h sec  (2cos2 ) = /2  h = 2 cos  13.24 Å/ok/kZjlsdks.kijvkrhgqbZjsfM;ksrjaxksdkslehifLFkrikuhlsijkofrZrgksdjolh/ksgh,djsM+kj}kjkxzg.k fd;ktkrkgSA,.Vhusdhikuh dhlrglsÅ¡pkbZD;kgksldrhgSrkfdmPprerhozrk(,dmfPp"B)ntZdjldsA (rjaxnS/;Z = )(ekuk ijkorZu ds ckn dykUrj  gks tkrk gS (A)  2 cos  (B)  2 sin  (C)  4 sin  (D*)  4 cos  Sol. (D) jsfM;ks rjaxs fo|qr pqEcdh; rjaxs gS blfy, ;gk¡ ijkorZu ds ckn vfrfjDr dykUrj izkIr dj ysrs gSA C h sec   h  A (h sec ) cos 2 B dqy iFkkUrj= AB + BC + /2 =  mfPp"B ds fy, h sec  cos 2 + h sec  = /2 h sec  (2cos2 ) = /2  h = 2 cos  13.25 Microwaves from a transmitter are directed normally towards a plane reflector. A detector moves along the normal to the reflector. Between positions of 14 successive maxima, the detector travels a distance 0.14 m. If the velocity of light is 3 × 108 m/s, find the frequency of the transmitter - (A*) 1.5 × 1010 Hz (B) 1010 Hz (C) 3 × 1010 Hz (D) 6 × 1010 H Sol. If detector moves x distance , distance from direct sound increases by x and distance from reflected sound decreases by x so path difference created = 2x 2(0.14) = 14 = 14 c/f 14  3  108 f = 0.14  2 = 1.5 × 1010 Hz. 13.25 ,dVªkalehVjls]lw{erjaxslh/ksgh,dleryijkorZddhvksjvfHkyEcor~HksthtkrhgSA,dlalwpdijkorZd dsvfHkyEcdsvuqfn'kxfrdjrkgSA14mÙkjksrjmfPp"BfLFkfr;ksadse/;]lalwpd0.14m nwjhr;djrkgSA;fn izdk'k dk osx 3 × 108 m/s gS] VªkalehVj dh vko`fÙk Kkr dhft, - (A*) 1.5 × 1010 Hz (B) 1010 Hz (C) 3 × 1010 Hz (D) 6 × 1010 Hz Sol. ;fnlalwpdx nwjhxfrdjrkgS] lh/ks/ofulsnwjhx }kjkc<+tkrhgSrFk ijkorZu/ofulsnwjhx }kjk?kVtkrhgSblfy,mRiUuiFk Urj=2x 2(0.14) = 14 = 14 c/f 14  3  108 f = 0.14  2 = 1.5 × 1010 Hz. 13.26 A man standing in front of a mountain at a certain distance beats a drum at regular intervals. The drumming rate is gradually increased and he finds that the echo is not heard distinctly when the rate becomes 40 per minute. He then moves nearer to the mountain by 90 meters and finds that the echo is again not heard when the drumming rate becomes 60 per minute (a) The distance between the mountain and the initial position of the man is : (A) 330 m (B) 300 m (C) 240 m (D*) 270 m (b) the velocity of sound is (A) 330 m/s (B*) 360 m/s (C) 300 m/s (D) 270 m/s 13.26 ,d igkM+ ds lkeus [kM+k ,d vkneh ,d fuf'pr nwjh ij yxkrkj vUrjky ij ,d Mªe ij foLian mRiUu djrk gSA Mªe dks ctkus dh nj yxkrkj c<+rh tkrh gS rFkk og Kkr djrk gS fd izfr /ofu Li"V lqukbZ ugha nsrh gS tc nj 40 izfr feuV gks tkrh gSA og rc igkM+ dh vksj 90m ikl vkrk gS rFkk izfr/ofu dks nqckjk lqurk gS tc Mªe ctkus dh nj 60 izfr feuV gks tkrh gS (a) igkM+ovknehdhizkjfEHkdfLFkfrdse/;nwjhgS: (A) 330 m (B) 300 m (C) 240 m (D*) 270 m (b) /ofudkosxgS (A) 330 m/s (B*) 360 m/s (C) 300 m/s (D) 270 m/s Sol. Drumming frequency = 40 cycle/min = 40 cycle/ 60 sec 1 60 sec 3 Drumming time period = f = 40 cycle = 2 sec/cycle (time duration between consecutive drumming) During this time interval, if sound goes to mountain and comes back then echo will not be heard distinctly. 