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1 M Dynmics - Dmped nd forced hrmonic motion. A P α B A ight estic spring hs ntur ength nd moduus of esticity mg. One end of the spring is ttched to point A on pne tht is incined to the horizont t n nge α, where tn α = 3. The other end of the spring is ttched to prtice P of mss m. The pne is rough nd the coefficient of friction between P nd the pne is. The prtice P is hed t point B on the pne where B is beow A nd AB =, with the spring ying ong ine of the pne, s shown in the figure bove. At time t = 0, the prtice is projected up the pne towrds A with speed (g). At time t, the compression of the spring is. () Show tht d d g. + ω = g, where ω = t (6) (b) Find in terms of, ω nd t. (7) (c) Find the distnce tht P trves up the pne before first coming to rest. () (Tot 7 mrks). A prtice P of mss m is suspended from fied point by ight estic spring. The spring hs ntur ength nd moduus of esticity mω, where ω is positive constnt. At time t = 0 the prtice is projected verticy downwrds with speed U from its equiibrium position. The motion of the prtice is resisted by force of mgnitude mωv where v is the speed of the prtice. At time t, the dispcement of P downwrds from its equiibrium position is. d d () Show tht + ω + ω = 0 (5) Edece Intern Review

2 M Dynmics - Dmped nd forced hrmonic motion Given tht the soution of this differenti eqution is = e ωt (A cos ωt + B sin ωt), where A nd B re constnts, (b) find A nd B. () (c) Find n epression for the time t which P first comes to rest. (3) (Tot mrks) 3. A ight estic string, of ntur ength nd moduus of esticity mg, hs prtice P of mss m ttched to its mid-point. One end of the string is ttched to fied point A nd the other end is ttched to fied point B which is t distnce verticy beow A. () Show tht P hngs in equiibrium t the point E where AE = 5. (5) The prtice P is hed t distnce 3 verticy beow A nd is reesed from rest t time t = 0. When the speed of the prtice is v, there is resistnce to motion of mgnitude mkv, where k = g. At time t the prtice is t distnce ( 5 + ) from A. (b) Show tht d d + k + k = 0. d t (c) Hence find in terms of t. (5) (7) (Tot 7 mrks) Edece Intern Review

3 M Dynmics - Dmped nd forced hrmonic motion. A prtice P moves in stright ine. At time t seconds, its dispcement from fied point O on the ine is metres. The motion of P is modeed by the differenti eqution d d + d t + = cos t 6 sin t. When t = 0, P is t rest t O. () Find, in terms of t, the dispcement of P from O. () (b) Show tht P comes to instntneous rest when t =. () (c) Find, in metres to 3 significnt figures, the dispcement of P from O when t =. () (d) Find the pproimte period of the motion for rge vues of t. () (Tot 7 mrks) Edece Intern Review 3

4 M Dynmics - Dmped nd forced hrmonic motion. () F= R R = mg cosα T = mg ( ): F mg sin α T = m 3 mg. mg mg = m 5 5 d w A + = g A 6 (b) w = s L m + w = 0 m =± wi C.F. = Asin wt + Bcoswt P.I. g = = w G.S. = Asin wt + Bcoswt t = 0, = 0 B = = wacoswt wbsin wt A g g t = 0, = g : = A A = = (sin wt + coswt ) A 7 (c) = 0 wacos wt wbsin wt = 0 A tnwt = = B wt = (first vue) A Edece Intern Review

5 M Dynmics - Dmped nd forced hrmonic motion = ( + ) = ( ) A [7]. () T d e mg R(φ) m d = mg T mω d ( terms) A m d = mg mω (e + ) mω d d d + ω + ω = 0 A 5 d (b) = e ωt (A cos ωt + b sin ωt) t = 0, = 0 A = 0 d = ωe ωt.b sin ωt + e ωt.bω cos ωt = 0 (use of product rue) t = 0, d = U : U = Bω. B = U A ω (c) d = Ue ωt sin ωt + Ue ωt cos ωt = 0 tn ωt = (sove for tn ωt) t = ω A 3 [] Edece Intern Review 5

