11-Simple Harmonic Motion

SECTION - I : STRAIGHT OBJECTIVE TYPE [k.M-I: lh/ks oLrqfu"B izdkj 11.1 A block of mass ‘ m ‘ is attached to a spring in natural length of spring constant ‘ k ‘ . The other end A of the spring is moved with a constant velocity v away from the block . Find the maximum extension in the spring. m æO;ekudk,dxqVdk,d fLçaxls tksM+kx;k gSftldkfLçaxfu;rkadkgSAfLçaxdknwljkfljk Avpjosx v lsxqVdslsnwj[khapktkjgkgSAfLçaxdhyEckbZesavf/kdreo`f)D;kgksxh& (A) 1 4 (B*) 1 (C) 2 (D) 2 Sol. Consider an observer moving with speed v with point A in the same direction. //////////////////////////////////////////////////////// In the frame of observer, block will have initial velocity v towards left. v //////////////////////////////////////////////////////// During maximum extension, the block will come to rest with respect to the observer. Now, by energy conservation, 1 mv2 = 1 2 2 2 max max = Sol. ekuk]çs{kdosxv lsfcUnqAdslkFkmlhfn'k esapyjgkgSA v //////////////////////////////////////////////////////// çs{kddsÝselsxqVdkck;havksjizkjfEHkdpkyv lspyrkçrhrgksxkA v //////////////////////////////////////////////////////// egÙkeo`f)dsnkSjkuxqVdkçs{kddslkis{kfLFkjvoLFk esagksxk]vr%ÅtkZlaj{k.kds}kjk 1 mv2 = 1 k x 2 2 2 max  xmax = 11.2 A system is shown in the figure. The time period for small oscillations of the two blocks will be. fp=kesafn;sgq,fudk;esanksuksaxqVdksadkNksVsnksyuksadsfy,vkorZdkyD;kgksxk& (A) 2  (B) 2  (C*) 2  (D) 2  Sol. Both the spring are in series eq = K(2K) 2K K  2K = 3 Time period T = 2 where  = m1m2 m1  m2 Here  = m  T = 2 2 = 2 Method II  mx = mx  x = x 2 force equation for first block; 2k 3 (x1 + x2 ) = –m d2 x1 dt2 d2 x1 Put x1 = x2  dt2 4k + 3m × x = 0  2 = 4k 3m  T = 2 Sol. nksuksafLçaxJs.khØeesagS eq vkorZdky K(2K) 2K = K  2K = 3 T = 2 tgk¡  = m1m2 m1  m2 ;gk¡  = m  T = 2 2 = 2 f}rh;fof/k  mx = mx2  x = x igys xqVds ds fy, cy dk lehdj.k 2k 3 (x1 + x2 ) = –m d2 x1 dt2 d2 x1 x1 = x2 j[kus ij  dt2 4k + 3m × x = 0  2 = 4k 3m  T = 2 11.3 Two particles P and Q describe S.H.M. of same amplitude a, same frequency f along the same straight line from the same mean positon. The maximum distance between the two particles is a phase difference between the particle is: . The nks d.k P rFkk Q leku vk;ke a rFkk leku vko`fÙk f ds lkFk leku ljy js[kk ds vuqfn'k leku ek/;fLFkfr ls ljy vkorZ xfr djrs gSaA d.kksa ds e/; vf/kdre nwjh a jgrh gSaA d.kksa ds e/; dykUrj gksxk & (A) zero (B*) /2 (C) /6 (D) /3 Sol. (B) x = a sin(t +  ) x = a sin(t +  )  |x – x | = 2a sin(t + 1  2 ) cos  1  2  2  2  To maximize |x1 – x2| : sin(t + 1  2 ) = 1 2  1  2   1  2   = 2a × 1 × cos    2   = cos    2    4 = 1  2 2   –  =  . 2 Hence (B). Sol. (B) x = A sin(t +  ) x = A sin(t +  )  |x – x | = 2A sin(2t + 1  2 ) cos  1  2  2  2  |x1 – x2| dks vf/kdre djus ds fy, sin(2t + 1  2 ) = 1 2  1  2   1  2   = 2a × 1 × cos    2   = cos    2    4 = 1  2 2   –  =  . 2 blfy, (B). 