12-String Wave

TOPIC STRING WAVE 12 SECTION - I : STRAIGHT OBJECTIVE TYPE [k.M-I: lh/ks oLrqfu"B izdkj 12.1 A travelling wave y = A sin (k x  t + ) passes from a heavier string to a lighter string. The reflected wave has amplitude 0.5 A. The junction of the strings is at x = 0. The equation of the reflected wave is: (A) y  = 0.5 A sin (k x + t + ) (B) y  =  0.5 A sin (k x + t + ) (C) y  =  0.5 A sin (t  k x  ) (D*) y  = – 0.5 A sin (k x + t  ) Sol. As wave has been reflected from a rarer medium, therefore there is no change in phase. Hence equation for the reflected waves can be written as y = 0.5A sin (–kx – t + ) = – 0.5A sin (kx + t – ) ,d çxkehrjax y=Asin(kxt+),dHkkjh Mksjhls gYdhMksjh dhvksj xqtjrhgSA ijkofrZrrjax dkvk;ke 0.5 A gSA Mksfj;ksa dk laf/k x = 0 ij gSA ijkofrZr rjax dk lehdj.k gSA (A) y  = 0.5 A sin (k x + t + ) (B) y  =  0.5 A sin (k x + t + ) (C) y  =  0.5 A sin (t  k x  ) (D*) y  = – 0.5 A sin (k x + t  ) Sol. pw¡fdrjaxfojyek/;elsijkofrZrgq;hgSvr%dykesaifjorZuughagksxkA blfy, ijkofrZr rjax ds fy, lehdj.k bl çdkj fy[kk tk ldrk gSA y = 0.5A sin (–kx – t + ) = – 0.5A sin (kx + t – ) 12.2 Which of the following travelling wave will produce standing wave, with node at x = 0, when superimposed on y = A sin (t  k x) fuEueasaalsdkSu&lhçxkehrjaxy=Asin(tkx)lsv/;kjksfirgksusij,slhvizxkehrjaxsmRiUudjsxhftudk fuLian x = 0 ij gksxkA (A) A sin (t + k x) (B*) A sin (t + k x + ) (C) A cos (t + k x) (D) A cos (t + k x + ) Sol. Substituting x = 0 we have given wave y = A sin t at x = 0 other should have y = – A sin t equation so displacement may be zero at all the time Hence (B) is correct option. Sol. fn;s gq, lehdj.k esa x=0 j[kus ij gesa y=Asintfeyrk gS vr% x=0ij nwljh rjax y=–Asintgksuk pkfg, ftl sfoLFkkiuges'kk'kwU;gksldsvr%fodYi(B)lghgSA 12.3 A wire of length ‘ 𝑙 ‘ having tension T and radius ‘ r ‘ vibrates with fundamental frequency ‘ f ‘. Another wire of the same metal with length ‘ 2𝑙 ‘ having tension 2 T and radius 2 r will vibrate with fundamental frequency: ‘𝑙 ‘ yEckbZ vkSj ‘r‘ f=kT;k dk ,d rkj ftleas ruko T gS] ewyvko`fÙk ‘f‘ ds lkFk dEié djrk gSA mlh /kkrqds,dvksjrkj ftldhyEckbZ‘2𝑙‘ruko2TvkSjf=kT;k2r gS]dhdEiudh ewyvko`fÙkgSa– (A) f (B) 2 f (C*) 1 f (D) 2 Sol. f = 2𝑙  =  r2 If radius is doubled and length is doubled, mass per unit length will become four times. Hence f’ = 1 2 x 2𝑙 = Sol. f = 1 2𝑙  =  r2 ;fnf=kT;kdksnksxqukdjfn;ktk;svkSjyEckbZdksHkhnqxukdjfn;ktk,rksçfr,dkadyEckbZæO;ekupkjxquk gkstk;sxkAvr% 1 f’ = 2 x 2𝑙 = 12.4 A string of length 1.5 m with its two ends clamped is vibrating in fundamental mode. Amplitude at the centre of the string is 4 mm. Distances between the two points having amplitude 2 mm is: 1.5eh0yEchMksjhtksnksuksafljksaijc¡/khgS]ewyfo/k esadEiédjjghgSAMksjhdse/;fcUnq¼dsUæ½ijvk;ke4 eh0eh0 gSA mu nks fcUnqvksa ds chp dh nwjh ftudk vk;ke 2 eh0eh0 gS] gksxh (A*) 1 m (B) 75 cm (C) 60 cm (D) 50 cm Sol.  = 2𝑙 = 3m Equation of standing wave (As x = 0 is taken as a node) y = 2A sin kx cos t y = A as amplitude is 2A. A = 2A sin kx  kx = 6 2   x = 6 2  5 or 6  x = 1 m 4  and  . x = 2 + 3  x = 1.25 m  x – x1 = 1m (A*) 1 m (B) 75 cm (C) 60 cm (D) 50 cm gy%  = 2𝑙 = 3m vizxkeh rjax dk lehdj.k (x=0ij fuLiUn ysrs gq,) y = 2A sin kx cos t y =Apqfd vk;ke2AgSa A = 2A sin kx  kx = 6 2  5 ;k 6 1  x = 6  x = 4 m vkSj 2  . x =  +   x 2 3 2 = 1.25 m  x – x1 = 1m 12.5 Two particles of medium disturbed by the wave propagation are at x1 = 0 and x2 = 1cm. The respective displacements (in cm) of the particles can be given by the equations : y = 2sin3t y = 2sin(3t – /8) The wave velocity is : (A) 16 cm/sec (B*) 24 cm/sec (C) 12 cm/sec (D) 8 cm/sec. Sol. Given  = 3  f =  2 = 1.5, Also x = 1.0 cm Given,  = 2 x   2 8 =  × 1   = 16 cm  v = f  = 16 × 1.5 = 24 cm/sec 12.5 x1= 0rFk x2=1cm ijek/;edsnksd.krjaxlapj.k}kjkçHk