6 M Dynmics - Dmped nd forced hrmonic motion A 3. () T + e mg T T = T + mg mge mg = ( e) + mg A 3 5 e = AE = (*) A A cso 5 mg (b) mg + ( ) ( + ) mg 3 g m = m A3 ( eeoo) + k + k = 0 (*) A 5 (c) AE: m + km + k = 0 m = k ± ki A GS: = e kt (A cos kt + B sin kt) A ft t = 0, = A = = ke kt (A cos kt + B sin kt) + e kt ( ka sin kt + kb cos kt) t = 0, = 0 ka + kb = 0 B = A = = e kt (cos kt + sin kt) A 7 [7] Edece Intern Review 6

7 M Dynmics - Dmped nd forced hrmonic motion. () Auiiry Eqution.: m + m + = 0, m = ± i, A Compementry. Function is: = e t (A cos t + B sin t) ft Let = p cos t + q sin t, = p sin t + q cos t, = Sub. in D.E. p cos t q sin t p sin t + q cos t = cos t 6 sin t p + q =, p q = 6 A 0p = 0 p = 0, q = 3 = 3 sin t + e t (A cos t + B sin t) A t = 0, = 0 0 = A = 6 cos t e t B sin t + e t B cos t t = 0, = 0 0 = 6 + B B = 6 = 3 sin t 6 e t sin t A (b) = 6[cos t + e t sin t e t cos t] Sub t = = 6[cos t + e t 6 e t cos t] = e e = 0 P comes to instntneous rest when t = A (c) sub t = in = 3 sin 6 e, =.07, A (d) t 3 sin t, pproimte period is, A [7] Edece Intern Review 7

8 M Dynmics - Dmped nd forced hrmonic motion. () Most cndidtes were be to mke resonbe ttempt though there were some sign errors in Newton s second w. Some weker cndidtes missed out the component of the weight. (b) (c) The uiiry eqution ws usuy soved correcty. Subsequenty common error ws either not to find prticur integr or to ttempt to find it hving redy used the initi conditions on the compementry function this ws heviy penised. Athough the strightforwrd method of equting the veocity to zero ws usuy known, those cndidtes who hd not simpified their nswer to prt (b) were often unbe to compete this prt. There were very sm number who ttempted to use n energy method, occsiony correcty.. In prt (), mny cndidtes fied to reise tht they needed to consider the equiibrium position nd obtined the printed nswer by ignoring the mg term in F = m. In (b), few wsted time by soving the differenti eqution when the soution ws given but fu mrks were often obtined in this prt. A mrk ws sometimes ost in (c) becuse cndidtes gve the gener soution of tnωt = insted of the time t which P first comes to rest. 3. Prt () ws resonby we done, though with so fir number of fudges to produce the given nswer. A cery drwn digrm here woud hve heped. The derivtion of the differenti eqution in prt (b) ws not we done: mny fied to see tht there were four forces cting on the prtice nd the correct etensions in the two prts of the string were rrey correct. Agin cer digrm woud hve heped mny here. Prt (c) ws fmiir to mny though by no mens wys producing fu mrks. Sever mde sips in soving the uiiry eqution, nd mny gve the wrong initi vue of to find one of the constnts of integrtion.. This ws nother question tht ws nswered we. The method for soving the differenti eqution ws we known nd, prt from few cndidtes who mde errors in finding the prticur integr, the mjority of soutions were correct. Most cndidtes knew wht ws required in prts (b) nd (c) nd t = ws substituted in the pproprite formu. In the fin prt ony the best cndidtes reised tht for rge vues of t the dispcement is pproimtey 3 sin t nd therefore the period is. Some tried to rgue tht the period ws times the vue used in prts (b) nd (c). Edece Intern Review 8