وزارة التربية والتعلين العالي الوديرية العاهة للتربية دائرة االهتحانات اهتحانات الشهادة الثانىية العاهة الفرع : علىم عاهة مسابقة في مادة الفيزياء المدة ثالث ساعات االسن: الرقن: الدورة اإلستثنائية للعام This exam is formed of four exercises in four pages. The use of non-programmable calculaor is allowed. Firs Exercise (6 poins) Inerference of Ligh The objec of his exercise is o show how o use Young's double sli apparaus o measure very small displacemens. A source pu a a poin S, emiing a monochromaic radiaion of wavelengh in air, illuminaes he wo slis S and S ha are separaed by a disance a. The screen of observaion is placed a a disance D from he plane of he slis.. Describe he aspec of he inerference fringes observed on he screen.. A a poin M of abscissa x = OM, he opical pah difference is given by he relaion: ax = MS MS = D a) A he poin O, we observe a brigh fringe, called cenral brigh fringe. Why? b) Wha condiion mus saisfy in order o observe, a M, a dark fringe? c) Give he expression of x in erms of a, D and, so ha M is he cener of a brigh fringe. d) Given: λ.55 m ; a =. mm; D =.5 m ; d = cm. We ake x =.65 cm. Are he waves inerfering a M in phase or ou of phase? Jusify your answer. 3. We move he source from S o poin S' verically up on he axis y'y perpendicular o he horizonal axis of symmery SO, by he disance b = SS'. In his case,we can wrie S'S S'S = d ab. a) The cenral brigh fringe is no longer a O bu a poin O'. i) Jusify his displacemen. ii) Specify, wih jusificaion, he direcion of his displacemen. b) Deermine he value of b, knowing ha OO' = cm. Second Exercise (8 poins) Moion of a conducor in wo fields Two verical rails ' and DD' are conneced by a resisor of resisance. A conducing rod MN, of mass m and lengh, can slide wihou fricion along hese rails and remains horizonally in conac wih hese rails. The whole se-up is placed wihin a uniform and horizonal magneic field B ha is perpendicular o he plane of he rails. The rod MN, released from res a he insan =, is found a an insan a a disance x from D, moving wih a velociy whose algebraic value is v ( v > ) ( adjacen figure).
. Deermine, a he insan, he expression of he magneic flux due o B hrough he circui MND in erms of B, and x, aking ino consideraion he arbirary posiive direcion as shown on he figure.. a) Deermine he expression of: i) The e.m.f e induced across he rod MN, in erms of v, B and. ii) The induced curren i in erms of, B, and v. b) Indicae, wih jusificaion, he direcion of he curren. B 3. Show ha he elecric power dissipaed by he resisor, a he insan, is given by : P el = v. 4. The rod MN is aced upon by wo forces: is weigh mg and he Laplace s force F of magniude F = i B. a) Applying Newon s second law, show ha he differenial equaion in v is given by: dv B v g. d m b) The soluion of his differenial equaion is: v = A( e ). Show ha: mg m A = and = B B. c) Show ha v would aain a limiing value V lim. d) i) Give he expression of v as a funcion of V lim a he insan = ii)deduce he ime a he end of which v aains pracically is limiing value. e) alculae he value of V lim and ha of, knowing ha: = cm, m = g, =., B =.5 T and g = m/s. 5. In he seady sae, saring from he insan when v = V lim, he mechanical energy of he sysem (MN in he field B, Earh) decreases. a) Explain his decrease. b) In wha form is his energy dissipaed? c) alculae he power dissipaed. Third Exercise (8 poins) Sudy of harging and Discharging of a apacior The adjacen circui allows o sudy he variaion of he volage u = u BM across a capacior of capaciance during charging and discharging. We consider a generaor delivering a consan volage E, a resisor of resisance = 5 and a coil of inducance L and of resisance r. Iniially, he swich K is in posiion () and he capacior is uncharged. An oscilloscope allows displaying he variaion of u as a funcion of ime. A harging of he capacior A he insan =, he swich is in posiion () and he capacior sars charging. A an insan, he circui carries a curren i and he capacior carries he charge q.. a) edraw he diagram of he circui indicaing on i he real direcion of i. b) Wrie down he relaion beween i and u... a) Derive he differenial equaion in u. b) The soluion of his differenial equaion is of he form: u = A+B e. Deermine he expressions of he consans A, B and. Documen