11.4 The time taken by a particle performing SHM on a straight line to pass from point A to B where its veloci- ties are same is 2 seconds. After another 2 seconds it returns to B. The time period of oscillation is (in seconds): ljyjs[k esaljyvkorZxfrdjjgs,dd.k}kjkAlsBfcUnqtgk¡bldsosxlekugS]rdigqapusesayxkle;2lSd.M gSAvxys2lSd.Mckn;gBijykSVrkgSAnksyudkvkorZdky¼lSd.M+esa½gS& (A) 2 (B) 4 (C) 6 (D*) 8 Sol. From the given information it can be inferred that points A and B are equidistant from mean position Hence from diagram it is clear that time period of oscillation is = 2 + 2 × 2 + 2 = 8 second. nhxbZlwpukls;gfu"d"kZfudyrkgSfdArFk BfcUnqek/;fLFkfrlslekunwjhijgSvr%fp=klsLi"VgSfd nksyu dk vkorZdky = 2 + 2 × 2 + 2 = 8 lsd.MA 11.5 A particle performs S.H.M. on xaxis with amplitude A and time period T. The time taken by the particle to travel a distance A/5 starting from rest is: ,dd.k x- v{kij ljyvkorZ xfrdj jgkgS ftldkvk;ke A rFkk vkorZdky TgSAd.k }kjkfojke lspyrs gq, A/5nwjhr;djuses ayxkle;gSA T T  4  T  1  T  1  (A) 20 (B*) 2  cos   5 (C) 2  cos   5 (D) sin   2  5       Sol. Method : Ι Particle is starting from rest, i.e. from one of its extreme position. A As particle moves a distance 5 , we can represent it on a circle as shown. d.kfLFkjvoLFk lspyukçkjEHkdjrkgSvFk ZrviusfdlhpjelhekUrfcUnqlspyukçkjEHkdjrkgSA A tSlsfdd.k] 5 nwjhr;djrkgS]bls,do`Ùkijn'kkZ;ktkldrkgSA cos  = 4A / 5  4 A 5  = cos–1  4    5   4 1 4 x –1 –1 t = cos   t = cos   T = 2  5   4  cos–1  5    5    Method : ΙΙ As starts from rest i.e. from extreme position x = A sin (t +  ) f}rh; fof/k% pwafd d.k fLFkjkoLFkk ls xfr djrk gS vFkkZr lhekUr fcUnq ls xfr djrk gSA x = A sin (t +  )  At t = 0 ij ; x = A   = 2  A – A 5 = A cos t 4 4 5 = cos t  t = cos–1 5 T t = 2  4  cos–1  5    11.6 A rod of length 𝑙 is in motion such that its ends A and B are moving along x-axis and y-axis respec- d tively. It is given that dt = 2 rad/s always. P is a fixed point on the rod. Let M be the projection of P  on x-axis. For the time interval in which  changes from 0 to 2 , choose the correct statement, 𝑙 yEckbZ dh NM+ bl izdkj xfr'khy gS fd blds fljs A rFkk B Øe'k% x-rFkk d y-v{k ds vuqfn'k xfr'khy gSA fn;k gS fd ges'kk dt = 2 jsfM;u@lsd.M gSA P NM+ ij fLFkj fcUnq gSA ekuk x- v{k ij fcUnq P dk iz{ksi M gSA  esa 0 ls  rd ifjorZu ds vUrjky ds nkSjku lgh dFku NkaVks & (A) The acceleration of M is always directed towards right (B*) M executes SHM (C) M moves with constant speed (D) M moves with constant acceleration (A)M dkRoj.kges'k nka;hrjQgksxkA (B*) M ljy vkorZ xfr djsxkA (C) M fu;r pky ls xfr djsxkA (D) M fu;r Roj.k ls xfr djsxkA y Sol. gy% d dt = 2   = 2t Let BP = a  x = OM = a sin = a sin(2t) Hence M executes SHM within the given time period and its acceleration is opposite to ‘x’ that means towards left x y d dt = 2   = 2t ekuk BP = a  x = OM = a sin = a sin(2t) pwafd M,fn;sx;s vkorZ dky dsfy, SHM djsxkrFkk bldk Roj.kxdsfoijhrgksxkvFk Zrcka;hrjQgksxkA x 11.7 The coefficient of friction between blocks of mass m and 2m is  = 2 tan . There is no friction between block of mass 2m and inclined plane. The maximum amplitude of two block system for which there is no relative motion between the blocks. m rFkk 2m nzO;eku ds