UNIT # 06 (PART I) SIMPLE HARMONIC MOTION EXERCISE I

Size: px

Start display at page:

Download "UNIT # 06 (PART I) SIMPLE HARMONIC MOTION EXERCISE I"

Robyn Evans
6 years ago
Views:

1 UNIT # 6 (PRT I) SIMPLE HRMONIC MOTION EXERCISE I. f Hz. - - O O. sin t t 5 or 6 6 Phse difference () 5 O R JEE-Physics or 6 6 / cos fro phser () Node6\E : \Dt\\Kot\JEE-dvnced\SMP\Phy\Solution\Unit 5 & 6\.SHM.p65. / / / cos Totl phse difference between the () 5 Fro fiure () 5 iu plitude () Position of prticle t t, (t) 5 let eqution of SHM is () sin(t+) t t, 5 5 sin /6 nd T Thus, eqution of SHM sin t 6 () / cos 6 phse difference () 5. sin t sin t T t t T 8, sin tt/ T 8 T O R t tt tt/ T t; T 8 s cos so tt/8 6. sin t, sin(t + ) Gretest distnce () sin tt/ sin sin 7

2 Now ccordin to question () sin t sin (t + ) t t + t. sint & y sin t sin t cos t sin cos 7 7. Miniu phse difference between two position () Tie ten ( ) T T -5 O R s fro fiure or t t t 5 sec 8. sin (t + ) t t s, sin(t+) v cos (t +) 9. t ts, cos(+) cos( ) cos 6 cost cos nd +bcos +b[cos ] b +(+b) b b b 8 Note : Plese correct the nswer of the question b 8 y y which represent oblique ellipse (). sin t sin & + sin() T 8 & +. Therefore O R Suppose plitude be nd distnce trveled in sec be nd in sec be. (s s ) t Therefore 8 cos cos & t. Therefore. Miu possible vere velocity will be round en position. () vere velocity in tie () T / T / T T/ - O R M.P. T/8 T/8 T/ T/ Node6\E : \Dt\\Kot\JEE-dvnced\SMP\Phy\Solution\Unit 5 & 6\.SHM.p65

3 t T T 8 cos cos vere velocity () there is no loss of enery () de 8 v + v 8v v totl displceent totl tie / T / T / T T. Tie period () (+) 8 v Men Position v B s 8. Let the nturl lenth of the sprin () Fro fiure (). Miu KE () 5 J Totl enery () 5 + 5J 5. For iu displceent () M() M () FN F5N F9N Node6\E : \Dt\\Kot\JEE-dvnced\SMP\Phy\Solution\Unit 5 & 6\.SHM.p65 6. Both the sprin re in series () K eq K(K ) K K K Tie period () T where T. K v K O R eq K eq Here () Totl etension () By enery conservtion () E K eq v v E v v v de d dv v 9. eq. ( )...(i) 5 ( )...(ii) 9 ( )...(iii) iii i 5 iii ii 5 f f f / / f M f M M. T Now T eff...(i)...(ii) (upwrds) T 9

4 . nd They will be in phse if () ( ) t,, t 6 sec. sin cos. T where T f f f f so T n n n n. Center of ss of syste is t distnce fro pe P is is nd oent of inerti of the syste ( P P ) sin I-sin ( + ) (sin~) T T 6. For weihtlessness () (f) (.5) f f 7. ns. (C) Tie period for sprin bloc syste is T does not effected. ( T ) / C.M. r F sin sin cos I sin sin for sll T 8. ns. (C) U() + b 9. t F U So KE b for sll 8 KE is incresed by n ount of. Let now Node6\E : \Dt\\Kot\JEE-dvnced\SMP\Phy\Solution\Unit 5 & 6\.SHM.p65 5