c) eferring o he graph of documen, deermine: i) The values of E and. Deduce ha he value of is 4F. ii) The minimum duraion a he end of which he capacior is pracically compleely charged. B Discharging of he capacior hrough a coil K is moved from posiion () a he insan l =.6 ms and becomes in posiion () a he insan = ms. The documen shows he variaion of u beween he insans and 7ms... The volage u remains consan beween he insans l and.why?. Saring from he insan = ms, he circui is he sea of elecric oscillaions. eferring o he graph of documen, give he value of he pseudo period T of he elecric oscillaions. 3. a) Wrie down he expression of he proper period T in an L circui. r b) Knowing ha L =.56 H and, calculae r. T T L 4. a) Deermine, referring o documen, he value of u a he insan = 6 ms. b) alculae he value of he loss in elecric energy in he circui a he end of he firs oscillaion. Fourh Exercise (8 poins) Mechanical Oscillaors The pars A and B are independen. We neglec fricion in all his exercise. A- Simple pendulum A simple pendulum (P) is formed of a paricle G of mass m conneced o one end of an inexensible sring, of negligible mass and of lengh L ; he oher exremiy is conneced o a fixed poin A. The pendulum is shifed by an angle θ m from is equilibrium posiion AG o he posiion AG i, hen released from res a he insan = s ; hus i oscillaes wih he ampliude θ m. A an insan, he posiion of AG is defined by θ, he angular abscissa relaive o is equilibrium posiion, and v is he algebraic measure of he velociy of G (Fig.). Take he horizonal plane hrough G as a graviaional poenial energy reference. 3
. Find he expression of he mechanical energy of he sysem [(P), Earh] in erms of m, g, L, v and θ.. Derive he second order differenial equaion in θ ha governs he moion of his pendulum. 3. a) Wha condiion mus saisfy in order o have a simple harmonic moion? b) Deduce, in his case, he expression of he proper period T of he oscillaions. c) Wrie down he ime equaion of moion, in he case θ m =. rad. Take : g = m/s ; L = m and. B- Horizonal elasic pendulum A solid (S) of mass m may slide on a horizonal plane; i is conneced o a spring () of siffness k = 4 N/m. When (S) is in equilibrium, is cener of mass G is found verically above he poin O, aken as origin on he horizonal axis of abscissa. (S), shifed from is equilibrium posiion, is released from res a he insan =. A an insan, he abscissa of G is x and he algebraic value of is velociy is v. A convenien apparaus gives he variaion of x as a funcion of ime (Fig. 3). The horizonal plane conaining G is aken as a graviaional poenial energy reference.. Derive he second order differenial equaion in x ha governs he moion of G.. The soluion of his differenial equaion is of he form : x =X m cos( ), where X m, T and are T consans.eferring o he graph of figure (3), give he values of X m and T and deermine. 3. a) Deermine he expression of he proper period T in erms of m and k. b) Deduce m. 4. a) eferring o he graph of figure (3), give he insans a which he elasic poenial energy is maximum. b) alculae hen he value of he mechanical energy of he sysem [(S), ( ), Earh]. - Behavior of he pendulums on he Moon We suppose ha he wo preceding pendulums are now on he Moon. Tell, wih jusificaion, for each pendulum, which of he saemens in he following able is rue. Saemen Saemen Saemen 3 T does no vary T increases T decreases 4