3  2𝑙 ...... (1) 2 v Now if he moves 90 m. This situation arises at t = 60 cycle/min, T = 1 = 1 sec/cycle f so for this case 1 = 2(𝑙  90) v ......(2) Solving equation (1) and (2) set 𝑙 = 270 m v = 360 m/sec. Sol. Mªe ctkus ds vko`fr= 40 pØ/feuV= 40 pØ/ 60 lSd.M Mªe ctkus dk vkorZdky 1 60 sec 3 = lSd.M/pØ = f = 40 cycle 2 (Øekxr Mªe ctkus ds e/; dk le;) bl le;vUrjkydsnkSjku];fn/ofu igkM+rdtkrhgS ookil vkrhgSrcizfr/ofuLi"VlqukbZughnsrh gSA 3  2𝑙 ...... (1) 2 v vc ;fn og90 m xfr djrk gSA ;g fLFkfrt = 60 pØ/feuVij vkrh gS 1 lSd.M/pØ vr% bl fLFkfr ds fy,1 = 2(𝑙  90) lehdj.k(1) o(2) dks gy djus ij set 𝑙 = 270 m v = 360 m/sec. ......(2) , T = f = 1 13.27 Figure shown is a graph, at a certain time t, of the displacement function S(x,t) of three sound waves 1,2 and 3 as marked on the curves that travel along x–axis through air. If P1,P2 and P3 represent their pressure amplitudes respectively, then correct relation between them is : gokesax-v{kdsvuqfn'kxfrdjjghrhu/ofurjaxksa¼ftudksaoØksaij1,2,3}kjkn'k Z;kx;kgS½dkfdlhle; t ij] foLFkkiu Qyu S(x, t) oØ fp=k esa iznf'kZr gSA vxj P1, P2 rFkk P3 Øe'k% b u d s n k c v k ; k e g S a r k s b u d s e/; lgh lEcU/k gS & (A) P1 > P2 > P3 (B*) P3 > P2 > P1 (C) P1 = P2 = P3 (D) P2 > P3 > P1 Sol. (B) P0 = B.K.S0 =  2     0  P  1   Thus, pressure amplitude is highest for minimum wavelength, other parameters B and S0 being same for all. From given graphs. < 2 <  . Hence (B). blizdkjU;wurerjaxnS/;Zdsfy,nkcvk;kevf/kdregS]lHkhdsfy,nwljsizkpyBoS0lekugSAfn;sx;sxzkQls & 3 < 2 < 1 . vr% (B). 13.28 In Quincke's tube a detector detects minimum intensity. Now one of the tube is displaced by 5 cm. During displacement detector detects maximum intensity 10 times, then finally a minimum intensity (when displacement is complete). The wavelength of sound is: Dohads uyh eas lalqpd U;wure rho`rk izsf{kr djrk gSA vc ,d uyh dks 5 cm foLFkkfir djrs gS ftlds nkSjku lalqpd 10 ckj vf/kdre rhozrk izsf{kr djrk gSa rFkk vrar% U;wure rhozrk izsf{kr djrk gSA (tc foLFkkiu iw.kZ gks tk, rc) /ofu dh rjaxnS/;Z gksxhA {New} (A) 10/9 cm (B*) 1 cm (C) 1/2 cm (D) 5/9 cm Sol. Path difference introduced due to displacement of tube = 2x = 10 cm due to one wavelength change maxima / minima will be attained once hence for 10 maxima’s Sol. uyh ds foLFkkiu ls mRiUuiFkkUrj = 2x =10cm ftldsdkj.k 10mfp"B curs gS ,dmfp"B dsiFkkUrj = vr% pathdifference iFkkUrj P = 10  = 10 cm so  = 1 cm. Ans. 13.29 x = x1 + x2 (where x1 = 4 cos t and x = 3 sin t) is the equation of motion of a particle along x-axis. The phase difference between x1 and x is :{New} New (A*) 37° (B) 53° (C) 90° (D) none of these x = x1 + x2 (tgk¡ x1 = 4 cos t rFkk x2 = 3 sin t) x-v{k esa d.k ds lHkh gSA x1 rFkk x ds chp dykUrj gS : (A*) 37° (B) 53° (C) 90° (D) buesa ls dksbZ ugh Sol. x = x sin (t + ) = x sin t cos  + x cos t sin . Comparing with given equation. 