Value of Maximum Force - Physics - Solved Past Paper, Exams of Physics

Download Value of Maximum Force - Physics - Solved Past Paper and more Exams Physics in PDF only on Docsity! PHYSICS DEPARTMENT EXAM FALL 2007. PART I INSTRUCTIONS • Please take a few minutes to read through all problems before starting the exam. Ask the proctor if you are uncertain about the meaning of any part of any problem. The questions are grouped in five Sections: Mechanics, E&M, Quantum, StatMech, and General. You must attempt at least one problem from each Section. Credit will be assigned for seven (7) questions only. • You should not have anything close to you other than your pens, pencils, calculator, and food items. Please deposit your belongings (books, notes, backpacks, etc.) in a corner of the exam room. • Departmental examination paper is provided. Colored scratch paper is also provided. Please make sure you: a. Write the problem number and your ID number on each sheet; b. Write only on one side of the paper; c. Start each problem on the attached examination sheets; d. If multiple sheets are used for a problem, please make sure you staple the sheets together and make sure your ID number is written on each of your exam sheets. • At the conclusion of the examination period, please staple sheets from each problem together. Circle the seven problems you wish to be graded: Mechanics E & M Quantum Stat Mech General 1 2 3 4 5 6 7 8 9 10 • Submit this top sheet to one of the proctors, who will check that you have circled the correct problem numbers above. Then submit your completed exam, separated into stacks according to problem number. CODE NUMBER: SCORE: 1 #1 : UNDERGRADUATE CLASSICAL MECHANICS PROBLEM: A rope of mass M and length L is suspended in the earth’s grav- itational field, g, with the bottom end of the rope touching a surface. The rope is then released from rest and falls limply on the surface (i.e., without the elements bouncing upwards). Find the force F (t) on the surface as a function of time, 0 < t < !, and make a sketch of it. At what time does F (t) reach its maximum? What is the value of this maximum force? SOLUTION: All the elements of the rope are in free fall. It takes time T =! 2L/g for the last element to reach the surface. Hence, at t > T we have F (t) = Mg , t > T . Consider now 0 < t < T . Here F is a sum of two terms, F = F1 + F2. The first one is the the weight F1 = µlg of the part that has already fallen. Here l(t) = gt2/2 is the length of the fallen piece and µ = M/L is the mass per unit length. The second term is the transfer of momentum F2 = dP/dt from the element of length dl = vdt that comes to rest during the time (t, t + dt). The velocity of this segment is v = gt, and so F2 = µv(vdt)/dt = µg2t2. Accordingly, F (t) = 1 2 µg2t2 + µg2t2 = 3 2 M L g2t2 = 3Mg t2 T 2 , 0 < t < T . We see that F reaches a maximum F = 3Mg at t = T , then experiences a sudden drop to the three times smaller value, after which it remains constant. CODE NUMBER: SCORE: 4 #4 : UNDERGRADUATE ELECTROMAGNETISM PROBLEM: A spiral spring has N turns and initial length x0. How does its length changes if a small current I is made to flow through it? The spring has an elastic constant k for longitudinal deformations. Assume that the spring can be treated as a perfect solenoid and that its radius R remains fixed. SOLUTION: It is convenient to think that the current I was created by some external source while the spring was kept at the original length x0. The coil was then short-circuited leaving the current flowing. Finally, the spring was allowed to expand or contract freely. In this formulation, the magnetic flux " = LI remains constant since the circuit has zero resistance, which simplifies the derivation. Since the current flows in the same direction in the adjacent coils of the spring, these coils attract. Hence, the spring would shorten. Let x be the new length. To find the contraction #x = x" x0 we can minimize the sum of the magnetic and the elastic energy, E = "2 2c2L + k 2 (#x)2 . where L = 4&2N2R2/c2x is the inductace of the spring (the derivation of this known formula is ele- mentary). Since " = const, it is easy to take the derivative of E with respect CODE NUMBER: SCORE: 5 to x. Equating it to zero, one finds that the mimimum energy is reached at #x & "2& 2 c2 N2R2I2 kx20 . CODE NUMBER: SCORE: 6 #5 : UNDERGRADUATE QUANTUM MECHANICS PROBLEM: Consider a one-dimensional quantum particle with the Hamil- tonian H = T + V (x) , T = " ! 