forgksrsgSAd.k sadsfoLFk iu(lseheas)Øe'k%fuEu lehdj.k sa}kjkfn;stkldrsgSa: y = 2sin3t y = 2sin(3t – /8) rjaxosxgS: (A) 16 cm/sec (B*) 24 cm/sec (C) 12 cm/sec (D) 8 cm/sec. Sol. fn;k gS= 3  f =  2 = 1.5, vkSj x = 1.0 cm fn;kgS,  = 2 x   2 8 =  × 1   = 16 cm  v = f  = 16 × 1.5 = 24 cm/sec 12.6 The displacement Vs time graph for two waves A and B which travel along the same string are shown in the figure. Their intensity ratio A / B is nksrjaxsaAvkSjBtks,dghMksjhdsvuqfn'kxfrekugS]dkfoLFkiu≤xzkQfp=kesaçnf'kZrfd;kx;kgSAmudhrhozrk dk vuqikr A / B gS& (A) 9 4 I1  a2  2 81 (B*) 1 (C) 16 (3)2 (8)2 (D) 3 2 Sol. I2 a2  2 = (2)2 (12)2 = 1 2 2 12.7 At t = 0, a transverse wave pulse travelling in the positive x direction with a speed of 2 m/s in a wire is described by the function y = seconds is : 6 , given that x  0. Transverse velocity of a particle at x = 2m and t = 2 x2 t=0ij]rkjesa/kukRedxfn'k esa2eh-@lS-lsxfrdjjghvuqizLFkrjaxksadksy= fn;k x;k gS fd x  0, d.k dk vuqizLFk osx x = 2eh- rFkk t= 2lS- ij gksxk & 6 lsiz nf'kZrdjrsgSaA;g x (A) 3 m/s (B*) – 3 m/s (C) 8 m/s (D) – 8 m/s (A)3eh-@lS- (B*) – 3eh-@lS- (C)8eh-@lS- (D)– 8eh-@lS- 6 Sol. (B) y(x, t = 0) = x2 6 then rc y(x,t) = (x  2t)2 y  t 24 = (x  2t)3 24 at x = 2, t = 2 Vy = (2)3 = – 3 m/s. 12.8 Wave pulse on a string shown in figure is moving to the right without changing shape. Consider two particles at positions x1 = 1.5 m and x2 = 2.5 m. Their transverse velocities at the moment shown in figure are along directions : ,djLlhesarjaxLianfcukviuhvkd`frifjofrZrfd;snkW;hafn'k esaxfrdjjgkgStSlkfp=kesaiznf'kZrgSAnks d.kØe'k%x1=1.5eh-rFkkx2=2.5eh-fLFkfr;ksaijekusAfp=kesaiznf'kZr{k.kijmudkvuqizLFkosxfuEuds vuqfn'kgksxk& (A) positive y–axis and positive y–axis respectively Øe'k%/kukRedy–v{krFkk/kukRedy–v{k (B*) negative y–axis and positive y–axis respectively Øe'k% _.kkRed y–v{k rFkk /kukRedy–v{k (C) positive y–axis and negative y–axis respectively Øe'k% /kukRed y–v{k rFkk _.kkRedy–v{k (D) negative y–axis and negative y–axis respectively Øe'k% _.kkRed y–v{k rFkk _.kkRed y–v{k Sol. (B) Dotted shape shows pulse position after a short time interval. Direction of the velocities are decided according to direction of displacements of the particles. fcUnqorvkdkjNksVslekUrjkydscknLiandhfLFkfrdksn'k ZrkgSAosxksadhfn'k d.k sadsfoLFk iudhfn'k ds vuqlkjfu/kZfjrgksrhgSA 12.9 A wave pulse is generated in a string that lies along x-axis. At the points A and B, as shown in figure, if RA and RB are ratio of wave speed to the particle speed respectively then : ,djLlhesa,drjaxLianmRiUufd;ktkrkgStksx-v{kdsvuqfn'kfLFkrgSAfp=kesafn[k ;svuqlkjfcUnqArFk B ijrjax osxrFkk d.kosx dkvuqikr Øe'k%RA rFkkRB gSArks & (A) RA > RB (B*) RB > RA (C) RA = RB (D) Information is not sufficient to decide. fu.kZ; ds fy, lwpuk vi;kZIr gSA V R  V Sol. RA = VA , B as(pqfd) VA > VB , RA < RB 12.10 Sinusoidal waves 5.00 cm in amplitude are to be transmitted along a string having a linear mass density equal to 4.00 × 10–2 kg/m. If the source can deliver a average power of 90 W and the string is under a tension of 100 N, then the highest frequency at which the source can operate is (take 2 = 10) : T;koØh;rjaxsftudkvk;ke5.00lseh-gS]dksmljLlhdsvuqfn'klapfjrfd;ktkrkgSftldkjs[kh;nzO;eku ?kuRo4.00×10–2 fd-xzk-@eh-gSAvxjL=kksrlsvkSlr90okVÅtkZiznkudhtkldrhgSrFkkjLlhesaruko100 U;wVu gSrksogmPprevko`fÙkftl ijL=kksrdk;Zdj lds]gksxh &(2 = 10ekusa): (A) 45.3 Hz (B) 50 Hz (C*) 30 Hz (D) 62.3 Hz (A) 45.3 gV~Zt (B)50gV~Zt (C*) 30gV~Zt (D) 62.3gV~Zt 1Sol. (C) P = 2A2V using V = 2 P = 1 2 A2 2  =  f = 2 1 = 2 using the given data, we get f = 30 Hz. 1 Sol. (C) P = 2A2V V= 2 dkiz;ksx djusij P = 1 2 A2 2  =  f = 2 1 = 2 vk¡dM+ksdsmi;ksxls f= 30Hz. 12.11 The figure shows four progressive waves A, B, C & D. It can be concluded from the