5 plitude be then totl KE ( ) KE Here v ( ) c. Totl enery (). E E T I T. sin t + cos t 5 sin(t+) 5, v (5)(). For () : t t T, prticle t etree position () F For (B) t t T/, prticle t en position (tt/ For (C) : For (D) : v (iu) () t tt, prticle t en position (tt) t t T/, prticle t en position (tt/) so U For nd condition eff ()() T f / f / f T 8. In n rtificil stellite ( ) eff T 9. <ccelertion()> T / sin t T /. Required tie () T T T. KE t centre () f KE t distnce () f Difference () f 6 f. Fro the rph, eqution of ccelertion cn be written s () cos t velocity cn be written is () v v sin t. KE v v sin t Hence the rph is s shown in ( ) Node6\E : \Dt\\Kot\JEE-dvnced\SMP\Phy\Solution\Unit 5 & 6\.SHM.p sin sin sin Now sin(t) T T 8 7. For st condition eff. 8 ( ) Here so en position t () Let sin(t + ) s prticle is t rest t (etree position) nd pliutde s prticle strt fro etree position. Therefore ( ) cost cost 5

6 . (+ cos t) sin ( t) [sin t + cos t + sin t] [sin t + sin ( ( + )t sin (( )t)] Required rtio () : ( ) : ( + ) 5. y sin t cos t sin t To bres off () in in oent it occurs first fter t (t) sin t t t , ( ) t,,,... t t ( ) N /. When cylindricl bloc is prtilly iersed () F B y (6 ) y c Miu plitude () c c c EXERCISE II. v cos t, sint v Striht line in v nd (v ). Initilly F B / + F B V...(i) finlly (+ y) F B (/+ y) Let cylinder be displced throuh y then restorin force, (y ) f net [(+y) + F B ] f net y y f net [y + y] [y + y] f Restorin force when it is slihtly depressed by n ount of. () F (V) () h h T s 7 5. t equilibriu () (.)(). c plitude of SHM () c Frequency (). 5Hz 6. The position of oentry rest in S.H.M. is etree position where velocity of prticle is zero. ( ) Motion Position / / / / N Node6\E : \Dt\\Kot\JEE-dvnced\SMP\Phy\Solution\Unit 5 & 6\.SHM.p65 5

7 Node6\E : \Dt\\Kot\JEE-dvnced\SMP\Phy\Solution\Unit 5 & 6\.SHM.p65 s the bloc loses contct with the pln t this position i.e. norl force becoes zero, it hs to be the upper etree where ccelertion of the bloc will be downwrds. ( ) 5 5 rd/s. Therefore period ( )T s 5 ccelertion in S.H.M. is iven by () Fro the fiure we cn see tht,t lower etree, ccelertion is upwrds ( ) N N ( +) t hlfwy up, ccelertion is / downwrds (/) N N ( ) t hlfwy down ccelertion is / upwrds (/) N N ( + ) 7. sin ( t), y sin(t) y Motion of prticle will be on striht line with slope /. (/) s r y 5 sin ( t) so otion of prticle will be SHM with plitude 5. (5) 8. r i yi ( cos t)i ( cos t) j cos t, y cos t y The otion of prticle is on striht line, periodic nd siple hronic () 9. t iu copression v v B & inetic enery of B syste will be iniu ( v v -B) B so v v B v K B v Fro enery conservtion ( ) v v v v. Let sll nulr displceent of cylinder be then restorin torque () I (R)R where I MR d R d M MR. s < so T L.. h (M )v h h f M. y sin (t + / Miu KE () 6 6 ( ) ens 75% of enery v 5. Miu speed () v v sin t So eqution of otion () 6. sint + 8cos 5t sin(t)+ (+cost) + 5sin(t + 7 ) plitude () 5 Miu displceent () c 5

8 7. Velocity of bloc just before collision () ( ) 9 ( ) / s Velocity of cobined sses ieditely fter the collision () ()() /s 6 New nulr frequency ( ) ' 9 9 Therefore ()v' ' ' ' 8. Fro enery conservtion ( ) 9. y nd t t T, y is iu so ccelertion is iu t t T. ( y t T y tt ) lso y t t T (t T y) t, so force is zero t t T T t,v PE oscilltion enery () Fill in the blns. EXERCISE III v ( ).5 nd nt (n+)t 8 n9 T 8.8s. L ( )L L f f f ( ) ( ) ( + ) <v> (n) sin t v 8 (n) (n ) v ( ), v ( ) /s If > ( > Now If the lower dis will bounce up. ) then iu norl rection fro round on lower dis ( ) N + ( + ) 6 v v & v v v v 7. sint, sin (t + ) (cos )sin t + sin cos t Miu vlue of (cos ) sin sin or 5 5 or 6 6 Node6\E : \Dt\\Kot\JEE-dvnced\SMP\Phy\Solution\Unit 5 & 6\.SHM.p65 5