0 0 0 Thus x0 cos  = 3 and x sin  = 4 4 Dividing we get tan  = 3 or  = 53° x = 4 cost = 4sin (t + 90°)  = 90° – 53° = 37° Sol. x = x0sin (t + ) = x0sin t cos + x0 cos t sin . fn;s x;s leh0 ls rqyuk djus ij . vr% x0 cos  = 3 rFk x0sin  = 4 4 foHkkftrdjuslsgeikrsgStan= 3 or  = 53° x = 4 cost = 4sin (t + 90°)  = 90° – 53° = 37° 13.30 S1 and S2 are two coherent sources of sound of frequency 110Hz each. They have no initial phase difference. The intensity at a point P due to S1 s then the resultant intensity at P is is  and due to S2 is 4 . If the velocity of sound is 330 m/ S1rFk S2dyklEc)izR;sd110Hzvko`frdsnks/ofuL=k srgSAmuesaizkjfEHkddykUrjughagSAfcUnqPijS1dsdkj.k rhozrk0rFk S2dsdkj.krhozrk40gSA;fn/ofudkosx330m/sgSrksfcUnqPijifj.k ehrhozrkgS& (A) I0 (B) 9I0 (C*) 3I0 (D) 8I0 330 Sol. The wavelength of sound source = 110 = 3 metre. The phase difference betwen interfering waves at P is =  = 2 2 (S P – S P) = 2 (5 – 4) =  2 1 3 3  Resultant intensity at P = I + 4I0 + 2 cos 2 3 = 3 I0 330 /ofuL=kksr dhrjaxnS/;Z = 110 =3metre. fcUnqPijv/;kjksfirgksusokyhrjaxksadse/;dykUrj =  = 2 2 (S P – S P) = 2 (5 – 4) =  2 1 3 3  fcUnq P ij ifj.kkeh rhozrk = I + 4I + 2 cos 2 = 3I 0 0 3 0 13.31 A conveyor belt moves to the right with speed v = 300 m/min. A pieman puts pies on the belt at a rate of 20 per minute while walking with speed 30 m/min towards a receiver at the other end. The frequency with which they are received by the stationary receiver is: ,dlkeku ƒ and constant rFkk fu;r 1  v – vs  Whenitrecedsfrom theobserver. tc isz{kd ls nwj tk jgk gS  v  ƒ = ƒ   < ƒ and constant. rFkk fu;r 2  v  vs  SECTION - II : MULTIPLE CORRECT ANSWER TYPE [k.M- II: cgqy lgh fodYi izdkj 13.35 In a resonance tube experiment, a closed organ pipe of length 120 cm resonates when tuned with a tuning fork of frequency 340 Hz. If water is poured in the pipe then (given vair = 340 m/sec.) : vuqukn uyhdsiz;ksxesatc,d120lseh0yEchvkWxZuikbi]340gVZ~t dhvko`frdsLofj=kdslkFkdfEir djkusijvuqukfnrgksrhgSA;fnikbiesaikuhHkjfn;ktk;srks& ¼fn;k gSV =340eh0/lS0½ (A*) minimum length of water column to have the resonance is 45 cm. (B*) the distance between two successive nodes is 50 cm. (C*) the maximum length of water column to create the resonance is 95 cm. (D) none of these. (A*)vuqukfnrgksusdsfy,ikuhdsLrEHkdhU;wure~yEckbZ45lseh0gSA (B*) nks Øekxr fuLinksa ds e/; nwjh 50 lseh0 gSA (C*) vuqukfnrgksus dsfy,ikuhdsLrEHkdhvf/kdreyEckbZ95lseh0gSA (D)mi;qZDresalsdksbZughA Sol. As V =   = V = 340 340 = 1m first Resonance depth (from upper end)  1 R1 = 4 = 4 m = 25 cm  R = 3 = 4  R = 5 = 4 3 m = 75 cm 4 5 m = 125 cm 4 i.e. third resonance does not establish Now Water is poured,  Minimum length of water column to have the resonance = 45 cm  Distance between two successive nodes =  = 1 m = 50 cm 2 2 & maximum length of water column to create resonance i.e. 120 – 25 = 95 cm. Sol. As V =   = V = 340 340 = 1m izFkevuqokfnrxgjkbZ ¼Åijhfljsls½  1 R1 = 4 = 4 m = 25 cm  R = 3 = 4  R = 5 = 4 3 m = 75 cm 4 5 m = 125 cm 4 i.e. rhljk vuqukn ugh curk