قسم: العلوم. This test includes three mandatory exercises. The use of non-programmable calculators is allowed.

Transcription

1 الهيئة األكاديمي ة المشتركة قسم: العلوم نموذج مسابقة )يراعي تعليق الدروس والتوصيف المعد ل للعام الدراسي المادة: الفيزياء الشهادة: الثانوية العام ة الفرع: علوم الحياة نموذج رقم 1 المد ة: ساعتان وحتى صدور المناهج المطو رة( This es includes hree andaory exercises. The use of non-prograable calculaors is allowed. xercise 1 (6 poins) Young s slis onsider he Young s slis device (Doc 1) ade up of wo very hin and horizonal slis S1 and S separaed by a disance a = 1, a screen () parallel o he plane conaining S1 and S and a onochroaic ligh source S. The screen () is a a disance D = fro he idpoin I of [S1S]. The ligh source (S) is on he perpendicular bisecor of [S1S]. This bisecor ees he screen () a a poin O. The wavelengh in air of he onochroaic ligh is = 650 n. (Doc 1) (P) 1) A paern is observed on he screen (). Indicae he nae of he corresponden phenoenon. ) Sae he condiions ensured by S1 and S in order o obain his paern. 3) onsider a poin M of he paern observed on he screen () such as OM x. Tae d1 = S1M and d = SM. Wrie he relaion ha gives he opical pah difference = d d1 a M in ers of a, D and x. 4) Define he inerfringe disance i. 5) Give he expression of i in ers of, D and a, hen calculae is value. 6) The poin O coincides wih he cenre of a fringe called cenral fringe. 6-1) alculae he opical pah difference a O. 6-) Specify wheher his fringe is brigh or dar. 7) Le N be he cenre of a fringe where =,75. Specify wheher his fringe is brigh or dar. 8) S is a a disance d = 10 c fro I. We displace S verically of a disance y = 1 c o he side of ax ay S1. The new opical pah difference is hen: '. Specify he direcion of he displaceen D d of he cenre of he cenral fringe (o he side of S1 or S) and calculae he displaceen. xercise (6 poins) () series circui The elecric circui of he docuen (Doc ) is fored of: A generaor delivering across is erinals a consan volage = 8 V; A resisor of unnown resisance ; A capacior of capaciance = 100 µf, iniially discharged; A swich K. K (Doc ) A B q -q 1/3

2 A he insan 0 = 0, we close he swich K. A an insan, he capacior is charged by q and he circui carries a curren i. 1) edraw he figure of he docuen (Doc ) and show he connecions of an oscilloscope ha allows o display he volage ug = across he generaor and he volage u = uab across he capacior. ) Wrie he expression of he curren i in ers of q. 3) Deduce he expression of i in ers of he capaciance and he volage u. 4) Deerine he differenial equaion ha describes he variaion of u as a funcion of ie. 5) The soluion of his differenial equaion is: u D 1 e. Deerine he expressions of he consans D and in ers of, and. 6) Deerine, a he insan =, he expression of he volage u in ers of. 7) eferring o he graph of u = f() of he docuen (Doc 3) below: 7-1) Deerine he value of. 7-) Deduce he value of he resisance. 8) Deerine he expression of he curren i as a funcion of ie. 9) Deduce he value of he curren i in seady sae. xercise 3 (7 poins) Horizonal elasic pendulu An air puc (S) of ass = 709 g is aached o he free end of a spring () of un-joined urns, of negligible ass and of siffness = 7 N. -l. This puc, of cenre of ass G, ay slide wihou fricion on a horizonal rail (Doc 4). The docuen (Doc 4) shows a horizonal axis Ox of origin O. A equilibriu, G coincides wih O. (S) is shifed 3 c fro O ( OG 0 = x 0 i = 3 i) in he posiive direcion and released wihou velociy a he insan 0 = 0. A an insan, x is he abscissa of G and dx v is he algebraic easure of is velociy. d /3