9 Mtch the colun. y sin t + sin t cos + cost sin. sin t + Etree en Etree 6 F 8 ( ) cos t sin t. is positive in II & IV; v is positive if f et > (IIIIV Copreshension #. M. ccelertion of () Coprehension #. Both the blocs reins in contct until the sprin is in copression. In this tie syste coplete hlf oscilltion. By reduced ss concept tie period of syste ( ) T s Required tie () T s. v v Node6\E : \Dt\\Kot\JEE-dvnced\SMP\Phy\Solution\Unit 5 & 6\.SHM.p65 (.6)()() 6s M. ccelertion of () (.)()() s Therefore iu ccelertion of syste cn be /s ( /s ) ( / ) 5 / 6 9. constnt Coprehension #.. R f, fr f R R f But cos t ( the cylinder is strtin fro ) () So f cos t v v v R ()() v s () Let velocity of rer be v nd front be v then v v v v ( v v ) Now by conservtion of echnicl enery () ()() ()v + ()v v + v But v + v v v v v v s v Coprehension #. v T (R / v) R. t t,, vv R (,R) so cos(t + ) & sin(t + ) + y v Coprehension # 5. s is t its netive etree t t (, t) so sin (t + /) cos (t) y B (,) (,). s B is t its equilibriu position nd ovin towrds netive etree t t (Bt ) so y sin (t + ) y sin(t) 55

10 . Distnce between & B (B) y ( cos t) ( sin t) 9 cos t cost 6 sin t 6 sin t 9 cos t sin t sin(t 7 ) Miu distnce () 9 9 7c Miniu distnce () 9 9 c EXERCISE IV(). The iu velocity of the prticle t the en position () v (n) v n..5. If t the instnt t, displceent be zero so displceent eqution is (t) y sint sinnt.5 sin (t). S cos t sin t cos t+ sin t 8 cos t 8 sin t 5 cos t sin t ' 5 8, 7. sin (t + ) 5 cos( t 7 ) 8 t t c sin...(i) Velocity v d cos(t + ) t t cos cos...(ii) fro (i) & (ii) (sin + cos ) + c nd tn rd. y sint sin T t sin T T F I HG K J F sin I HG T K J T s 5. Miu distnce( ) sin ()(.9).8 6. Velocity () v v v c 9 lenth of pth () 9.5 c 7. Enery t ll the three points re equl () v 8 + v 7 v + v v (i) v v (ii) fro (i) & (ii) & / then, totl enery is equl for the iu inetic enery () 6 v v 65 /s 9 O R v 8 7 ( ), ( ) fter solvin the eqution () V 65 / s Node6\E : \Dt\\Kot\JEE-dvnced\SMP\Phy\Solution\Unit 5 & 6\.SHM.p65 56

11 8. s v so v n nd v n n 5 c 9. sint 6sin t sint (sin t) ( sin sin sin) sint (sint sint) sint sint + sintsint otion is SHM with nulr frequency So 6 n. (i) t equilibriu position ( ) F du d (ii) F du ( ) ( ) d F SHM Here T (iii) 6 6. Frequency of oscilltion ( ) s Node6\E : \Dt\\Kot\JEE-dvnced\SMP\Phy\Solution\Unit 5 & 6\.SHM.p65. v 8 8 rd/s Therefore eqution of SHM (). sin t. (i) Miu speed of oscilltin body () v T Here etre, T.57 s v. /s 57. (ii) Miu inetic enery () K v () 8J (iii) Totl enery of prticle will be equl to iu inetic enery. () (iv) Tie period of ss suspended by sprin ( T K so force constnt ( ) K T b b. 6 N/ f.5 Hz Let iu plitude be then () v where diference in equilibriu position ( ) nd.5 v /s.5 Therefore () c 6 6. Coon velocity fter collision be v then by COLM (v ) u Mv Mu v Hence, inetic enery ( ) (M) u F I HG K J Mu It is lso the totl enery of vibrtion becuse the sprin is unstretched t this oent, hence if is the plitude, then K Mu F HG I KJ M K u 57