gS vc ikuhMkyusij]  vuqukn ds fy,ikuh LrEHk dh U;wure yEckbZ= 45cm  nks mÙkjksrj foLiUnks ds e/; nwjh=  = 1 m =50cm 2 2 rFkkvuquknmRiUu djus dhikuhdsLrEHkdh vf/kdreyEckbZ i.e. 120 – 25 = 95 cm. 13.36 Two identical stretched wires are vibrated together. They produce 8 beats per second. When tension in one wire is changed then the beat frequency is increased. If T1and T2 denote the tensions in the two wires at any instant and T1 > T2 (initially) then how the change may be performed. (A*) T1 decreased (B*) T1 increased (C*) T2 decreased (D*) T2 increased nksrkU;¼f[kps½,dlekurkj,dlkFknksfyrfd;stkrsgSrks8foLian/ls0mRiédjrsgS;fn,drkjdkruko cnyrsgSrksfoLiandhvko`fÙkc<+tkrhgS;fnfdlhle;rkjksaesarukoØe'k%T1rFk T2gS(izkjEHkesa]T1 >T2) rksrukoesaifjorZudSlsfd;ktkrkgS& (A*) T1 dks?kVkdj (B*) T1 dksc<+dkj (C*) T2 dks?kVkdj (D*) T2 dksc<+kdj Sol. (A,B,C,D) n = on increasing or decreasing (T1 & T2) significantly we can get result of higher beats. Sol. n = T1;kT2dks?kVkdj;kc<+kdjnksuksarjglsvf/kdfoLianizkIrfd;stkldrsgSA SECTION - III : ASSERTION AND REASON TYPE [k.M- III: dFku vkSj dkj.k izdkj 13.37 Statement 1 : Doppler formula for sound wave is symmetric with respect to the speed of source and speed of observer Statement 2 : Motion of source with respect to stationary observer is not equivalent to the motion of an observer with respect to a stationary source. oDrO;1:/ofudsfy,MkIyjlw=k]lzksrdhpkyrFk izs{kddhpkydslkis{k,dlekugksrkgSA oDrO;2:fLFkjizs{kddslkis{klzksrdhxfr]fLFkjlzksrdslkis{kizs{kddhxfrdslekuughagksrhgSA (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C) Statement-1 is True, Statement-2 is False (D*) Statement-1 is False, Statement-2 is True (A) oDrO;-1lR; gS] oDrO;-2lR; gS ;oDrO;-2,oDrO;-1 dk lgh Li"Vhdj.k gSA (B)oDrO;-1lR; gS] oDrO;-2lR; gS ; oDrO;-2,oDrO;-1 dk lgh Li"Vhdj.k ugha gSA (C) oDrO;-1lR; gS] oDrO;-2vlR; gS ; (D*) oDrO;-1vlR; gS] oDrO;-2lR; gS Sol. (D) Doppler formula for sound a wave is not symmetric w.r.t speed of source and speed of observer. /ofudsfy,MkIyjlw=k]lzksrdhpkyrFk izs{kddhpkydslkis{k,dlekuughagksrkgSA 13.38 Statement 1 : The base of Laplace correction was that exchange of heat between the region of compression and rarefaction in air is negligible. Statement 2 : Air is bad conductor of heat and velocity of sound in air is quite large. oDrO;1:ykIyklla'k s/kudkvk/k j]ek/;eesalaihMuofojyudse/;m"ekvknkuiznkugSA oDrO;2:gokm"ekdsfy,[kjkcpkydgSrFk gokesa/ofudkosxT;knkgksrkgS (A*) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C) Statement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True (A*) oDrO;-1lR; gS] oDrO;-2lR; gS ;oDrO;-2, oDrO;-1 dk lgh Li"Vhdj.k gSA (B)oDrO;-1lR; gS] oDrO;-2lR; gS ; oDrO;-2,oDrO;-1 dk lgh Li"Vhdj.k ugha gSA (C) oDrO;-1lR; gS] oDrO;-2vlR; gS ; (D) oDrO;-1vlR; gS] oDrO;-2lR; gS Sol. (A) Propagation of sound in air is an adiabatic process. ok;qesa/ofudklapj.k:)ks"eizØedsrgrgksrsgSA SECTION - IV : COMPREHENSION TYPE [k.M- IV:cks/ku izdkj Comprehension