xqVdksa ds e/; ?k"kZ.k xq.kkad = 2 tan  gSA urry rFkk 2m nzO;eku ds xqVds ds e/; ?k"kZ.k ugha gSA tc nkasukas xqVdksa ds e/ ; dksbZ lkis{k xfr ugha gks rks f}sxqVdk fudk; dk vf/kdre vk;ke gSA (A) g sin  Sol. (C) mgsin (B) k 3mgsin (C*) k (D) Noneof thesebuesa ls dksbZ ugh The maximum static frictional force is vf/kdreLFkSfrd?k"kZ.kcygS f = µmg cos = 2 tan  mg cos  = 2 mg sin Applying Newton’s second law to upper block at lower extreme position mPpCykWd dhfuEupjefLFkfrijU;wVudkf}rh; fu;eyxkus ij f – mgsin = m2A  f = m2A + mg sin  3mgsin or;k 2A = g sin  or A = k 11.8 A block of mass ‘m’ is suspended from a spring and executes vertical SHM of time period T as shown in figure. The amplitude of the SHM is A and spring is never in compressed state during the oscillation. The magnitude of minimum force exerted by spring on the block is ,dCykWd ftldknzO;eku ‘m’gSbldks ,dfLizax lsyVdk;k tkrkgS o ;g Å/okZ/kj ljy vkorZxfr dj jgk gS ftldkvkorZdkyfp=k uqlkjTgSAnksyudkvk;keAgSrFk fLizaxnksyudsnkSjkudHkhHkhlaihfMrvoLFk esaugh gksrhgSAfLizax}kjkCykWdijvkjksfirU;wurecydkifjek.kgS& (A*) mg  42 m A T2 (B) mg  42 m A T2 2 (C) mg  T2 m A 2 (D) mg  T2 m A Sol. The spring is never compressed. Hence spring shall exert least force on the block when the block is at topmost position. fLizaxdHkhHkhlaihfMrvoLFk esaughvkrhgSAvr%fLizaxCykWdijU;wurecyrHkhyxk;sxhtcCykWdviuhmPpre fLFkfrijgksA Fleast = kx0 – kA = mg – m2A = mg – 4  mA T 2 11.9 Graph shows the x(t) curves for three experiments involving a particular spring- block system oscillat- ing in SHM. The kinetic energy of the system is maximum at t = 4 sec. for the situation : rhuiz;ksxksaftuesafLizaxCykWdfudk;ljyvkorZxfresanksyudjjgkgSdsfy,x(t)oØdksxzkQ¼ys[kkfp=k½ esa iznf'kZrfd;k x;k gSAfdl fLFkfr esale; t= 4lSd.Mijfudk;dh xfrtÅtkZvf/kdregksxhA (A*) 1 (B) 2 (C) 3 (D) Sameinall lHkh esa leku Sol. At t = 4 x = 0 i.e. particle must pass through the mean position so curve (3) is not correct The slope of x–t graph is the speed Among curves (1) and (2), slope of (1) greater at t = 4 so (A) is correct t = 4, x = 0 gSA vFk Zr~d.kek/;fLFkfrlsxqtjrkgSvr%oØ(3)lghughagSA x - t oØ dk wyjgh,dehVjNM+f0vko`fÙklsNksVsnksyufp=k uqlkjdjrhgSAvxj NM+ dk uhps dk vk/kk Hkkx dkV fn;k tk, rks NksVs nksyuksa dh u;h vko`fÙk D;k gksxh & (A) f0 (B*) f0 (C) 2f0 (D) 2 f0 1 Sol. f0 = 2 where, 𝑙 is distance between point of suspension and centre of mass of the body. Thus, for the stick of length L and mass m : tgk¡𝑙yVdufcUnqonzO;ekudsUnzdse/;nwjhgSA vr% NM+ dh yEckbZ Lo nzO;eku m ds fy, 1 f0 = 2 1 = 2 when bottom half of the stick is cut off tc NM+dk fupykvk/kk fgLlk dkV fn;ktkrk gSA 1 f0’ = 2 = 1  2 = f0 Ans. 11.14 A straight rod of negligible mass is mounted on a frictionless pivot and masses 2.5 kg and 1 kg are sus- pended at distances 40 cm and 100 cm respectively from the pivot as shown. The rod is held at an angle  with the horizontal and released. ux.; nzO;eku dh lh/kh NzM+ ?k"kZ.k jfgr dhyfdr fcUnq ij j[kh gS rFkk fp=kkuqlkj 2.5 fdxzk0 rFkk 1 fdxzk ds nzO;eku dhyfdr fcUnq ls Øe'k% 40 lseh0 rFkk 100 lseh0 