12 5. dditionl force ( ).5 6. Centre of ss will be t rest s there is no eternl force ctin on the syste. ( 7. So effective lenth ( ) T eff eff I T 6 EXERCISE IV(B). Let bloc be displced by then displceent in sprins be ( ),, nd Such tht () Now let restorin force on be F then (F) f F F F F F T. (i) Let M initil copression in sprin () COME : ( + b) ( ) + (+M) (b+) 8. T I where () b I L L L 7L & L (Distnce of centre of ss fro hine) () 7L T ((L / )) 7L 8 9. Moent of inerti bout hined point ( I R + R R R f I R R R (ii) b M b f b M (iii) Let h initil heiht of over the pn (h) v coon velocity of (+M) fter collision (v(+m)) COLM : h ( M)v...(i) COME : + (+M)v + (+M)b ( +b) M b h b Node6\E : \Dt\\Kot\JEE-dvnced\SMP\Phy\Solution\Unit 5 & 6\.SHM.p65 58

13 . () Let the liquid of density.5 occupy the left portion B nd the liquid of density occupy the riht portion BC of the tube. The pressure t the lowest point D due to the liquid on the left is (B.5 BC D) P (R Rsin).5 s tn., sin..., cos. dp R [ ] R [.6 ].6 where y R, the liner displceent. (y R, ) Restorin force () F.6 y R C This shows tht () F y Hence the otion is siple hronic with force constnt ().6 Now, totl ss of the liquid ().5 D B R R.5p 5 R Node6\E : \Dt\\Kot\JEE-dvnced\SMP\Phy\Solution\Unit 5 & 6\.SHM.p65 The pressure due to the liquid on the riht is () P (Rsin +Rcos) + (R Rcos).5 Since the liquids re in equilibriu () P P or (R Rsin).5 (Rsin+Rcos) Solvin, we et () tn. or tn (.) + (R Rcos).5 (b) Let the whole liquid be iven sll nulr displceent towrds riht. Then the pressure difference between the riht nd the left libs is ( ) dp P P [Rsin(+) + Rcos(+) + [R Rcos().5 [R Rsin()].5 R [.5 sin().5 cos(] R [.5 (sin cos + cos sin).5 (cos cos sinsin)] For sll () sin,cos dp R [.5 sin +.5 cos.5 cos +.5 sin] Tie period () T. T sin I 5r.6.9R R seconds. d T 5. Given + b + s R O tn b y b T d Q tn P LetOT, TQ, PT d, TS y 59

14 () Fro eoetry d + y sin ( + ) nd d + cos ( + ) tn sin ( + ) T T t t T...(ii) d T d T totl 6. Fro enery eqution ( F ///// v + b E + b T b 8. (i) F sin /,, K f,, f 7. T L (Lift is sttionry) () T L (Lift is ccelerted upwrd) () T L (Lift is decelerted upwrd) () Let totl upwrd distnce trvelled () t t for upwrd ccelerted otion () T T t t T...(i) & for upwrd decelerted otion () L L F L F sin L KL L LKL 5 L KL K 5L (ii) If the rod is displced throuh n nle, then () L L L ML (L)L M 9 M M L 9. (i) This syste cn be reduced to ( ) Where M eq \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ \\\\\\\\\\\\ (.)(.).5.. nd eq N eq. f Hz.5 Node6\E : \Dt\\Kot\JEE-dvnced\SMP\Phy\Solution\Unit 5 & 6\.SHM.p65 6