3 1) The echanical energy of he syse ((S), (), arh) is conserved. 1-1) Deerine he second order differenial equaion in x. 1-) Verify ha x x cos is he soluion of his differenial equaion. 1-3) alculae he values of he consans x and. ) Wrie down he expression of he naural period T0 of he oion in ers of and hen calculae is value. 3) The docuen (Doc 5) below shows he curves giving he variaions of he ineic energy K of (S), of he elasic poenial energy Pe of () and of he echanical energy M of he syse ((S), (), arh). Idenify he curves K, Pe and M of he docuen (Doc 5). 4) ach of he curves A and is sinusoidal of a period T. eferring o he graph of docuen (Doc 5) : 4-1) Pic up he value of he period T; 4-) opare is value o he naural period T0 of he oion. 3/3

4 الهيئة األكاديمي ة المشتركة قسم: العلوم أسس التصحيح )تراعي تعليق الدروس والتوصيف المعد ل للعام الدراسي المادة: الفيزياء الشهادة: الثانوية العام ة الفرع: علوم الحياة نموذج رقم 1 المد ة: ساعتان وحتى صدور المناهج المطو رة( xercise 1 (6 poins) Young s slis Quesion Answer Mar 1 Inerference. The ligh sources us be synchronous (hey us have he sae frequency) and coheren (hey us eep a consan phase difference). 3 ax D 4 The inerfringe disance is he disance beween he ceners of wo consecuive fringes of he sae naure. 5 D i a i i d = d1 = d d1 = 0 or x = 0 ax 0 D 6- = 0 so = wih = 0 Z The inerference is consrucive and he fringe is brigh so wih = 1 Z The inerference is desrucive and he fringe is dar. ax O' ay y.d 0 x O' D d d 10 xo' The cenral fringe oves 0. owards S 1/3

5 xercise (6 poins) () series circui Quesion Answer Mar 1 K P M A B dq i d 3 du q u so i d 4 Law of addiion of volages: 5 upm = upa +uab +ubm upa = u ; uab = u and ubm = 0 So : u u Ɐ Oh s law: u i u du d The differenial equaion in ers of u is hen: u D 1 e u D De du d u du 1 D D e e d du eplace u and by heir expressions in he differenial equaion. d We ge: D e D De Ɐ D( 1)e D 0 Ɐ Idenifying, we ge: D-=0 D= 1 0 τ = 6 1 A = ; u 1 e 1 e 0, A = ; u = 0.63 = 0.63 x 8 = 5.04 V 5 V fro he graph we ge : = s /3

6 8 9 du i e e e d Peranen regie: = ; i e 0 0 A xercise 3 (7 poins) Horizonal elasic pendulu Quesion Answer Mar 1-1 dp g P g consan because he rail is horizonal 0 d M K P e P g The echanical energy of he syse (puc, spring, arh) is conserved M = v + x dm + Pg = consan Ɐ 0 Ɐ d x x xx 0 0 Ɐ x x x 0 Ɐ The produc of he wo quaniies is always nil. Bu x' is no always nil, we ge: x x 0 Ɐ 1- x x cos x' x sin x" x cos x eplace x" by is expression in he differenial equaion: The relaion x x 0 is rue. A 0 = 0 s ; v0 = x' 0 x sin 0 sin 0 0 or rd A = 0 s ; x0 x cos 0 For 0 rd : x x 3c (accepable because x 0) 0 For rd : x0 x 3c x 3 c (rejeced because x is always posiive) T T 0 π s 7 1 The curve A corresponds o Pe because a 0 = 0 s, x0 0 bu P e x so Pe(0) 0 J The curve B corresponds o M because i has a consan value 1 The curve corresponds o K because a = 0 s, v = 0 /s bu K v so K(0) = 0 J 4-1 Fro he graph we ge : T = 1 s 4- T = T0/ 3/3