vuqPNsn # 1 Paragraph for Question Nos. 26 to 27 iz'u 26 ls 27 ds fy, vuqPNsn A driver is riding a car with velocity vC between two vertical walls on a horizontal surface as shown in figure. A source of sound of frequency 'f' is situated on the car. (vc << v, where v is the speed of sound in air) ,dMªkbojfp=k uqlkj]{kSfrtlrgijnksÅ/okZ/kjnhokjksadse/;vC osxlsdkjpykjgkgSA'f'vko`fÙkdk,d/ ofu dk L=kksr dkj ij fLFkr gSA (vc << v, tgk¡ v gok esa /ofu dh pky gSA) 13.39 Beat frequency observed by the driver corresponding to sound waves reflected from wall-1 and wall-2 (reflected waves corresponding to waves directly coming from source) : nhokj-1rFk nhokj-2lsijkofrZr/ofurjaxksdslaxrMªkboj}kjkçsf"krfoLiUnvko`fÙk(rajxksadslaxrijkofrZrrjax lh/ksL=k srlsvkjghgS): (A) vc f v (B) 2vc f v (C) vc f 2v (D*) 4vc f v 13.40 Consider the sound wave observed by the driver directly from car has a wavelength  and the sound wave 1  2 after reflection from wall-1 observed by the driver has wavelength 2 then 1  2 is : Mªkboj}kjkdkjlslh/ksçsf"kr/ofurjaxdhrjaxnS/;Z1 rFk nhokj-1lsijkorZudsi'pkr~Mªkboj}kjkçsf"kr/ 1  2 ofurjaxdhrjaxnS/;Z2 gSrc 1  2 gksxkA (A*) vc (B) 2vc (C) vc (D) 4vc v Sol.(26& 27) v 4v v (26) f1i f2i = f1r = = f2r = v f v  vc v f v  vc Now, for driver vc Mªkbojdsfy, f = v  vc f dr1 v 1r and rFkk f dr2 = v  vc f v 2r So, beat frequency vr% foLiUn vko`fÙk= | fdr1 – fdr2 | v  v v  v (v  vc )2  (v  vc )2  = c f1r  c f2r =   f v  4vv c  f v  4vc  f  (v  vc )(v  vc )  =  v2  =   .   (27)  = v  vc f = v  vc 2 f –  = +  = 2vc f 2v f 1  2 1  2 = vc . v Comprehension # 2 An Indian submarine is moving in “Arab Sagar” with a constant velocity. To detect enemy it sends out sonar waves which travel with velocity 1050 m/s in water. Initially the waves are getting reflected from a fixed island and the reflected waves are coming back to submarine. The frequency of reflected waves are detected by the submarine and found to be 10% greater than the sent waves. Sonar waves Fixed Island Now an enemy ship comes in front, due to which the frequency of reflected waves detected by subma- rine becomes 21% greater than the sent waves. vuqPNsn- 1 ,dfgUnqLrkuhiuMwCchvjclkxjesafu;rosxlsxfr'khygSA'k=kqdkirkyxkusdsfy,;glksukjrjaxksdksHkstrk gStksikuhesa1050m/slslapfjrgSArjaxsizkjEHkesa,dfLFkjVkiwlsijkofrZrgksrhgSrFk ijkofrZrrjaxsokil iuMwCchijvkrhgSAijkofrZrrajxksdhvko`fÙkiuMwCch}kjklalwfprdhtkrhrFkkHksthxbZrjaxksls10% vf/kdfeyrhgS l ksukj r j axs fLFkj Vki w vc'k=kqdk,dtgktlkeusvktkrkgS]ftldsdkj.kiuMwch}kjklalqfprijkofrZrrajxksadhvko`frHksthxbZ vko`fr;ksals21%vf/kdgkstkrhgSA 13.41 The speed of Indian submarine is fgUnqLrkuhiuMwCchdhpkygS (A) 10 m/sec (B*) 50 m/sec (C) 100 m/sec (D) 20 m/sec.   f = f   Sol. 1 0  v  v  v1 f0 f1 Island  v  v1   f"–fo  f´´ = f   , v = 1050   0.1 0  v  v1  f"–fo  2V1 = 0.1  fo  fo V – V1 v1 = 50 m/sec. 13.42 The velocity of enemy ship should be : (A*) 50 m/sec. toward Indian submarine. (B) 50 m/sec. away from Indian submarine. (C) 100 m/sec. toward Indian submarine. (D) 100 m/sec. away from Indian submarine. 