15 (ii) Copression in one sprin is equl to etension in other sprin ( ) R (.6) 6 5 Totl enery of the syste ( ) E (R ) (R ) (R) (.) 5 5 J (iii) Fro echnicl enery conservtion () v + v E.v 5 v s EXERCISE V-. Tie period of ss loded sprin () T So T...(i) Sprin constnt () is inversely proportionl to the lenth of the sprin, i.e. (()) coplete sprin / cut sprin / n n cut sprin n coplete sprin Node6\E : \Dt\\Kot\JEE-dvnced\SMP\Phy\Solution\Unit 5 & 6\.SHM.p65. K b K For sll nulr displceent, net torque towrds en position is ( ) ( ) + ( b) b I ( + K b ) (L + b ML ) ( + b ) b b M M L ( ) L ( ) Hence frequency () f b M L ( ) Tcut sprin coplete sprin T [Fro (i) n T coplete sprin cut sprin T coplete sprin n cut sprin. In siple hronic oscilltor the potentil enery is directly proportionl to the squre of displceent of the body fro the en position; t the en position the displceent is zero so the PE is zero but speed is iu; hence KE is iu. ( ). The tie period of the swin is (). T eff Where l eff is the distnce fro point of suspension to the centre of ss of child. s the child stnds up; the l eff decrese hence T decreses. ( l eff l eff T M T nd 5T M On dividin () 5T M T 5 M 6 9 M M 9 M 6

16 5. s iu vlue of ( ) sin B cos B so plitude of the prticle ( ) 6. hronic oscilltor crosses the en position with iu speed hence inetic enery is iu t en position (i.e., ) ( ) Totl enery of the hronic oscilltor is constnt. PE is iu t the etree position. ( ) 7. Miu velocity () v Given () (v ) (v ) 8. The tie period of siple pendulu of density when held in surroundin of density is ( ) t ediu t ir t wter t t t 9. T t But + & t T t + t + T t + t T. The totl enery of hronic oscilltor is constnt nd it is epressed s ( ) E (plitude) E is independen to instntneous displceent. (E). Nturl frequency of oscilltor () Frequency of the pplied force () Net force ctin on oscilltor t displceent () ( )...(i) Given tht () F cost...(ii) Fro eqs. (i) nd (ii) we et ((i) (ii)) ( ) cost...(iii) lso, cost...(iv) Fro eqs. (iii) nd (iv), we et ( )cost cost. Initil ccelertion (). ns. () 5. F s. y sin t cos t cost otion is SHM with tie period (). y.sin t dy v. cos t cos t y. cost v. sin t. cos t Phse difference between v nd v (v v ) t t 6 d 5. On coprin the bove eqution with the eqution of SHM( ) d T Node6\E : \Dt\\Kot\JEE-dvnced\SMP\Phy\Solution\Unit 5 & 6\.SHM.p65 6

17 Node6\E : \Dt\\Kot\JEE-dvnced\SMP\Phy\Solution\Unit 5 & 6\.SHM.p65 6. The epression of tie period of siple pendulu is () T eff eff Where eff l is the distnce between point of suspension nd centre of rvity of bob. s the hole is suddenly unplued, eff l first increses then decrese becuse of shiftin of CM due to which the tie period first increses nd then decreses to the oriinl vlue. (l eff l eff 7. Miu velocity of prticle durin SHM is () v 8. KE. 7 T.s T 75 TE Now fro sint we hve sin t t t T s cost v ( ) () ( sint) sint Speed will be iu if t t.5s. cos t v d sin dv t cos t cos t cos t cos(t + ) nd. W.D f K f +U f K i U i f. + v c /s \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\. The oriinl frequency of oscilltion () f On incresin the nd by ties, then f' becoes ( ) f f. Mss, plitude, frequency (,, (KE) v., v T + v T + ( ) T 5. n M T T + ( ) constnt () T T constnt ()... (i) nd n... (ii) M ccordin to conservtion of liner oentu () Mv (M + )v M (M + ) Fro eqution (i) & (ii) M M M. M M M M M 6