nwjh ij yVdk;s x;s gSA NM+ dks {kSfrt ls  dks.k ij ys tkdj eqDr fd;k tkrk gS& (A) The rod executes periodic motion about horizontal position after the release. eqDrdjusdsi'pkrNM+]{kSfrtfLFkfrdsifjr%nksyuxfrdjsxhA (B*) The rod remains stationary after the release. eqDrdjusdsi'pkr~]NM+fLFkjjgsxhA (C) The rod comes to rest in vertical position with 2.5 kg mass at the lowest point NM+Å/okZ/kjfn'k esafLFkjvoLFk esavk;sxhrFk 2.5fdxzk0nzO;ekufuEurefcUnqijgksxkA (D) The rod executes periodic motion about vertical position after the release. eqDrdjusdsi'pkr~NM+]Å/okZ/kjfLFkfrdsifjr%vkorZxfrdjsxhA Sol. Torque about hinge 2.5 g. (0.4)cos – 1g.(1) cos = 0 So rod remains stationary after the release. Sol. dhyfdarfcUnqdsifjr%vk?kw.kZ 2.5 g. (0.4)cos – 1g.(1) cos = 0 blfy,NksM+usdscknNM+fLFkjgksxhA 11.15 A particle of mass m = 2 kg executes SHM in xy-plane between points A and B as shown in figure, under action of force → F  Fx i + Fy ˆj . Minimum time taken by particle to move from A to B is 1 sec. At t = 0 the particle is at x = 2 and y = 2. Then Fx as function of time t is fp=kkuqlkj nzO;eku m=2kgdk ,dd.k ArFkkBdschp xy-ry esa]cy →  F ˆi + F ˆj dsvUrxZr ljyvkorZ xfr F x y djrk gSAAlsBrd xfr djus esad.k }kjk fy;k x;k U;wure le; 1sec.gSAt=0ijd.k x=2rFkk y=2ijgS rks le;t ds Qyu ds :i esa Fx gS – (A) – 42 sin t (B*) – 42 cos t (C) 42 cos t (D) NoneofthesebuesalsdksbZugha Sol. Let the line joining AB represents axis ‘r’. By the conditions given ‘r’ coordinate of the particle at time t is r = 2 cos t 2  = T = 2 2 =   r = 2 cos t r x = r cos 45° = 2 = 2 cos t  a = – 2 x = – 2 2 cos t  F = max = – 42 cos t Sol. ekukArFk Bdksfeykusokyhjs[k v{k‘r’gSAfn;sx;sçfrcU/kls]tle;ijd.kdkfunsZ'k ad‘r’gSA r = 2 cos t 2  = T = 2 2 =   r = 2 cos t x = r cos 45° = = 2 cos t  a = – 2 x = – 2 2 cos t  F = max = – 42 cos t 11.16 Two masses m1 and m2 (m1 > m2) are suspended by two springs vertically and are in equilibrium, extensions in the springs were same. Both the masses are displaced in the vertical direction by same distance and released. In subsequent motion T1, T2 are their time periods and E1, E2 are the energies of oscillations respectively then : nks nzO;eku m1 rFkk m2 (m1 > m2) nks fLizaxksa ls Å/oZ yVdk;s x;s gS rFkk lkE;koLFkk esa gSaA nksuksa fLizax esa f[kapko leku gSA nksuksa nzO;ekuksa dks Å/oZ fn'kk esa leku nwjh rd foLFkkfir djds eqDr NksM+ fn;k tkrk gSA vkxkeh xfr esa T1, T2 muds vkorZdky rFkk E1, E2 Øe'k% nksyuksa dh ÅtkZ gS rks & (A) T1 = T2 ; E1 < E2 (B) T1 > T2 ; E1 > E2 (C) T1 < T2 ; E1 > E2 (D*) T1 = T2 ; E1 > E2 Sol. K1 = m g ; k 1 x 2 g = m2 x Thus, Using T = 2  T = 2   T m1 (m1 g/ x) = T T = 2  1 2 Energy of oscillation : 1 E = 2 m 2 A2 Since  and A are same for both and m > m  E > E . Sol. K1 = m g ; k 1 x 2 g = m2 x bl izdkj, T = 2 mi;ksx djus ij T = 2   T = T2 T = 2  nksyudhÅtkZ 1 E = 2 m  A 2 2 ,oaAnksuksadsfy,lekugS,oam1 >m2  E1 > E2. SECTION - II : MULTIPLE CORRECT ANSWER TYPE [k.M- II: cgqy lgh fodYi izdkj 11.17 A particle is subjected to two simple harmonic motions along x and y directions according to, x = 3 