Question

1. (30pt) LC Circuit and Simple Harmonic Oscillator (From $23.12 RLC Series AC Circuits) Let us first consider a point mass m > 0 with a spring k> 0 (see Figure 23.52). This system is sometimes called a simple harmonic oscillator. The equation of motion (EMI) is given by ma= -kr (1) where the acceleration a is given by the second derivative of the coordinate r with respect to time t, namely dr(t) (2) dt de(t) (6) at) (3) dt Hence, the above EoM is indeed the following second order ordinary i... one variable) differential equation: Fut) Om + ker(t). (4) where we have explicitly showed the time dependence of Ir(t). Let us consider two extreme initial conditions: • When t = 0, the point mass is at the origin = = 0 and initial velocity is given by <> : 1.(0) = 0, .(0) = (5) With these initial condition, we can solve this De and obtain: .(t) sin (6) • Whent -0, the point mass is at 19 >0 at rest: (0) - 0,01(0)-0. (7) wip://openste/bok/collegepl/pp/23-12-rise 1 With these initial condition, we can solve this DE and obtain: (t) = 7. Cos (a) Find the (SI) unit for the spring constant k and check that the prefactor in r.(O) (4.6), namely has the appropriate unit. (You can also check that the argument Vt of the trigonometric functions is unit-less.) (b) Check that both. () and (t) satisfies eq.. That is, first, show that they satisfy the corresponding initial conditions. Then compute the second derivative explicitly, substitute the expression in eq.), and show that the right hand side is indeed zero. You can use the following facts: dr 0088) --sin(sr) (9) d dr sin(sr) - 3008(er) (10) d

Answer 1

+ dt? Spring - Mass system d²alt) Kalt): 0 with initial condition, (t=0) no 7o and at rest, Mo(t): com (1) And it alt=0) = 0 and N (+20) Evos voto au 220 CD 8 nalt) = Nom sin (J +) i (a) for unit of spring constant, from equation dex dt² KE- 26) .. SI unit of ki 2 m kg. (m gr?) N/m [ Newton 61 Z ms CScanned with CamScanner

ST No has the unit in rk Ya (m 15) kg [As N/m = kg /8? kg ls² van * m. S .8 (11 s :, K (Answer), which has the unit of meter in is the unit of amplitude, ta kg ca K [ kg / 3² Y2 is m 1 xs :. The phase term is unit-less, b d² dt sin K + ) -2 K - No к 17 Hall) = no va sin ( +) nalt) of at levo o [V retro com (JK)] vo v sme Creme sim (JE 1] E-VOJ Sim (JE) to lo han sin ( JE 1)] malt) d² kalt) + k na (t) 20 salt) satisfies equa tion 4 K к i. m dt equation ESScanned with CamScanner

Initial condition at for nalt) tis 0, , sin (0) co and Noyo riat=0) No Tomo CO valt) dhalt) dt K [sin 1 12 'बैन No someone on (SE) to cos (Tm ) Nalt=0) No. Co (0) which is greater " Halt) satisfies the boundary than zero. initial boundar conditions, for Rolt a Initial boundary condition no(t) = d.o.com at t=0, no (2-0) R 00 en (0) at t=0, No (t=0) d. XD() lezo dt d at [ no pat) =(-20 og han sin () loco [as sim (o) o i kolt) satisfies the initial boundary conditions. CSScanned with CamScanner

d² ID zt P no Con m 19 d dt 61 ?_n no K (**) - K. one u tkab dt² @) kb (t) (t) = [? [0 (-15) sin (Vit)] ) [no Co (tt) Ek xo(t) Bao Lt) (+) = 0 3. no(t) satisfies the equation 4 energy KE - Żm (v (4)]" Im Et nalt)]' * T Sn (931 en una ventana Co (Vit) ] Im va cos? (kt Potential energy energy = fof. = 2K [na (4)] -Ex [ non sin t) 3 - Ik wanna sin ( E moo Kinetic m 2 12 2 . 2 Sima (en) 2 Total energy Kif. + P E. K imre [sin (it ont £mwa :) CSScanned with CamScanner

... Total energy is independent of time to For ab.lt) Kof. sh Į ml dan (10)] at 두 d m " ؟ 20 co dt 2 2 mao > TUIS RX q (t)}]" [- Van sim ( t)]" = tma tha sin ( t) (VE) P.E. I %%p ()]? = 2 x [no es. (1 t) ] czka cona ( 1 ) energy & Kn [sim ant) It) - kah 3. In both cases kalt) and right), total energies are independent of independent of time and the values are, tmm and km² respectively, (Answer) Total Reft Pi €. & cas ? CSScanned with CamScanner

Period 7 (t + t) - Vik • 2 1 > K T = 20 m => [T Ta 21] m. R (Answer) Le Circuit AE man 6 Cr 16 @ voltage on capacitor. We (t)= 944) (ea capacitance of the & Induced emf on L a=-2 dilt) capacitor) dt g Ve s. À ve enero > alt) [-2 di it dt 7 alt) + L d с dt + doo cale [ As Ilt) *dt & L ďq (t) at² 91t), o () Equation A similar as equation (4) where alt) andre refers to alt) I refers to a refers to is ta K CSScanned with CamScanner

(1) functional forms of nalt) and no(t) → malt) Nom sin (Shah and no(t) = no con tt) 9. No comparing with Le- cirevit, Ye t 9 Lt) = 1. in > Palt) = to the sin (ze t) Qo(t) = 0o CD 7 ablt): % cos lite co (the +) daa (t) :. Ia lt) colitiet dt Todce. The 7 Ia It) = To con Gtet) Io(t) = dy(t) dt do te sin the } ( 9 110 (5) e le sin CSS.canned with CamScanner

21 + (i) For time period to T, phase difference Here phase part the :. tee 14+)- the ts 21 7 the T = 21 (Answers) > T = 2x120 (3) spring + f.dalt) m at 2.3 If some frictional force is present in Oscillator, p286 (t) къtt) that it has solution of the form Ht12m) tlfunk/2m + Bēt V4e-umK [em] We see rotte alt) - to Ae .. we see the that amplitude is longer constant, it with time. decays type of decay depends on the value of frictional constant to Similarly for Le-Oscillator, if such frictional frictional fored foren is present, the amplitude of te oscillation will with time. Here the resistance in decay circuit dissi dissipate the rgy "Noltage (Ve) ener Le .t A CSScanned with CamScanner