18 6. X sin t X X + sin (t + ) X X X + sin (t + ) sin t Becoe X X X + sin (t + ) sin t) 7. By usin T Where d nd T d T d 9. Eqution of dped siple pendulu () d d bv + sin bv +. U() EXERCISE V-B [U] [ML T ] [] [ML T ] [ ] [L ] Now, tie period y depend on () T (ss) (plitude) y () z [M L T] [M] [L] y [ML T ] z [M +z L y z T z ] Equtin the powers, we et ( ) z or z / y z or y z / Hence, T (plitude) / T () / T. By solvin bove eqution b t e sin (t + ) t t, so b. FBD of piston t equilibriu (FBD) Force constnt lenth of sprin P t + P...(i) FBD of piston when piston is pushed down distnce ( FBD). U() ( e ) It is n eponentilly incresin rph of potentil enery (U) with. Therefore, U versus rph will be s shown. ((U) U ) U K d P t + (P +dp) Process is dibtic () PdV PV C dp V P Usin,, we et f MV...(ii) X Fro the rph it is cler tht t oriin.potentil enery U is iniu (therefore, inetic enery will be iu) nd force ctin on the prticle is lso zero becuse ( ) Node6\E : \Dt\\Kot\JEE-dvnced\SMP\Phy\Solution\Unit 5 & 6\.SHM.p65 6

19 JEE-Physics Node6\E : \Dt\\Kot\JEE-dvnced\SMP\Phy\Solution\Unit 5 & 6\.SHM.p65 F du (slope of U rph). d Therefore, oriin is the stble equilibriu position. Hence, prticle will oscillte siple hroniclly bout for sll displceents. Therefore, correct option is (d). (), (b) nd (c) options re wron due to followin resons : ( (d)(, (b) (c) ) () t equilibriu positoin F du i.e, slope of d U rph should be zero nd fro the rph we cn see tht slope is zero t nd ±. (b) (c) Now on these equilibrius stble equilibriu position is tht where U is iniu (Here ). Unstble equilibriu position is tht where U is iu (Here none). Neutrl equilibriu position is tht where U is constnt (Here ± ). Therefore, option () is wron. (F du d U ± U() U U( ± ). () For ny finite non zero vlue of, force is directed towrds the oriin becuse oriin is in stble equilibriu position. Therefore, option (b) is incorrect. ( (b)) t oriin, potentil enery is iniu, hence inetic enery will be iu. Therefore, option (c) is lso wron. ( (c)). Free body dir of bob of the pendulu with respect to the ccelertin fre of reference is s follows: (FB ) sin sin cos Net force on the bob is () F net cos (fiure b) o r sin Net ccelertion of the bob is () T eff cos L eff T L cos 5. In SHM, velocity of prticle lso oscilltes siple hroniclly. Speed is ore ner the en position nd less ner the etree positions. Therefore, the tie ten for the prticle to o fro O to / will be less thn the tie ten to o it fro / to, or T < T. ( / / T < T.) 6. Potentil enery is iniu (in this cse zero) t en position () nd iu t etree positions ( ±). (() (±)) t tie t,. Hence, PE should be iu. Therefore, rph I is correct. Further is rph III, PE is iniu t. Hence, this is lso correct. (t IIII ) 7. Bloc Q oscilltes but does not slip on P. It ens tht ccelertion is se for Q nd P both. There is force of friction between the two blocs while the horizontl plne is frictinless. The sprin is connected to upper bloc. The (P Q) syste oscilltes with nulr 65