'k=kqdstgktdhpkygksuhpkfg, (A*)50 m/sec. fgUnqLrkuhiuMwCch dhvksj (B)50m/sec. fgUnqLrkuhiuMwCchlsnwj (C)100 m/sec. fgUnqLrkuhiuMwCch dhvksj (D)100m/sec. fgUnqLrkuhiuMwCchlsnwj v Sol.  v  50  2  v  v2  f´´ = f´  v  v    0 v  50  2     (v  v2 )(v  50)  f´´ = f  0  (v  v2  = 1.21f [21% greater then sent waves] )(v  50)  0 get v2 = 50 m/sec toward Indian submarine Sol.  v  50  v2  v  v2  f´´ = f´  v  v    0 v  50  2     (v  v2 )(v  50)  f´´ = f  0  (v  v2 )(v  50)  = 1.21f0 [21% Hksthx;hrajxksls] v2 =50m/secfgUnqLrkuhiuMwCchdhvksj 13.43 If the wavelength received by enemy ship is ´ and wavelength of reflected waves received by submarine  ´  is ´´ then  ´´ equals   (A) 1 (B*) 1.1 (C) 1.2 (D) 2  ´  ;fn'k=kqdstgktdksizkIrrjaxnS/;Z´ gSrFk iuMwCch}kjkizkIrijkofrZrrjaxdkrjaxnS/;Z ´´ gS]rc  ´´   cjkcj gS (A) 1 (B*) 1.1 (C) 1.2 (D) 2 Sol. ´ = v wrt to observer f´ v  v 2 = (v  v 2 ) f0 (v  50)  v  50 f0 ´´ = v  50 = (v  v 2 )(v  50) f (v  v 2 )(v  50) 0 (v  v 2 )(v  50) f0 (v  v 2 ) ' ´´ = v  v 2 v  v2 1050  50 = 1050  50 = 1.1 Sol. ´ = v izs{kd dlskis{k = v  v 2  v  50 f´ f (v  v 2 ) f0 0 (v  50) ´´ = v  50 f (v  v 2 )(v  50) 0 (v  v 2 )(v  50) = (v  v 2 )(v  50) f0 (v  v 2 ) ' ´´ = v  v 2 v  v2 1050  50 = 1050  50 = 1.1 13.44 Bulk modulus of sea water should be approximately ( = 1000 kg/m 3) leqnzhikuh dkvk;ru izR;kLFkrk xq.kkad yxHkxgS -(water = 1000kg/m3) (A) 108 N/m2 (B*) 109 N/m2 (C) 1010 N/m2 (D) 1011 N/m2 Sol. v =  1050 = B  109 N/m2 Comprehension # 3  In sound wave, y (x, t) equation and P (x, t) equation have a phase difference of 2 .  /ofurjaxeas]y(x,t)rFkkP(x,t)lehdj.kesadykUrj 2 gksrkgSA Pressure amplitude in P (x, t) equation is equal to BAK. rFkknkcdkvk;keP(x,t)lehdj.kesaBAKdscjkcj gSA y (x, t) equation of a longitudinal wave is given as: vuqnS/;Zrjaxdklehdj.ky(x,t)gSA  50  y = 10–2 sin 2 1000 t  17 x (AIISI units lHkh SI ek=kd)   13.45 At t = 0, change in pressure is maximum at x = m. t= 0ij vf/kdre nkc ifjorZu x = eh- ij gSA (A*) 0.34 (B) 0.255 (C) 0.085 (D) all of these;s lHkh  50  Sol. At t = 0, ij y = 10–2 sin 2  17 x   Change in pressure will be maximum where y = 0 at t = 0, vf/kdre nkcifjorZugksxk] tgkWt=0ijy=0 2  50 17 x = 0, , 2.... or;k x = 0, 0.17 m, 0.34m... 13.46 If density of the gas is 10–2 kg/m3, find the pressure amplitude : ;fnxSldk?kuRo10–2 kg/m3, gSrksnkcvk;kecrkvksa (A) 200.62 N/m2 (B) 421.24 N/m2 (C) 100.26 N/m2 (D*) 21.36 N/m2  Sol. v = k    2  B =   k   (P) = BAK = 2 K2  AK =  A2 K Substituting the values, weget ;s eku j[ksu ij 10–2 10–2 (21000)2 (P) = (2 50/17) = 21.36 N/m2 SECTION - V : MATRIX - MATCH TYPE [k.M-V:eSfVªDl&lqesyizdkj 13.47 Match the columns I & II. Column I Column II (A) Pitch (p) Number of harmonics present in the sound (B) Loudness (q) Intensity (C) Quality (r) Frequency (D) wave front (s) wave form (t) locus of points vribrating in a phase LrEHk A o B