figure that with respect to wave A: fp=kesapkjizxkehrjaxsA,B, CrFk Diznf'kZrgSafp=kls;gfu"d"kZfudkyldrsgSafdrjaxAdslkis{k (A) the wave C is ahead by a phase angle of /2 & the wave B lags behind by a phase angle /2 (B*) the wave C lags behind by a phase angle of /2 & the wave B is ahead by a phase angle of /2 (C) the wave C is ahead by a phase angle of  & the wave B lags behind by the phase angle of  (D) the wave C lags behind by a phase angle of  & the wave B is ahead by a phase angle of  (A) rjax C, /2dyk dks.k ls vkxs rFkkrjax B,/2dyk dks.k ls ihNs gS (B*) rjaxC,/2dyk dks.k lsihNs rFkkrjax B,/2dyk dks.k lsvkxs gS (C) rjax C,  dyk dks.k ls vkxs rFkk rjax B,  dyk dks.k ls ihNs gS (D) rjaxC,dykdks.klsihNs rFkkrjaxB,dykdks.klsvkxs gS Sol. In figure, 'C' reaches the position where 'A' already reaches after t =  2 and 'A' reaches the position where 'B' already reaches after t =  2 Hence (B). Sol. fp=kesa'C'ml fLFkfrij gS]tgk¡ 'A't=  le;cknigysghigqWppqdkgSarFkk 2 'A' ml fLFkfr ijt =  le; ckn igWqprk gSa tgkW 'B' igys igqp pqdk gSA 2 vr% (B). 12.12 A 75 cm string fixed at both ends produces resonant frequencies 384 Hz and 288 Hz without there being any other resonant frequency between these two. Wave speed for the string is : ,d75lseh-dhjLlhnksuksafljksaijca/khgqbZgSrFk 384Hzo288Hzdhvuquknvko`fÙkblizdkjmRiUudjrh gSafdbunksuksavko`fÙk;ksadse/;vU;dksbZvuquknvko`fÙkughagSrksjLlhesarjaxdhpkygksxh& (A*) 144 m/s (B) 216 m/s (C) 108 m/s (D) 72 m/s (A*)144eh-@lS- (B) 216eh-@lS- (C) 108eh-@lS- (D)72eh-@lS- Sol. 384 = 288 = nv 2𝑙 mv 2𝑙 ...(i) ...(ii) from equation (i) & (ii)  n    4       m   3  so n = 4 from equation (i) 4v 10v 384 = 2  3 / 4 = 3 v = 144 m/s nv 384 = 288 = 2𝑙 mv 2𝑙 ...(i) ...(ii) lehdj.k (i) o(ii) ls  n    4       m   3  blfy, n = 4 lehdj.k (i) ls 4v 10v 384 = 2  3 / 4 = 3 v = 144 m/s 12.13 A string of length ‘𝑙’ is fixed at both ends. It is vibrating in its 3rd overtone with maximum amplitude ‘a’. The 𝑙 amplitude at a distance 3 from one end is : ,d‘𝑙’yEckbZdhjLlhnksuksafljksalsca/khgqbZgSA;gbldsrhljsvf/kLojdesavf/kdrevk;ke'a’dslkFkdEiUu 𝑙 dj jgh gSA blds ,d fljs ls 3 nwjh ij vk;ke gksxk & (A) a (B) 0 (C*) 3 a 2 (D) a 2 Sol. For a string vibrating in its nth overtone ((n + 1)th harmonic)  (n  1)  x  y = 2A sin   L  cos t  𝑙   4 𝑙  For x = 3 , 2A = a and n = 3; y = asin 𝑙 . 3  cos t    4  3  = a. sin cos t = – a.  cos t i.e. at x = 3 𝑙 3 ; the amplitude is     3 a . 2 nth vf/kLojd ;k ((n+1)th lauknh )esas dfEir jLlh ds fy,  (n  1)  x  y = 2A sin   L  cos t  𝑙   4 𝑙  x = 3 ds fy, 2A = a vkSj n = 3; y = asin 𝑙 . 3  cos t    4  3  = a. sin cos t = – a.  cos t 3 𝑙 vFkkZrx= 3 ijvk;ke     3 a gksxk 2 12.14 What is the percentage change in the tension necessary in a sonometer of fixed length to produce a note one octave lower (half of original frequency) than before fu;ryEckbZdslksuksehVjdsrukoesafdrukizfr'krifjorZufd;ktk;sfdigysls,dv"Vddeds¼ewyvko`fÙk dkvk/k ½/ofumRiUugkslds (A) 25% (B) 50% (C) 67% (D*)75% Sol. (D) In a sonometer,    1 2 = 2 = T  T T  T2 = 4 T  T1  % change will be : (D)lksuksehVjesa 1 2 T1 × 100 = 4 × 100 = 75% Ans. T1   1  2 = 2 = T  T2 = 4  %ifjorZugksxk--- T  T T  T1 1 2 T1 × 100 = 4 × 100 = 75% mÙkj T1 12.15 A chord attached about an end to a vibrating fork divides it into 6 loops, when its tension is 36 N. The tension at which it will vibrate in 4 loops is: (A) 24 N (B) 36 N (C) 64 N (D*) 81 N Sol. (D) For waves along a string :      T Now, for 6 loops : 3 = L   = L/3 & for 4 loops : 2 = L   = L/2  1  2  T = 9 xT = 9 × 36 2 3 2 4 1 4 = 81 N. Ans. ,ddEiédjrkLofj=kblds,dfljsijtqMs+,d/k xsdks6ywiesafoHkkftrdjrkgS]tc/k xsesaruko36NgSA og ruko ftldsfy, ;g 4ywiesa dEié djsxk] gS( (A) 24 N (B) 36 N (C) 64 N (D*) 81 N Sol. (D) Mksjhdsvuqfn'krjaxdsfy,   T    T vc 6 ywids fy, : 31 = L vkSj 4 ywids fy, : 22 = L    = L/3 1  = L/2 2  1  2  T = 9 xT = 9 × 36 2 3 2 4 1 4 = 81 N. Ans. 12.16 Two vibrating strings of same length, same cross section area and stretched to same tension are made of materials with densities  & 2 . Each string is fixed at both ends. If v represents the fundamental mode of vibration of the one made with density  and v for another, then v /v 1 2 is: nkslekuyEckbZ,oaleku vuqizLFkdkV{ks=kQydhdfEirjfLl;ksadkslekurukols[khapkx;kgSrFk buds inkFk asZds?kuRorFk 2gSAizR;sdjLlhnksuksafljksaijfLFkjgSA;fn ?kuRolscuhjLlhesadEiudhewyfo/ k dks v1lsn'k ZrsgSrFk v2vU;jLlhesan'k ZrkgS]rksv1/v2 gS& 1 (A) (B) 2 (C*) 2 (D) Sol. Velocity of sound is inversely proportional to the square root of density of the medium. i.e., V = constant V1  V2 = = = Ans. Sol. /ofu dk osx ek/;e ds ?kuRo ds oxZewy ds O;qRØekuqikrh gksrk gS vFkZr~, V = fu;r V1  V2 = = = Ans. 12.17 Which of the following function correctly represents the wave equation for finite values of x and t : x rFkktdsfuf'prekuksadsfy,fuEuesalsdkSulkQyurjaxlehdj.kdksiznf'kZrdjrkgS& (A) y = x2 – t2 (B) y = cosx2 sint (C*) y = log (x2 – t2) – log(x – t) (D) y = e2x sint x2  t2 a Sol. (C) y = log y x  t 1 = log(x + t) ( log a – log b = log b ) x = (x  t) 2y x2 =  (x  t)2 and y  t = (x / t) (x  t) v = (x  t) 2y t2 v 2 =  (x  t)2 2 y 1  x2 = v 2 2 y t2 Which is the general form of wave equation. x2  t2 a Sol. (C) y = log y x  t 1 = log(x + t) ( log a – log b = log b ) x = (x  t) 2y x2 =  (x  t)2 y ;k  t (x / t) = (x  t) v = (x  t) 2y t2 v 2 =  (x  t)2 2y  x2 1 = v 2 2 y t2 tksrjax xfrds lehdj.k dk O;kid :i gSA 12.18 The figure shows at time t = 0 second, a rectangular and triangular pulse on a uniform wire are approaching each other. The pulse speed is 0.5 cm/s. The resultant pulse at t = 2 second is le; t= 0lsd.M ij fp=k ij ,d vk;rkdkj rFkk f=kHkqtkdkj rjaxs (pulse) ,d le:i rkj ij ,d nwljs dh vksj vk jgh gSA rjax dh pky 0.5cm/sgSA t= 2lsd.M ij ifj.kkeh rjax gksxhA (A) (B) (C) (D*) Sol. At t = 2 second, the position of both pulses are separately given by fig.(a) and fig. (b); the superposition of both pulses is given by fig. (c) t=2lsd.MijnksuksaLianksdksfp=k(a)ofp=k(b)esav/;kjksi.kdscknfp=k(c)esacrk;kx;kgSA 12.19 A loop of a string of mass per unit length  and radius R is rotated about an axis passing through centre perpendicular to the plane with an angular velocity . A small disturbance is created in the loop having the same sense of rotation. The linear speed of the disturbance for a stationary observer is : jLlhdk,dywiftldkizfr,dkadyEckbZnzO;eku]f=kT;kRgS]bldsrydsyEcor~rFkkdsUnzlsxqtjusokyh v{kdslkis{kdks.kh;osxls?kw.kZudjjgkgSAywiesa,dy?kqfo{k sHkmRiUufd;ktkrkgSftlds?kw.kZudhfn'k ywi ds leku gSA fo{kksHk dh js[kh; pky ¼,d fLFkj izs"kd ds fy,½ gksxh & (A) R (B*) 2R (C) 3R (D) zero 'kwU; Sol. dm.2R = 2T sin d 2 Rd 2R = 2T d 2  2R2 = T  v = Also speed of string is R = 2R2 = R  The velocity of disturbance w.r.t. ground = R + R = 2R. jLlh dh pky R gS  vr% fo{kksHkdh /kjkryds lkis{kpky = R + R =2R. 12.20 A uniform rope having mass m hangs vertically from a rigid support. A transverse wave pulse is produced at the lower end. The speed u of wave pulse varies with height h from the lower end as : amnzO;ekudh,dlekujLlhn` 1  V > V . 