وزارة التربية والتعلين العالي الوديرية العاهة للتربية دائرة االهتحانات اهتحانات الشهادة الثانىية العاهة الفرع : علىم عاهة هشروع هعيار التصحيح الدورة اإلستثنائية للعام Firs exercise (6 poins) Par of he Q Answer Fringes are: alernaing brigh and dark- parallel o each oher and o he slis- recilinear- equidisan. Mark.a A poin O, =, all he waves arriving o O are in phase: we observe a O a brigh fringe..b A poin M of abscissa x, we observe a dark fringe if he pah difference λ a his poin is such ha: δ (k) wih k an ineger.c The abscissa x of he poin M is obained from he expression of he ax kd difference of pah: k = x = D a.d ax.6.5 = =. 3 3. mm. 4, hus is a 3 3 D.5.55 muliple ineger of wavelenghs: he waves ha inerfere a M are in phase. 3.a.i If O remains he cener of he cenral brigh fringe : ' =(S'S + S O) (S'S + S O) = (S'S S'S ) Bu for he.b.f we have : ' = O displaced o O'. 3.a.ii ' =(S'S + S O') (S'S + S O') = (S'S S'S ) + (S O' S O') = (S'S S'S ) = (S O' S O') ; (S'S S'S ) > (S O' S O') < S O' < S O' O displaced downward. Anoher meh. ab ax b x bd ' = x = The.B.F displaced d D d D d downward. 3.b The cenral fringe on he screen corresponds o a zero pah difference a his poin M: ' = SS + S M (SS + S M) S ' = (SS SS ) + ( S M S M) ' = d ab + D ax. ¾ The cenral fringe corresponds o ' = ; is posiion is defined by: ab ax + = d D d d () so: b = x b = x =.667 cm =.667 mm D D.5 ¾
Second exercise (8 poins) Par of he Q Answer Mark The magneic flux = B. S n = BS = Bx..a.i The induced emf e = d dx = B d d.a.ii The value of he curren i = e = Bv.b i >, hen i passes from M o N in he rod. 3 The dissipaed elecric power: P el = i = ( Bv ) B = v. 4.a dp B mg F ; wih F = ibl = v d Projec verically downward: B dp mg v = = m dv d d. 4.b 4.c dv B vg d m dv = A e ; A e +A B B - A d m m and A = A B m mg m A = and = B B = Bv.. e = g. A B m = g When increases, he erm e ends owards zero and v ends owards A. mg Thus V lim = A =. B 4.d.i For =, v = V lim ( e - ) =.63 V lim. 4.d.ii The conducor MN reaches pracically is limiing speed for = 5 =.5 s. 4.e V lim =. = m/s and =..5..5. 5.a The kineic energy does no vary since he speed of MN remains consan while he graviaional poenial energy decreases 5.b I is dissipaed as Joule s effec in he resisor. 5.c P = B v lim =.5.. =. s =. W.
Third exercise (8 poins) Par of he Q A..a ircui and direcion of i Answer i e - Mark + E - q B A M + - A..b A..a A..b du i c d From he law of addiion of volages: u AM = u AB + u BM E = i + u du u E d e u = A + B e A = u = A + B = A = - B u = A A du d A e A e A Ae E ; A = E and. A..c.i (u ) = E = 6 V. For =, u =.63 E = 3.78 V. From he graph: =. ms. = = 4-6 F. A..c.ii min = 5 =.5 ms = 5-4 s. B. During he passage from () o (), he circui is open and he volage u does no vary and reains he value of 6 V beween =.6 ms and = ms B. T = 5 ms = 5-3 s T = L. B.3.a B.3.b T = L 4.96-3 s, replace in he given equaion o ge r 3 B.4.a A he insan = 6 ms, u =.5 V B.4.b c W = W W = (E u ) = 4-6 (36.5 ) = 6.89-5 J 3
Fourh exercise (8 poins) Par of he Q Answer Mark A. ME = KE k +PE g = mv + mgl ( cosθ ). A. A.3.a A.3.b A.3.c dme ' '' ' g mvx' ' mglθgsinθ ml θ θ mglθ sinθ ' ' sin. d L m should be < o.. m < o < sin = in rad ; in his case: he differenial equaion g is: '' L g g The proper angular frequency is such ha ( = =. L L The proper period is T = L =. g ) Le = m sin( + ) ; wih m =. rad and = a = we have: = m = rad =.sin( + ) sin = m sin = B. ME = KE + PE elasic + PE g = mv + kx + = cons. dme k mvx'' kxv x'' + x. d m g = rad/s ; L B. X m =. m ; T = s ; if =, x = X m cos = X m cos = =. B.3.a x = X m cos( ) v = X m sin( ) T T x''= X m cos( ) ; T T T replacing in he differenial equaion, we obain: X m T T T k cos( ) + Xm cos( ) = T m T = T m k m T =. k B.3.b T = s and k = 4 N/kg we obain: m =. kg B.4.a B.4.b E pe = kx, E pe is max. if x is max. x = ±. m a he insans,,5s, s,,5s, s. For P.E g max, KE = ME = P.E g = k(x m ) = J. L For he simple pendulum T = ; g bu g moon < g earh = g so T increases. Saemen () m For he elasic pendulum T = ; k T does no vary on he Moon.(Saemen ) 4