dk feyku dhft, LrEHk I LrEHk II (A) rkjRo (p) /ofu esa mifLFkr lauknh;ksa dh la[;k (B) izcyrk (q) rhozrk (C) xq.kork (r) vko`fÙk (D) rjaxkxz (s) rajx :i (t) lekudyk esadEiu djjgsa fcUnqdk fcUniFk Ans. a  r, b  q, c  p,s, d  t 13.48 Match the Column: (A) y = 4 sin (5x – 4 t) + 3 cos (4t – 5x + /6) (p) Particles at every position are performing SHM  t  x   t  x  (B) y = 10 cos    330  sin (100)    330  (q) Equation of travelling wave (C) y = 10 sin (2x – 120t) + 10 cos (120t + 2x) (r) Equation of standing wave (D) y = 10 sin (2x – 120 t) + 8 cos (118t – 59/30x) (s) Equation of Beats (t) Initial displacement of particle at origin is zero Ans. (A) p, q (B) q, s, t (C) r (D) s,q Sol. (A) y = 4 sin (5x – 4 t) + 3 cos (4t – 5x + /6) is super position of two coherent waves moving in positive direction, so their equivalent will be an another travelling wave.  t  x   t  x  (B) y = 10 cos    330  sin (100)    330  lets check at any point, say at x = 0, y = (10 cos t) sin (100 t) at any point amplitude is changing sinusoidally. so this is equation of beats. (C) y = 10 sin (2x – 120t) + 10 cos (120t + 2x) = superposition of two coherent waves travelling in opposite direction.  equation of standing waves. (d) y = 10 sin (2x – 120 t) + 8 cos (118t – 59/30x) = superposition of two waves whose frequencies are slightly different ( = 120,  = 118)  equation of Beats. fuEu dk feyku dhft, & (A) y = 4 sin (5x – 4 t) + 3 cos (4t – 5x + /6) (p)d.kizR;sdfLFkfrijSHMiznf'kZrdjrkgS  t  x   t  x  (B) y = 10 cos    330  sin (100)    330  (q) izxkehrjaxdhlehdj.k (C) y = 10 sin (2x – 120t) + 10 cos (120t + 2x) (r) vizxkehrjax dklehdj.k (D) y = 10 sin(2x – 120t) + 8 cos(118t – 59/30x) (s) foLiUn dh lehdj.k (t) ewy fcUnq ij d.k dk çkjfEHkd foLFkkiu 'kwU; gSA Ans. (A) p, q (B) q, s, t (C) r (D) s,q Sol. (a) y = 4 sin (5x – 4 t) + 3 cos (4t – 5x + /6) nks/kukRedfn'k esaxfrekudyklEc)rajxksdkv/;kjksi.kgS]blizdkjbudkrqY;Hkh,dvU;izxkehrjaxgSA  t  x   t  x  (b) y = 10 cos    330  sin (100)    330  fdlh fcUnq ij tk¡p djrs gS]x = 0ij, y = (10 cos t) sin (100 t) fdlhHkhfcUnqijvk;keT;koØh;ifjofrZrgksrkgSAblizdkj;gfoLiUnksadhlehdj.kgS (c) y = 10 sin (2x – 120t) + 10 cos (120t + 2x) =foijhrfn'k esalapkfjrnksdyklEcU/krjaxksdk v/;kjksi.k vizxkehrjxksadklehdj.k (d) y = 10 sin (2x – 120 t) + 8 cos(118t – 59/30x) = nks rjaxks dk v/;kjksi.k ftudh vko`fr;ksa esa tjk lk vUrjgS (1 = 120, 2 = 118) foLiUn dh lehdj.k SECTION - VI : INTEGER TYPE [k.M-VI:iw.kkZadizdkj 13.49 A 3 m long organ pipe open at both ends is driven to third harmonic standing wave. If the amplitude of pressure oscillation is 0.1 % of the mean atmospheric pressure (P = 105 N/m2). if the amplitude of: x Particle oscillation is 1089 m then x is nksuksafljksals[kqyk,d3myEckvkxZuikbZivizxkehrjaxdhrhljslauknhijctrkgSA;fnnkcnksyudkvk;ke ek/; ok;qe.Myh; nkc (P = 105 N/m2) dk 0.1% gSA x d.k nksyu Ans. 1 1089 gks rks x gksxkA Sol. 3 = 3.  