12.29 For a certain stretched string, three consecutive resonance frequencies are observed as 105, 175, 245 Hz respectively. Then select the correct alternative (s) : (A) The string is fixed at both ends (B*) The string is fixed at one end only (C*) The fundamental frequency is 35 Hz (D) The fundamental frequency is 52.5 Hz fdlh[khphgqbZjLlhesarhuØekxrvuquknhvko`frØe'k%105,175,245Hz izsf{krdhtkrhgS]rkslghrF;gksxsaA: (A)jLlhnksuksrjQlsca/khgqbZgS (B*)jLlh,dfljslsaca/khgqbZgS (C*)ewyvko`fr35HzgSA (D)ewyvko`fr52.5HzgSA Sol. As ƒ : ƒ : ƒ = 3 : 5 : 7, string is fixed at one end, Its fundamental frequency is ƒ = ƒ1  105 = 35 Hz 1 2 3 0 3 3 jLlh ,d fljs ls ca/kh gqbZ gS vr% ƒ :ƒ :ƒ =3:5: 7,o ewy vko`fr ƒ = ƒ1  105 =35Hz 1 2 3 0 3 3 SECTION - III : ASSERTION AND REASON TYPE [k.M- III: dFku vkSj dkj.k izdkj 12.30 STATEMENT-1 : In a small segment of string carrying sinusoidal wave, total energy is conserved. STATEMENT-2 : Every small part moves in SHM and in SHM total energy is conserved. dFku : jLlh ds NksVs ls [k.M ftlesa T;k rjaxs xfreku gS] dqy ÅtkZ lajf{kr gSA dkj.k :çR;sd NksVk Hkkx ljyvkorZ xfrdjrk gSvkSj ljyvkorZ xfresa dqyÅtkZ fu;rjgrh gSA (A) If both Assertion and Reason are true and the Reason is correct explanation of the Assertion. (B) If both Assertion and Reason are true, but Reason is not correct explanation of the Assertion. (C) if Assertion is true, but the Reason is false. (D*) if Assertion is false, but the Reason is true. (A) ;fnnksuksdFkurFkkdkj.klR;gSrFkkdkj.k]dFkudhlghO;k[;kdjrkgSA (B) ;fnnksuksdkj.krFkk dFkulR;gSijUrq dkj.kdFkudhlghO;k[;kugh djrkA (C) ;fn dFku lR; gS rFkk dkj.k vlR; gSA (D*) ;fn dFku vlR; gS ijUrq dkj.k lR; gSA Sol. Every small segment is acted upon by forces from both sides of it hence energy is not conserved, rather it is transmitted by the element. vYiHk xdsnksuksarjQvkjksfircydsdkj.kÅtkZlajf{krughjgrhgSvkSj;gblHk xlsikjxfergkstkrhgSA 12.31 STATEMENT-1 : Two waves moving in a uniform string having uniform tension cannot have different velocities. STATEMENT-2 : Elastic and inertial properties of string are same for all waves in same string. Moreover speed of wave in a string depends on its elastic and inertial properties only. oDrO;-1:nksrjaxs,dlekujLlhesaxfrdjjghgSjLlhesaruko,dlekugSrksmudsosxvyx&vyxughgksldrsgSA oDrO;-2:,dlekujLlhesajLlhdsizR;kLFkotM+Rodsxq.klHkhrjaxksadsfy,lekugksrsgSrFk jLlhesarjaxdhpky dsoyizR;kLFkrFk tM+Roxq.kijghfuHkZjdjrhgSA (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) Two waves moving in uniform string with uniform tension shall have same speed and may be moving in opposite directions. Hence both waves may have velocities in opposite direction. Hence statement- 1 is false (D) le:irukookyhle:iMksjhesaxfrdjrhgqbZnksrjaxksadhlekupkygksxhrFk foijhrfn'k vksaesaxfrdjldrh gSAblfy;snksuksarjaxksadsosxfoijhrfn'k esagksldrsgSaAblfy;soDrO;-1vlR;gSaA 12.32 STATEMENT-1 : A standing wave pattern is formed in a string. The power transfer through a point (other than node and antinode) is zero always. STATEMENT-2 : At antinode tension is perpendicular to the velocity. oDrO;-1:,djLlhesavizxkehrjaxdkizfr:icujgkgSAfdlhfcUnq¼fuLiUnoizLiUndsvykok½ls'kfDrlapj.k ges'k 'kwU;gksrkgSA oDrO;-2:izLianijrukoosxdsyEcor~gksrkgSA (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. (False) at node v = 0, at antinode Tension  to velocity  at the points power = 0 (P = At other points P  0. fuLiUn ij v =0, izLiUnksa ijruko osxds yEcor~gSA fcUnqvksaij 'kfä= 0 (P= → → ) → → F.V nwljsfcUnqvksaijP0. SECTION - IV : COMPREHENSION TYPE [k.M-IV:cks/kuizdkj Comprehension - 1 old rrb 11.33, 11.34, 11.35 In a standing wave experiment, a 1.2 kg horizontal rope is fixed in place at its two ends (x = 0 and x = 2.0 m) and made to oscillate up and down in the fundamental mode, at frequency 5.0 Hz. At t = 0, the point at x = 1.0 m has zero displacement and is moving upward in the positive direction of y axis with a transverse velocity 3.14 m/s. ,dvizxkehrjaxç;ksxesa,d1.2fdxzk0dh{kSfrtjLlhnksuksafljksaij(x=0vkSjx =2.0eh0)ck¡/khxbZgSrFk ewy fo/kk esa Åij uhps nksyu dj jgh gS] ftldh vko`fÙk 5.0 gV~tZ gSA t = 0 ij, x = 1.0 eh0 ij fLFkr fcUnq dk foLFkkiu 'kwU; gS vkSj ;g Åij dh vksj / kukRed y fn'kk esa xfreku gS vkSj ftldk vuqçLFk osx 3.14 eh0@ls0 gSA 12.33 Tension in the rope is : (A) 60 N jLlhesarukogS (B) 100 N (C) 120 N (D*) 240 N (A) 60 N (B) 100 N (C) 120 N (D*) 240 N 12.34 Speed of the participating travelling wave on the rope is (A) 6 m/s (B) 15 m/s (C*) 20 m/s (D) 24 m/s jLlhij çxkehrjaxdhpkygS& (A) 6 m/s (B) 15 m/s (C*) 20 m/s (D) 24 m/s 12.35 What is the correct expression of the standing wave equation ? (A*) (0.1) sin (/2)x sin (10 ) t (B) (0.1) sin ()x sin (10 ) t (C) (0.05) sin (/2)x cos (10 ) t (D) (0.04) sin ()x sin (10 ) t vizxkehrjaxdslehdj.kdklghO;atdD;kgS\ (A*) (0.1) sin (/2)x sin (10 ) t (B) (0.1) sin ()x sin (10 ) t (C) (0.05) sin (/2)x cos (10 ) t (D) (0.04) sin ()x sin (10 ) t Sol. µ = 1.2 2 = 0.6 kg/m f = 5 Hz  = 2𝑙 = 4m V = f = 5 x 4 = 20 m/s Ans. 11.34 using v =  y   t   T = 202 x 0.6 = 240 N Ans. 11.33 = 3.14 m/s  max (2A) = 3.14 Amplitude 2A = 3.14 2 x (3.14) x 5 = 0.1 m Equation of standing wave is y = (0.1) sin (/2)x sin (10 ) t Ans. 11.35 Sol. µ = 1.2 2 = 0.6 kg/m f = 5 Hz  = 2𝑙 = 4m V = f = 5 x 4 = 20 m/s Ans. 11.34 v =  T = 202 x 0.6 = 240 N dk mi;ksx djus ij Ans. 11.33  y   t  = 3.14 m/s  max (2A) = 3.14 3.14 vk;ke 2A = 2 x (3.14) x 5 vizxkehrjaxdklehdj.k = 0.1 m y = (0.1) sin (/2)x sin (10 ) t Ans. 11.35 Comprehension - 2 A sinusoidal wave is propagating in negative x–direction in a string stretched along x-axis. A particle of string at x = 2m is found at its mean position and it is moving in positive y direction at t = 1 sec. The amplitude of the wave, the wavelength and the angular frequency of the wave are 0.1meter, /4 meter and 4 rad/sec respectively. x–v{k ds vuqfn'k f[kaph gqbZ ,d Mksjh esa ,d T;k&oØh; (sinusoidal) rjax _.kkRed x-fn'kk esa lapfjr gksrh gSA Mksjh dk ,d d.k le; t = 1 sec ij x = 2m ij viuh ek/; fLFkfr ij ik;k tkrk gS rFkk t = 1 sec ij ;g /kukRed y-fn'kk esa xfr djrk gSA rjax dk vk;ke] rjaxnS/;Z rFkk dks.kh; vko`fÙk Øe'k% 0.1 m, /4 m rFkk 4 rad/sec gSaA 12.36 The equation of the wave is rjaxdklehdj.kgS& (A*) y = 0.1 sin (4t –1)+ 8(x – 2)) (B) y = 0.1 sin (t–1)– (x – 2)) (C) y = 0.1 sin (4t –1)–8(x – 2)) (D) none of these buesa ls dksbZ ugh Sol. The equation of wave moving in negative x-direction, assuming origin of position at x = 2 and origin of time (i.e. initial time) at t = 1 sec. y = 0.1 sin (4 t – 1) + 8 (x – 2)) Shifting the origin of position to left by 2m, that is, to x = 0. Also shifting the origin of time backwards by 1 sec, that is to t = 0 sec. y = 0.1 sin (4 t – 1) + 8 (x – 2)) gy% fLFkfrdkewyfcUnqx=2ijrFk le;dkewyfcUnq(vFkkZr~çkjfEHkdle;)t=1sec.ijekursgq,]_.k Redx–fn'k esaxfrdjrhgqbZrjaxdklehdj.k& y = 0.1 sin (4t + 8x) fLFkfrdsewyfcUnqdkscka;hvksj2m,foLFk firdjusij]vFk Zr~x=0dsfy,Ale;dsewyfcUnqdksHkh1sec,ihNs foLFk firdjusij]vFk Zr~t=0sec.dsfy,] y = 0.1 sin [(4t + 8(x – 2)] 12.37 The speed of particle at x = 2 m and t = 1sec is x = 2 m rFkkt = 1sec ij d.k dh pky gSa & (A) 0.2 m/s (B) 0.6 m/s (C*) 0.4 m/s (D) 0 Sol. As given the particle at x = 2 is at mean position at t = 1 sec.  its velocity v = A = 4 × 0.1 = 0.4  m/s. fn;k x;k d.k t= 1sec.ij x=2ij ek/; fLFkfr ij gSA  bldk osx v = A = 4 × 0.1 = 0.4  m/s. 12.38 The instantaneous power transfer through x=2 m and t= 1.125 sec, is t=1.125secijx=2m ijrkR{kf.kd 'kfDr dk LFkkukUrj.kgS & 4 (A) 10 J/s (B) 3 2 J/s (C) 3 J/s (D*) 0 Sol. Time period of oscillation T = 2  2  1 sec.  4 2 Hence at t = 1.125 sec, that is, at T seconds after t = 1 second, the particle is at rest at extreme position. 