sin 100  t; y = 4 sin 100  t. (A) Motion of particle will be on ellipse traversing it in clockwise direction. (B*) Motion of particle will be on a straight line with slope 4/3. (C*) Motion will be a simple harmonic motion with amplitude 5. (D) Phase difference between two motions is /2. Sol. x = 3 sin 100 t y = 4 sin 100 t Equation of path is y  4 x 3 i.e. y  4 x 3 4 which is equation of a straight line having slope 3 Equation of resulting motion is → r  xi  yj Amplitude is = (3ˆi  4ˆj) = 5 sin 100t ,dd.kdksx- v{kvkSjy-v{kdhfn'k esanksljyvkorZxfrdjk;htkrhgSftldklehdj.kx = 3sin100t vkSj y = 4 vkSj sin 100  t gS& (A) d.k nh?kZ o`Ùkh; iFk ij nf{k.kkorZ fn'kk esa xfr djsxkA (B*) d.kljyjs[kkij]ftldhço.krk4/3gS]xfrdjsxkA (C*) ;g,dljyvkorZ xfrgksxhftldkvk;ke5gksxkA (D) nksuksaxfr;ksadkdykUrj/2gSA Sol. x = 3 sin 100 t y = 4 sin 100 t iFkdklehdj.kgS y  4 vFkkZr~ y  4 x x 3 3 4 tks ,d ljy js[kk dk lehdj.k gS ftldh ço.krk 3 lehdj.k gksxk gSA ifj.kkeh xfr dk ifj.k ehxfrdklehdj.kgksxk → r  xi  yj vk;ke = (3ˆi  4ˆj) = 5 sin 100t 11.18 The speed v of a particle moving along a straight line, when it is at a distance (x) from a fixed point of the line is given by v2 = 108 – 9x2 (all quantities are in cgs units) : ,d d.k tks fd ,d ljy js[kk ds vuqfn'k xfr dj jgk gS dh pky 2 = 108– 9x2 ls iznf'kZr dh tkrh gS tc ;g d.kljyjs[k ijfLFkrfdlhfuf'prfcUnqlsx nwjhijgSA(leLrjkf'k;k¡cgsiz.k yheasgSa) (A) the motion is uniformly accelerated along the straight line ljyjs[k dsvuqfn'kxfrRofjrgksxhA (B*) the magnitude of the acceleration at a distance 3cm from the fixed point is 27 cm/sec2 fuf'prfcUnqls3lsehdhnwjhijRoj.kdkifjek.k27lseh@lS2 gksxk (C*) the motion is simple harmonic about the given fixed point. fn;sx;sfuf'prfcUnqdslkis{kxfrljyvkorZxfrgksxhA (D) the maximum displacement from the fixed point is 4 cm. fuf'prfcUnqlsvf/kdrefoLFk iu4lsehgksxkA Sol. v2 = 108 – 9x2 2vdv dx = – 18 x ⇒ acc. A = –9x (non-uniform) vleku at x = 3cm ij a = –27 or |a| = 27cm/s2 also a = – 9x is a S.H.M equation so particle perform S.H.M about the give fixed point rFk a=–9x,dl-vk-x-lehdj.kgSvr%d.kfn;sx;sfLFkjfcUnqdslkis{kl-vk-x-djsxkA V is maximum at x = 0 V, x= 0ij vf/kdre gSA and V is Zero at x = rFk V,x= ij'kwU;gSA So Amplitude = 2 cm vr% vk;ke = 2 3 cm 11.19 A horizontal plank has a rectangular block placed on it. The plank starts oscillating vertically and simple harmonically with an amplitude of 40 cm. The block just loses contact with the plank when the latter is at momentary rest. Then : ,d{kSfrtiV~Vsij,dvk;rkdkjCykWdj[k gSAiV~VkÅ/oZfn'kkesa40lseh-dsvk;kelsljyvkorhZ:ils nksyudjukçkjEHkdjrkgSAtcxfrdsnkSjkuiV~Vk{kf.kd:ilsfLFkjkoLFk esavkrkgSrksfi.MdkiV~Vsls laca/kgVtkrkgSrks:  2  (A*) the period of oscillation is  5   2  nksyuksadkvkorZdky  5  gksxkA     (B*) the block weighs double its weight, when the plank is at one of the positions of momentary rest. tcfi.Mbldhfdlh,d{kf.kdfojkekoLFk dhfLFkfresagksrkgSrksfi.MdkHk jnqxqukgkstk,xkA (C*) the block weighs 0.5 times its weight on the plank halfway up fi.M dk Hkkj vk/kk gksxk tc iV~Vk vk/kh ÅapkbZ ij gksxkA (D*) the block weighs 1.5 times its weight on the plank halfway down fi.MdkHk jMs<+xqukgksxktciV~Vkuhpsdhvksjvk/khnwjhijgksxkA (E*) the block weighs its true weight on the plank when the latter moves fastest fi.MdkHkkj bldkokLrfodlR;HkkjgksxktciV~Vkrhozreosxls pyjgkgksxkA Sol. Given fn;k gS A= 0.4m , and vkSj a= g 10 so vr% 2A = g ⇒ 2 = 0.4 = 25   = 5 T = 2 = 2/5 sec.lSDM+ At lowest position acceleration.fuEure fcUnq ij Roj.k = 2A + g = g + g = 2g So weight vr% Hkkj = m (2g) = 2mg athalf distance vk/kh nwjh ij a=g/2 So weight at upper half distance vr% Hkkj Åijh vk/kh nwjh ij =m(g-g/2) =mg/2 andweightatlower halfdistance rFkk fupyh vk/kh nwjh ij Hkkj 3mg = m(g + g/2) = 2 actual weight at equilibrium position (maximum v) lghHk jlkE;oLFk fLFkfrij¼vf/kdreosx½ 11.20 A 20 gm particle is subjected to two simple harmonic motions ,d20xzkedsd.kijfuEunksljyvkorZxfr;k¡v/;kjksfirgSa& x1 = 2 sin 10 t,  x2 = 4 sin (10 t + 3 ). where x1 & x2 are in metre & t is in sec.  x2 =4sin(10t+ 3 ).tgk¡x1 ox2 ehVj esa rFkktlsd.M esa gS (A*) The displacement of the particle at t = 0 will be 2 m. t=0ijd.k dkfoLFkkiu 2 eh-gksxkA (B*) Maximum speed of the particle will be 20 m/s. d.k dhegRre pky 20 eh-@ls-gksxhA (C*) Magnitude of maximum acceleration of the particle will be 200 d.kdsegRreRoj.kdkifjek.k200 7 eh-@ls-2gksxkA (D*) Energy of the resultant motion will be 28 J. ifj.k ehxfrdhÅtkZ28JgksxhA Sol. At t = 0 m/s2.  Displacement x = x1 + x2 = 4 sin 3 Resulting Amplitude A = Maximum speed = A = 20 m/s = 2 m.   = 2 m Maximum acceleration = A2 = 200 m/s2 1 Energy of the motion = 2 m2 A2 = 28 J Ans. gy% t = 0 ij  foLFkkiu x = x1 + x2 = 4 sin 3 = ifj.kkeh vk;ke A= egRre pky = A = 20 m/s egRre Roj.k = A2 = 200 m/s2 1 2 m.   = 2 m xfr dh ÅtkZ = 2 m2 A2 = 28 J Ans. 11.21 The position of a particle w.r.t. origin varies according to the relation x = 3 sin 100t + 8 cos2 50t. Which of the following is/are correct about this motion. ,d d.k dh fLFkfr ewy fcUnq ds lkis{k lEcU/k x = 3 sin 100t + 8 cos2 50t ds vuqlkj ifjofrZr gksrh gSA bl xfr ds ckjs esa dkSuls dFku lgh gSA (A) themotionof theparticle isnot S.H.M. (d.k dh xfr ljy vkorZ xfr ugha gSA) (B*) the amplitude of the S.H.M. of the particle is 5 units d.kdhljy vkorZ xfr dkvk;ke 5bdkbZgSA (C) the amplitude of the resultant S.H. M. is 73 units ifj.kkehljyvkorZxfrdkvk;ke bdkbZgSA (D*) the maximum displacement of the particle from the origin is 9 units. ewyfcUnqlsd.kdkvf/kdrefoLFk iu9bdkbZgSA Sol. x = 3 sin 100 t + 8 cos2 50 t 8[1 cos100t] = 3 sin 100 t + 2 x = 4 + 3 sin 100 t + 4 cos 100 t tan   4  (x – 4) = 5 sin (100t + )     Amplitude = 5 units Maximum displacement = 9 units. vk;ke=5bdkbZ vf/kdrefoLFkkiu =9bdkbZ 11.22 Two blocks A (5kg) and B(2kg) attached to the ends of a spring constant 1120N/m are placed on a smooth horizontal plane with the spring undeformed. Simultaneously velocities of 3m/s and 10m/s along the line of the spring in the same direction are imparted to A and B then nksxqVdsA(5kg)oB(2kg)fLizaxfu;rkad1120N/mds,dfLizaxdsfljksaijtqM+sgq,,dfpdus{kSfrtryijj[ksgSaA