Thus D ëc y + c y ë 6= c D ëy ë+c D ëy ë. Note that expressions èè and èè are not equal because sinèc y + c y è 6= c sin y + c sin y. sin is a nonline

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1 The following integral may be useful: Z dx a + bx =. Linear Independence and Linearity. Physics 4 Methods in Theoretical Physics Prof. Mike Ritzwoller May 4, 5 Final Exam p ab tan, xp ab a èaè è4 pts.è Consider two real functions y èxè and y èxè. State the mathematical condition that must be met for them to be linearly independent. For the functions to be considered to be linearly independent on some interval è; è, there must be some point x on the interval that the Wronskian is non-zero. That is W èy ;y è= y èxè yèxè y èxè yèxè 6=: èbè è4 pts.è Using this condition, determine if the two functions y èxè = e x and y èxè =xe x are linearly independent for all nite real values of x. W èe x ;xe x è= xe x e x e x + xe x = e x + xe x, xe x = e x 6=: è4è ex ècè è4 pts.è Consider a real-valued dierential operator Dëyë. State mathematically the condition that must be met for the operator to be determined to be linear. Dëc y + c y ë=c Dëy ë+c Dëy ë. èdè Using this condition, determine if the following dierential operators are linear: èiè è4 pts.è D ëyë =y + pèxèy + qèxèy, The one's linear. D ëc y + c y ë = èiiè è4 pts.è D ëyë =y +siny. This one's not linear. " è d dx + pèxè d dx + qèxè ëc y + c y ë = ëc y + c y ë + pèxèëc y + c y ë + qèxèëc y + c y ë è6è, = c y, + pèxèy + qèxèy + c y + pèxèy +èxèy è7è = c D ëy ë+c D ëy ë: è8è D ëc y + c y ë = " d dx +sin è ëc y + c y ë èè èè è3è è5è è9è = ëc y + c y ë +sinëc y + c y ë èè = c y + c y +sinèc y + c y è èè 6= c èy +siny è+c èy + sin y è èè = c D ëy ë+c D ëy ë: è3è

2 Thus D ëc y + c y ë 6= c D ëy ë+c D ëy ë. Note that expressions èè and èè are not equal because sinèc y + c y è 6= c sin y + c sin y. sin is a nonlinear function.. Potential Energy and Work. èaè è5 pts.è Consider the following potential function in D: Uèxè =x e,x, where x and é. Find the equilibrium points and show whether they are stable or unstable by examining the second-derivative conditions. èconsider only non-innite values of x.è du dx = xe,x è, xè =! x =;x= ;x= : è4è Ignoring innity, there are two equilibrium points; i.e., two points at which the force is zero. d U = e,x x, 4x + dx è5è x =! d U =é! stable dx è6è x =! d U dx =,e, é! unstable è7è èbè è5 pts.è Consider the following force eld: F =,kr where r = ix + jy + kz and k is a constant. Determine if F is conservative. If it is, nd the potential Uèrè such that F rf =,krèix + jy + kzè @x @x so F is conservative. Note that up to a constant, the potential at point r = èx; y; zè is equal to the integral of,f dr from the origin to that point: Uèrè =, = k = k Z r=èx;y;zè Z r=èx;y;zè Z x F dr = k Z r=èx;y;zè èix + jy + kzè èidx + jdy + kdzè è9è èxdx + ydy + zdzè èè Z z zdz = k x + y + z ; èè Z y xdx + k ydy + k where the integrals in equation èè are performed along the straight lines linking the limits on the integral. So, for the rst integral è = dy = dz = y = zè, for the second integral è = dx = dz = zè, aand for the third integral è = dx = dyè. 3. Force in D, no Resistance. è pts.è The force acting on a particle of mass m is given by F = kvx where v is velocity andk is a positive constant. The particle passes through the origin with speed v at time t =. Find xètè. Z x ma = m dv Z dv v = mv dt dx = kvx! dv = k xdx v m dx = v = v + k Z t Z x dt m x! dt èè dx = v + k m x èè è3è