4 Hence instantaneous power at x = 2 at t = 1.125 sec is zero. nksyu dk vkorZdky T = 2  2  1 sec.  4 2 blfy;s t=1.125sec,vFkkZr~ t=1sec.dsckn T secondsij] d.kpje (vfUre)fLFkfr ijfojke ijgksxkA blfy;s x=2ijt=1.125secijrkR{kf.kd'kfDr'kwU;gksxhA SECTION - V : MATRIX - MATCH TYPE [k.M- V:eSfVªDl&lqesy izdkj 12.39 Match the column : Column–I Column–II (A) (p) Speed of component travelling wave is portion Two strings each of length 𝑙 and linear mass AP will be density  and 9 are joined together and system is oscillated such that joint P is node T is tension in the strings. A and B are fixed ends. (B) (q) Speed of component travelling wave in the Two strings each of length 𝑙 and linear mass portion AP will be more than that in portion BP. density  and 9 are joined together and system is oscillated such that joint P is antinode. T is tension in each string.A and B are fixed ends. (C) (r) Frequency of oscillation of the system AB can P is the mid–point of the string fixed at both ends. be T is tension in the string and  is its linear mass density. (D) (s) Frequency of oscillation of the system AB can T is the tension in the string fixed at A and B is free be end. P is mid–point.  is its the linear mass density. (t) Wavelength of the wave in the portion PB can 2𝑙 fuEu dks lqesfyr dhft, : be 3 . LrEHk–I LrEHk–II (A) (p) AP Hkkx esa lapfyr rjax ?kVd dh pky 𝑙 yEckbZ rFkk js[kh; nzO;eku ?kuRo  o 9 dh nks Mksfj;k ,d nwljs ls tqM+h gqbZ gS rFk fudk;blçdkjnksyudjrkgSfdfcUnq PfuLiUngST,Mksfj;ksaesarukogSAArFk BfLFkj (tM+or~)fljsgSA gksxhA (B) (q)APHk xesalapfjrrjax?kVdhdhpkyHk xBP 𝑙yEckbZrFk js[kh;nzO;eku?kuRoo9dh esalapfjrrjax?kVddhpkylsvf/kdgksrhgSA nksMksfj;k,dnwljslstqM+hgqbZgSrFk fudk;bl çdkjnksyudjrkgSfdfcUnqPçLiUngST, Mksfj;ksaesarukogSAArFk BfLFkj(tM+or~)fljsgSA (C) (r)fudk;ABdsnksyudhvko`fÙk Pnksuksafljksalsca/khMksjhdke/;fcUnqgSA gksldrh gSA TMksjhesarukogSrFk bldkjs[kh;nzO;eku?kuRogSA (D) (s) fudk;ABdsnksyudhvko`fÙk Alsca/khrFk BfljslseqDrMksjhesarukoTgSAP gksldrhgSA e/;fcUnqgSAbldkjs[kh;nzO;eku?kuRogSA (t) PB Hkkx esa rjax dh rjaxnS/;Z 2𝑙 3 gks ldrh gSA Ans. (A) – p,q,r,t ; (B) – p,q,s ; (C) – p,r,s,t ; (D) – p,s 12.40 Match the statements in column- with the statements in column-. Column-I Column-II (A) A tight string is fixed at both ends and (p) At the middle, antinode is formed sustaining standing wave in odd harmonic (B) A tight string is fixed at one end and (q) At the middle, node is formed free at the other end in even harmonic (C) A tight string is fixed at both ends and (r) the frequency of vibration is 300% vibrating in four loops more than its fundamental frequency (D) A tight string is fixed at one end and (s) Phase difference between SHMs of any free at the other end, vibrating in 2nd two particles will be either  or zero. overtone (t) The frequency of vibration is 400% more than fundamental frequency. LrEHk-esafn;sx;sdFkuksadksLrEHk-lslqesfyrdfj;sA LrEHk-I LrEHk-II (A) ,d ruh gqbZ jLlh nksuksa fljks ij tM+or gS rFkk (p) fo"ke lauknh esa e/; ij (at the middle) blesa ,d vizxkeh rjax cuh gqbZ gSA izLiUn cusxkA (B) ,d ruh gqbZ jLlh ,d fljs ls tM+or gS o (q) le lauknh esa e/; ij fuLiUn curk gSA nwljs fljs ij eqDr gSA (C) ,d ruh gqbZ jLlh nksuks fljks ij tMor gS (r) d Eiu dh vko`fr bldh eqy vko`fr ls 300% vkSj pkj ywiks esa dEiu dj jgh gS vf/kd gksxh (D) ,d ruh gqbZ jLlh ,d fljs ij tMor gS vkSj nwljs (s) fdUgh nks d.kksa ds SHM¼ljy vkorZ xfr½ fljs ij eqDr gS rFkk ;g nwljs vf/kLojd esa dEiu e s a dykUrj  ;k 'kwU; gksxkA dj jgh gS Ans. (A) p,q,s (B) s (C) q,r,s (D) s,t (t) dEiu dh vko`fr] ewy vko`fr ls 400% vf/kdgksxhA Sol. (A) Number of loops (of length /2) will be even or odd and node or antinode will respectively be formed at the middle. Phase difference between two particle in same loop will be zero and that between two particles in adjacent loops will be . (B) and (D) Number of loops will not be integral. Hence neither a node nor an antinode will be formed in in the middle. Phase difference between two particle in same loop will be zero and that between two particles in adjacent loops will be . (A) rFkk(C) ;fnywiks (yEckbZ/2) dhla[;k le;k fo"kegksus ijØe'k% fuLiUno izLiUne/; ijcurk gSA ,dghlekuywiesad.k sadse/;dykUrj'kwU;rFk lehiorhZywidsd.k sadslkFkdykUrjgksxkA (B) rFkk(D)ywiksdhla[;kiw.k Zadughgksxhvr%e/;ijurksfoLiUn]ughizLiUncusxkA ,dghlekuywiesad.k sadse/;dykUrj'kwU;rFk lehiorhZywidsd.k sadslkFkdykUrjgksxkA 12.41 Match the following : Column-I Column-II (A) In refraction (p) Speed of wave does not change (B) In reflection (q) Wavelength is decreased (C) In refraction from rarer to denser medium (r) Frequency does not change (D) In reflection from a denser medium (s) Phase change of  takes place (t) Phase does not change fuEudkslqesafyrfdft,% LrEHk-I LrEHk-II (A) viorZuesa (p) rjax dh pky ugh cnyrh (B) ijkorZueas (q) rjaxnS/;Z?kVtkrhgSA (C) fojyek/;elsl?kuek/;eesviorZuls (r) vko`frughcnyrh (D) l?kuek/;elsijkorZuls (s) dkdykUrjgksrkgSA (t) dykifjofrZrughgSA Ans. (A) r, t (B) p,r (C) q, r, t (D) p, r, s Sol. for refraction medium changes so frequency does not change beacuse frequency depends on sources. for reflection medium does not change so speed of wave and wave length does not change refraction from rarer to denser medium wave length decreases beacuses density of denser medium high but frequency does not change. reflection from denser medium speed and frequency does not change beause medium is same but due to reflection change of . takes place viorZuesaek/;ecnytkrkgSAvr%vko`frughcnyrhD;ksfdvko`frL=k srijfuHkZjdjrhgSA ijkorZuesaek/;eughcnyrkblfy,pkyrFkkrjaxnS/;ZughcnyrhgSA fojyek/;elsl?kuek/;eeasviorZulsrjaxnS/;ZdegkstkrhgSD;ksfdl?kuek/;edk?kuRoT;knkgksrkgSA ijUrqvko`frughcnyrhgS l?ku ek/;e ls ijkorZu ls ek/;e ugh cnyrk blfy, pky rFkk vko`fr ugh cnyrh ijUrq ijkofrZr rFkk vifrr rjaxks ds chp  dk dkykUrj vk tkrk gSA SECTION - VI : INTEGER TYPE [k.M-VI:iw.kkZadizdkj 12.42 A 40 cm long wire having a mass 3.2 gm and area of cross section 1 mm2 is stretched between the support 40.05 cm apart. In its fundamental mode. It vibrate with a frequency 1000/64 Hz. the young’s modulus of the wire is 10x N/m2 then x is. ,d 40 lseh0 yEcs rkj] ftldk æO;eku 3.2 xzke gS vkSj vuqçLFk dkV dk {ks=kQy 1 eh0eh02 gS] dks 40.05 lseh0 nwj nks vkyEcuksa ds chp cka/kk x;k gSA ewy fo/kk esa ;g 1000/64 gVtZ dh vko`fÙk ds lkFk dEié djrk gSA rkj ds fy, ;ax xq.kkad 10x N/m2 gksrksxgksxkA Ans. 1 Sol. ewyfo/k dsfy;s µ = 3.2 gm 40 cm 3.2  103 = 40 10 2 3.2 = 40 32 = 4000 kg/m  𝑙 = 2   = 2𝑙 ...........(1) v f =  1 = 2𝑙 1000  64 = 1 2  40 102 1000   2 32   64  2  40 10 2  4000 = T 10000 64 32 × 4000 10 / 8 10 = T  T = 8 N vc y = 106 .05  102 40  102 107 = 8 40 (.05) = 109 N/m2. [ Ans. 1 × 109 Nm2 ] 12.43 A nonuniform string of mass 45 kg and length 1.5 m has a variable linear mass density given by  = kx, where x is the distance from one end of the string and k is a constant. Tension in the string is 15 N which is uniform. Find the time (in second) required for a pulse generated at one end of the string to travel to the other end. ,d vleku jLlh (nzO;eku 45 kg, yEckbZ 1.5m) ftldk js[kh; nzO;eku ?kuRo ifjorZu'khy gS rFkk  = kx, ls O;Dr fd;k tkrk gS tgk¡ x jLlh ds ,d fljs ls nwjh gS rFkk k ,d fLFkjkad gSA jLlh esa ruko 15 N gS tks ,dleku gSA jLlh ds ,d fljs ij mRiUu LiUn dks nwljs fljs rd tkus esa yxk le; ¼lSd.M esa½ Kkr djksA Ans. 2 dM Sol.  = Kx = dx M 𝑙 2M  dM = Kx dx and K = 𝑙2 O O V = = dx 𝑙 = dt O dx = t dt o ∴ t = = = = = 2. 12.44 The equation of a string wave is given by (all quantity expressed in S.I. units) Y = 5 sin10 (t – 0.01x) along the x-axis. The magnitude of phase difference between the points separated by a distance of 10 m along x- axis is x then x is ,d rjaxdh lehdj.k¼lHkhjkf'k;k¡S.I.bdkbZesagSa½Y =5sin10(t– 0.01x) ls x-v{kdsvuqfn'knh tkrhgSA x v{k ds vuqfn'k 10 m nwj fLFkr fcUnqvksa ds e/; dykUrj dk ifjek.k x gS rks x D;k& Ans. 1 Sol. The magnitude of phase difference between the points separated by distance 10 metres 10ehVjnwjfLFkrfcUnqvksadse/;dykUrjdkifjek.k = k × 10 = [10 × 0.] × 10 = 