fLizaxvfoÑrgSAAoBdks,dlkFklekufn'k esafLizaxdhjs[k dsvuqfn'kØe'k%3m/so10m/sdsosxfn;stkrs gSarks- (A) when the extension of the spring is maximum the velocities of A and B are zero. (B*) the maximum extension of the spring is 25cm. (C*) the first maximum compression occurs 3/56 seconds after start. (D*) maximum extension and maximum compression occur alternately. (A)tcfLizaxdkfoLrkjvf/kdregSrksAoBdsosx'kwU;gSA (B*)fLizaxdkvf/kdrefoLrkj25cmgSA (C*)izFkevf/kdrelEihMuizkjEHkgksusds3/56lsd.MckngksrkgSA (D*)vf/kdrefoLrkjovf/kdrelEihMu,ddsckngksrsgSaA Sol. At max. extension both should move with equal velocity.  By momentum conservation, (5 × 3) + (2 × 10) = (5 + 2)V V = 5 m/sec. Now, by energy conservation 1 5 × 32 + 2 Put V and k 1 × 2 × 102 = 2 1 (5 + 2)V2 + 2 1 kx2 2  xmax = 1 m = 25 cm. 4 Also first maximum compression occurs at ; 3T t = 4 = 3 2 4 = 3 2 4 3 = 56 sec. (where   reduced mass,  = m1m2 m1  m2 ). Sol. vf/kdrefoLrkjijnksuksadkslekuosxlspyukpkfg,  laosxlaj{k.kls (5 × 3) + (2 × 10) = (5 + 2)V V = 5 m/sec. vc,ÅtkZlaj{k.kls 1 5 × 32 + 2 1 × 2 × 102 = 2 1 (5 + 2)V2 + 2 1 kx2 2 V ok j[kusij  xmax = 1 m = 25 cm. 4 rFkkigykvf/kdrelEihM+ufuEule;ijgksrkgS 3T t = 4 = 3 2 4 = 3 2 4 3 = 56 sec. m1m2 (tgk¡ lekuhr nzO;eku ,  = m1  m2 ). SECTION - III : ASSERTION AND REASON TYPE [k.M- III: dFku vkSj dkj.k izdkj 11.23 Assertion : Kinetic energy of SHM at mean position is equal to potential energy at ends for a particle moving in SHM. Reason : Total energy in SHM is conserved. dFku : ,d ljyvkorZ xfr djrs gq,d.k dsfy, ek/;fLFkfr ijxfrt ÅtkZ fljksa ijfLFkfrt ÅtkZ ds cjkcj gksrhgSA dkj.k : ljy vkorZ xfr esa dqy ÅtkZ fu;r jgrh 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. PE is related to reference. Only when PE at mean position is taken zero, the assertion is true. fLFkfrtÅtkZfunsZ'kra=klslacaf/krgSAdsoytcek/;fLFkfrijfLFkfrtÅtkZ'kwU;yhtk;srksdFkulghgSA 11.24 STATEMENT-1 : A SHM may be assumed as composition of many SHM's. STATEMENT-2 : Superposition of many SHM's (along same line) of same frequency will be a SHM. oDrO;-1: ,d ljy vkorZ xfr dks dbZ lkjh ljy vkorZxfr;ksa ds la;qXeu ls cuk eku ldrs gSA oDrO;-2: ,dghlekujs[kkijlekuvko`fÙkokyhdbZljyvkorZxfr;ksa dsv/;kjksi.kls,d ljyvkorZxfr gksxhA (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. Statement-2 itself explains statement-1. dFku-2, dFku-1dh O;k[;k djrk gSA 11.25 STATEMENT-1 : A particle is moving along x-axis. The resultant force F acting on it is given by F = – ax – b. Where a and b are both positive constants. The motion of this particle is not SHM. STATEMENT-2 : In SHM resultant force must be proportional to the displacement from mean position. oDrO;-1 : ,d d.k x-v{k ds vuqfn'k xfr dj jgk gS bl ij dk;Zjr ifj.kkeh cy F bl izdkj gS fd F = – ax – b tgk¡ a rFkk b nksuksa /kukRed fu;rkad gSA bl d.k dh xfr ljy vkorZ xfr ugha gksxhA oDrO;-2:ljyvkorZxfresaifj.k ehcyek/;fLFkfrlsges'k foLFk iudslekuqikrhgksrkgSA (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) The mean position of the particle in statement-1 is x =  b a and the force is always proportional to