3 tèxè = xètè = s s è! m k v k tan, x mv r mv sv k t A k m è4è è5è 4. First order ODE, application to an RL-circuit. Consider an RL-circuit, a circuit containing a resistor of resistance R and an inductor of inductance L. Kircho's law for the current Iètè in this electrical circuit is: where Eètè is the driving voltage. L di + RI = Eètè dt è6è èaè è5 pts.è Solve equation è6è for Iètè ifeètè =E = constant. Write the ODE in standard form for a st order linear ODE, and then follow standard procedure to solve: di R dt + I = E è7è L L P ètè = R=L è8è Qètè = E =L è9è = e R e Rt=L Iètè = E L P ètèdt = e Rt=L Iètè = E R + ce,rt=l Z e Rt=L dt + c = E R ert=l + c I = Ièè! c = I, E R Iètè = E R + I, E e,rt=l : R èbè è5 pts.è Find the limiting value of Iètè ast!. As t!, Iètè! E =R. 5. Homogeneous nd order ODE with Constant Coecients, Application to the Damped Pendulum Equation. Consider the small amplitude unforced pendulum equation with friction: d dt + d dt +! =; where is the angle the bob makes with the vertical,! = p g=` is the frequency of the undamped oscillator, ` is the length of the pendulum, and is a frictional coecient. èaè è pts.è Write down the auxiliary equation for equation è35è and determine the condition for underdamping of this oscillator. The auxiliary equation is s + s +! =, whose roots are: s ; =,, q q, 4! = i!, =4=, i! : è36è Underdamping means that the square-root is imaginary. That is, 4! é,or! é=. èbè è pts.è Find the frequency of the underdamped oscillator,!. 3 è3è è3è è3è è33è è34è è35è

4 q! =!, =4. ècè è pts.è Derive the general solution to equation è35è for the underdamped oscillator in the following form: ètè =e,t= èc cos! t + c sin! tè: è37è where! is the frequency of the underdamped oscillator. Given the roots of the auxiliary equation, s ;,wehave as a general solution: ètè = Ae s t + A e s t è38è = e,t= hae i! t + A e,i! t i è39è = e hèa i,t= + ibèe i! t +èa, ibèe,i! t è4è " è = e,t= a ei! t + e,i! t +ib ei! t, e,i! t è4è i = e,t= èc cos! t + c sin! tè; è4è where c is twice the real part of A and c is i times the imaginary part of A. èdè è pts.è Assume that the pendulum is initially displaced from rest and is released such thatèè = and _ èè =. Apply these initial conditions to the solution given by è37è. From equation è37è: = èè = c! ètè =e,t= è cos! t + c sin! tè: è43è To apply the second initial condition, _ èè =, we take the time derivative of the previous equation: _ètè =! e,t= è, sin! t + c cos! tè, e,t= è cos! t + c sin! tè è44è = _ èè =! c,! c =! ètè = e,t=! cos! t +!! sin! t! è45è : è46è 6. Inhomogeneous nd order ODE with Constant Coecients, Application to the Forced Damped Pendulum Equation. è pts.è Now consider driving the pendulum in è5 sinusoidally: d dt + d dt +! = sin t: è47è Show that if the oscillator is underdamped the general solution can be written as: ètè =e,t= èc cos! t + c sin! tè, cost +è,! èsint è,! è + : è48è Consider the analogue problem: d dt + d dt +! = eit : è49è 4

5 The imaginary part of the particular solution to this problem will be the solution to the problem with sin t forcing. Substituting the trial solution, x p ètè =Ae it :, + i +! Ae t = e it ; è5è A = è!, è+i è5è which is the particular solution we seek. = è!, è, i è!, è + è5è h Im Ae iti = Im ëa ècos!t + i sin!tèë è53è = è!, è sin t, cos t è!, è + è54è =, cos t +è,! èsint è,! è + 7. Application of Initial Conditions for the -D String. Consider a string of length L clamped at both ends èx = ;x = Lè satisfying the following PDE for displacement yèx; tè: = The solution to this equation that satises the boundary conditions is: yèx; tè = X n= sin k n x èa n cos! n t + B n sin! n tè è56è è57è èaè è5 pts.è What are the allowed values of the discrete wavenumbers k n and frequencies! n? k n = n=l;! n = ck n = nc=l. èbè è5 pts.è Consider starting the string to oscillate by displacing it from rest with the following pattern yèx; è = y sinèx=lè. Apply these initial conditions and write down the solution for the oscillations of this string. The string is excited with a pattern of displacement the same as its second mode èn = è. Thus, only the second mode will be excited with A = y and all other A n = when n 6=. Because the string starts from rest, B n =. Thus, the solution is yèx; tè =y sin k n x cos! n t = y sin x L cos ct 8. Separation of Variables for a PDE, in -D: Application to Laplace's Equation. è pts.è Consider Laplace's equation in -D Cartesian coordinates: L è58è r uèx; yè u =; where x a and y b. Apply separation of variables, uèx; yè = XèxèY èyè, introduce a separation constant, and derive the following two ODEs: X èxè+ Xèxè = ; è6è Y èyè, Y èyè = : è6è 5

6 where the primes indicate a derivative with respect to the independent variable. Substituting uèx; yè = XèxèY èyè into equation è59è, we get: YX + XY = è6è X X + Y Y = è63è X =,Y X Y =, è64è where X = dx=dx and Y = dy=dy, after the rst equation we divided both sides by XY, and we introduced the separation constant in the nal equation as usual. From the nal equation, the two ODEs we seek emerge. 6