20 JEE-Physics frequency. The sprin is stretched by. (QP QP (P Q) ) Miu ccelertion in SHM ( Now consider the lower bloc. () Let the iu force of friction f () f f f 8. y Kt d y T nd T y T T (i) K y /s (s K /s ) y 6 5 y 9. ()y. Displceent eqution fro rph () X sin t T 8 ccelertion () t t s, c/s 6 sin t. In series sprin force rein se; if etension in nd re nd respectively. Then + +. plitude of point P will be the. et. in. (P ) So plitude of point P is (P) L F L L L L L L L net L 6 6 /. f M C Q. Fro superposition principle : () y y + y + y sin t + sin (t + 5 ) + sin (t +9 ) [sin t + sin(t + 9 )] + sin (t + 5 ) sin (t + 5 ) cos 5 + sin (t + 5 ) 6 ( ) sin( t 5 ) sin (t + 5 ) Therefore, resultnt otion is siple hronic of plitude. ( ) nd which differ in phse by 5 reltive to the first. ( ( ) ) Enery in SHM (plitude) E resul tn t ( ) E sin le E resultnt E sin le. For B nd C B X B cos t + B sin t B sin t Node6\E : \Dt\\Kot\JEE-dvnced\SMP\Phy\Solution\Unit 5 & 6\.SHM.p65 66

21 JEE-Physics This is eqution of SHM of plitude B ( B If B nd C B, then X B + B sin t This is lso eqution of SHM bout the point XB. Function oscilltes between X nd XB with plitude B. (B BB) P r r p h. For liner otion of disc () F net M + f where f frictionl force (f) For rollin otion () MR fr M R f M F et Therefore F et F et. Totl enery of syste () E Mv de + Mv + dv d M(v) dv M M. Usin enery conservtion lw () Mv ( Velocity of centre of ss v Loction/ coordinte of centre of ss t tie tv t (t-v t) v t [v t ( cos t)] ( + ) v t [v t ( cos t)] + v t + v t v t ( cos t)] + v t + ( cos t) v t + ( cos t)...(i) (ii) To epress in ters of. ( ) v t ( cos t) d d cos t...(ii) is displceent of t tie t. (t ) d ccelertion of t tie t. (t ) v sint Node6\E : \Dt\\Kot\JEE-dvnced\SMP\Phy\Solution\Unit 5 & 6\.SHM.p65 f M & f R MR But f µm Mv K K SUBJECTIVE M K 9 M v µ M K. (i) Two sses nd re connected by sprin of lenth. The sprin is in copressed position. It is held in this position by strin. When the strin snps, the sprin force is brouht into opertion. The sprin force is n internl force w.r.t. sses sprin syste. No eternl force is pplied on the syste. The velocity of centre of ss will not chne. When the sprin ttins its nturl lenth, then ccelertion is zero nd ( ) ( ( ) ). Put fro (i) v t ( cos t) [v t ( cos t)] When d. ( cos t), cos t fro (ii). 67

22 . sphere of rdius R is hlf suered in liquid of density. (R) For equilibriu of sphere () O R Weiht of sphere Upthrust of liquid on sphere. () V V () where denisty of sphere()...(i) Fro this position, the sphere is slihtly pushed down. Upthrust of liquid on the sphere will increse nd it will ct s the restorin force. ( ) Restorin force () F Upthrust due to etr iession () F (etr volue iersed) ( F () ) (ss of sphere ) cc() R F R R R R R, [fro (i)]. sll body of ss ttched to one end of verticlly hnin sprin perfors SHM. ( ) nulr frequency () plitude () Under SHM, velocity () v y fter detchin fro sprin, net downwrd ccelertion of the bloc. ( ) Heiht ttined by the bloc h (h) h y + v ( y ) h y For h to be iu (h) dh, y y*. dy dh dy + ( y*) y * y * y* Since > (iven) > > y*. y* fro en position < ( <) Hence y*. R is proportionl to.() Hence the otion is siple hronic. () Frequency of oscilltion (. R Node6\E : \Dt\\Kot\JEE-dvnced\SMP\Phy\Solution\Unit 5 & 6\.SHM.p65 68

SOLUTIONS TO CONCEPTS CHAPTER 10

SOLUTIONS TO CONCEPTS CHAPTER 10 SOLUTIONS TO CONCEPTS CHPTE 0. 0 0 ; 00 rev/s ; ; 00 rd/s 0 t t (00 )/4 50 rd /s or 5 rev/s 0 t + / t 8 50 400 rd 50 rd/s or 5 rev/s s 400 rd.. 00 ; t 5 sec / t 00 / 5 8 5 40 rd/s 0 rev/s 8 rd/s 4 rev/s