2   = 2 m P = 100 N/m2, V = 330 m/s,  = 1 kg/m3 Pm = B s0 k =  v2 s 2    s = Pm  v22 = 0 2 100 1 330  330  2  s0 = 1 1089  m x 13.50 In previous question density oscillation is 1089 kgm-3 then x is Speed of sound v = 330 m/s, density of air  = 1.0 kg/m3. x mijksDriz'uesa?kuRonksyudkvk;ke 1089 kgm-3 gksrksxgksxkA /ofu dh pky v = 330 m/s, ok;q dk ?kuRo  = 1.0 kg/m3. Ans. 1 Sol. 3 = 3.  2   = 2 m P = 100 N/m2, V = 330 m/s,  = 1 kg/m3 d p B = dv / v d P = d /  [ m =  v  O = d  dv + v ]  . d p d  = B  (d )  max = B (d p) max Pm = B (d ) max = .Pm v 2 1 = 1089 kg/m3 13.51 In a car race sound signals emitted by the two cars are detected by the detector on the straight track at the end point of the race. Frequency observed are 330 Hz & 360 Hz and the original frequency is 300 Hz of both cars. Race ends with the separation of 100 m between the cars. Assume both cars move with constant velocity and velocity of sound is 330 m/s. Find the time taken by winning car (in sec.) [2] ,d dkj jsl esa nks dkjksa }kjk mRlftZr /ofu ladsrksa dks lh/ks iFk ij] lalwpd }kjk jsl ds vfUre fcUnqvkas ij] lalwfpr fd;k tkrk gSA izsf{kr vko`fÙk 330 Hz o 360 Hz gS ,oa nksuks dkjksa dh ewy vko`fÙk 300 Hz gSA jsl dkjkas ds e/; 100m nwjh ds lkFk lekIr gksrh gSA ekfu;s fd nksuks dkjsa vpj osx ls xfr djrh gS o /ofu dk osx 330 m/s gSA thrus okyh dkj }kjk fy;k x;k le; Kkr (in sec.) dhft, [2] Ans. 4 Sol. Let the velocities of car 1 and car 2 be V1 m/s and V2 m/s.  Apparent frequencies of sound emitted by car 1 and car 2 as detected at end point are f1 = f0 V V  V1 , f2 = f0 330 V V  V2 330  330 = 300 330  V1 , 360 = 300 330  V2  V = 30 m/s and V = 55 m/s. The distance between both the cars just when the 2nd car reach and point B (as shown in figure is) 100m = V t – V t  t = 4 sec. 2 1 Sol. ekukfd dkj 1 o dkj 2 ds osx V1 m/s o V2 m/s gSA  vfUrefcUnqijdkj1odkj2}kjkmRlftZr/ofudhvkHkklhvko`fr;kagS f1 = f0 V V  V1 , f2 = f0 330 V V  V2 330  330 = 300 330  V1 , 360 = 300 330  V2  V = 30 m/s and V = 55 m/s. nksuksdkjksdse/;nwjhBhdtcnwljhdkjBfcUnqijigq¡pusokyhgS¼fp=kesan'k Z;svuqlkj½ 100m = V t – V t  t = 4 sec. 2 1 13.52 A bat emits ultrasonic sound of frequency 1000 kHz in air. If the sound meets a water surface, it gets partially reflected back and partially refracted (transmitted) in water. Difference of wavelength transmitted to wavelength reflected is x cm. then x is (speed of sound in air = 330 m/sec, Bulk modulus of water = 2.25 x 109,  = 1000 kg/m3) ,d pexknM+1000kHzvko`fÙkdhijkJO;/ofu ok;qesamRlftZrdjrhgSA;fn /ofuikuh dhlrg ijfxjrhgS] ;gikuhlsokilvkaf'kdijkofrZrovkaf'kdviofrZr¼ikjxr½gksrkgSAikjxferrjaxnS/;ZoijkofrZrrjaxnS/;Zdk vUrjx cm. gSrksx gksxk% (ok;q esa /ofu dh pky= 330 m/sec, ikuh dk vk;ru izR;kLFkrk xq.kkad= 2.25 x 109,  = 1000 kg/m3) Ans. 117 Sol.  air = Vair = f 330 1000 = 0.33 m Vwater = = = 1.5 x 103 = 1500 water = 1500 1000 = 1.5 m  –  = 1.5 – 0.33 = 1.17m. Sol.  air = Vair = f 330 1000 = 0.33 m Vwater = = = 1.5 x 103 = 1500 water = 1500 1000 = 1.5 m water – air = 1.5 – 0.33 = 1.17m.