displacement from this mean position. The particle executes SHM about this mean position. Hence statement-1 is false (D) oDrO;-1 esa d.k dh e/;eku fLFkfr x =  b gS rFkk cy ges'kk e/;eku fLFkfr ls foLFkkiu ds lekuqikrh gksrk gSA d.k bl e/;eku fLFkfr ds ifjr% ljy vkorZ xfr djrk gSA blfy;s oDrO;-1 vlR; gSA SECTION - IV : COMPREHENSION TYPE [k.M-IV:cks/kuizdkj A small block of mass m is fixed at upper end of a massless vertical spring of spring constant K = 4mg L and natural length '10L'. The lower end of spring is free and is at a height L from fixed horizontal floor as shown in the figure. The spring is initially unstressed and the spring-block system is released from rest from shown position. vuqPNsn # 1 ,dNksVkCykWdftldknzO;ekum gSbldks,dHkkjghuÅ/okZ/kjfLizaxdsÅijhfljsijla;ksftr(fixed)djnsrs gSA fLizax dk fLizax fu;rkad K = 4mg gS rFkk izkd`frd yEckbZ '10L' gS] L fLizax dk fupyk fljk eqä gS rFkk fLFkj {kSfrt ry ls fp=kkuqlkj L Å¡pkbZ ij gS fLizax izkjEHk esa fcuk [khaph gqbZ gSA fLizax CykWd fudk; dks fojke ls fp=kkuqlkj nh xbZ fLFkfr ls eqä fd;k tkrk gSA 11.26 At the instant speed of block is maximum, the magnitude of force exerted by spring on the block is ml{k.k tcCykWd dhpky vf/kdregSrcfLizax }kjkCykWd ijyxk;s x;scy dkifjek.k gksxk (A) (A) mg (B*) mg (C) Zero (D) None of these 2 mg (B*) mg (C)'kwU; (D)buesalsdksbZugh Sol. When speed of block is maximum, net force on block is zero. Hence at that instant spring exerts a force of magnitude 'mg' on block. tcCykWd dhpkyvf/kdregS CykWdij ifj.kkehcy 'kwU;gS vr%ml {k.kfLizax CykWdijmgifjek.kdk cy yxk;sxkA 11.27 As the block is coming down, the maximum speed attained by the block is tcCykWduhpsdhrjQvkjgkgksrkgSrksCykWd}kjkizkIrvf/kdrepkygksxhA (A) (B) (C*) 3 gL 2 (D) 3 gL 2 Sol. At the instant block is in equilibrium position, its speed is maximum and compression in spring is x given by kx = mg (1) From conservation of energy CykWd ds lkE;oLFkk ds {k.k] CykWd dh pky vf/kdre gksxh rFkk ml {k.k fLizax esa laihMu x ds fy, kx = mg (1) ÅtkZlaj{k.kls mg (L + x) = 1 kx2 + 2 1 mv2 2 max .... (2) from (1) and (2) we get v max = . (1) o (2) ls v 3 max = 2 gL . 11.28 Till the block reaches its lowest position for the first time, the time duration for which the spring remains compressed is tcrdCykWdviuhfuEurevoLFkkijigyhckji¡gqprkgSrksogle;kUrjkyD;kgksxkftldsfy,fLizaxlaihfMr jgrhgSA Sol. (A)  (C)  + + Vmax = 1 sin–1 3 2 sin–1 3 and  = = 2 (B*) (D) 1 + sin–1 3 2 + sin–1 3  A = Vmax  = 3 L 4 Hence time taken t, from start of compression till block reaches mean position is given by blfy;slaihMudsizkjEHklsCykWd}kjkek/;fLFkfrrdigq¡pusesafy;kx;kle;tfn;ktkldrkgS] x = A sin  t where tgk¡ x = L 1  t = sin–1 3 T 2 Time taken by block to reach from mean position to bottom most position is 4 = 1 4 = Hence the required time = + sin–1 3 2 CykWd}kjk ek/;fLFkfrlslclsfupyhfLFkfrrdigq¡pusesafy;kx;kle; 4 = 1 blfy;s vko';d le; = + sin–1 3 Comprehension - 2 [ A large tank of cross-section area A contains liquid of density  . A cylinder of density  / 4 and length 𝑙 and cross- section area a (a <