2 2m d2 dx2 . Suppose that m suddenly changes from m0 to m1 = m0/' at t = 0. As- suming the particle was in the ground state at t < 0, find: (i) probability to remain in the ground state at t > 0 and (ii) change in the energy expectation value (H). Consider two cases: (a) Infinite square well, i.e., V (x) = 0 for 0 < x < L and infinite otherwise. (b) Parabolic well, V (x) = Cx2/ 2. Hint: For any a > 0, "" #" dx exp("ax2) = # & a . SOLUTION: In a sudden perturbation the wavefunction does not change; hence, the probability in question is the squared overlap of the two ground- state wavefunctions, ((1|(0)2. The potential energy (V ) also does not change. Therefore, (H)|t=+0t=#0 = (T )| t=+0 t=#0 = ('" 1)(T )|t=#0 . (a) In this case the ground-state wavefunction (0 = sin(&x/L) at t < 0 is also the ground state (1 at t > 0; hence the probability to remain in the ground state is 1. The change in energy is (H)|t=+0t=#0 = '" 1 2 &2!2 m0L2 . (b) The normalized ground-state wavefunction is ( = &#1/4l#1/2 exp("x2/2l2), where l = (!2/mC)1/4. The overlap is ((1|(0) = 1$ & 1 (l0l1)1/2 "" #" dx exp $ "x 2 2 % 1 l20 + 1 l21 &' = ( 2l0l1 l20 + l21 . CODE NUMBER: SCORE: 9 #7 : UNDERGRADUATE STATISTICAL MECHANICS PROBLEM: Molecules of an ideal gas have internal energy levels that are equidistant, En = n+, where n = 0, 1, . . . and + is the level spacing. The degeneracy of nth level is n + 1. Find the contribution of these internal states to the energy of the gas of N molecules at temperature T . SOLUTION: For a non-interacting ideal gas, E = " $ $, N ln * , where * is the single-molecule partition function * = "+ n=0 (n + 1) exp(",n+) . This partition function can be evaluated as follows (x ' ,+): * = "ex d dx "+ n=0 exp , " (n + 1)x - = "ex d dx e#x 1" e#x = [1" exp(",+)] #2. Hence, the sought contribution to the energy is E = 2N+ exp(+/kT )" 1 . Alternatively, one can reproduce this result as follows. One can imagine that every molecule has two independent internal degrees of freedom of harmonic oscilator type, with energy spacing + each. It is easy to see that this model gives the same spectrum and degeneracies if the energy is counted from the ground state. With this convention, the average energy of a single harmonic oscillator is +nB(+), where nB(+) is the Bose-Einstein occupation number. Therefore, for the entire gas we get E = 2N+nB(+), in agreement with the first derivation. CODE NUMBER: SCORE: 10 #8 : UNDERGRADUATE STATISTICAL MECHANICS PROBLEM: What is the change of entropy that occurs when two moles of an ideal gas A and three moles of an ideal gas B, both at standard temperature and pressure are allowed to mix? What if the gases are the same, e.g., A and A? SOLUTION: As we will show, the entropy increases if the gases being mixed are not identical. According to the general principles of statistical mechan- ics, the entropy is S = k lnW , where W is the number of microstates that correspond to a given macrostate. For an ideal gas, we have W = WtrWint where Wtr and Wint are number of microstates due to translational and internal degrees of freedom, respectively. If this gas is non-degenerate, then Wtr = 1 N ! % V '3T &N + % e '3T V N &N , N * 1 , where N is the number of molecules, V is the volume, and '3T is the cube of the thermal wavelength (e!ectively, the “volume” occupied by a mole- cule at temperature T ). The important factorial term N ! eliminates the overcounting of states for indistinguishable particles. The only parameters that change as a result of the mixing are V and the number of moles n. Therefore, we can write S(n) = nR ln(V/n) + const , where R is the universal gas constant. Note that at standard temperature and pressure V and n are directly proportional, V = (22.4 l),n. Using this fact, the increase in entropy due to the mixing can be written as #S = #SA + #SB = n1R ln % V1 + V2 V1 & + n2R ln % V1 + V2 V2 & = n1R ln % n1 + n2 n1 & + n2R ln % n1 + n2 n2 & = R ln % 3125 108 & - 3.4R . On the other hand, if the gases were identical, e.g., A and A, then #S = (n1 + n2)R ln % V1 + V2 n1 + n2 & " n1R ln % V1 n1 & " n2R ln % V2 n2 & = 0 , as expected. CODE NUMBER: SCORE: 11 #9 : UNDERGRADUATE GENERAL PROBLEM: Lightning is known to release a large amount of energy in a form of a short burst. Let W be the energy output per unit length of the lightning and f be the dominant acoustic frequency of the thunder it emits. (a) Use dimensional analysis to express W in terms of f and physical para- meters of the surrounding air, e.g., the speed of sound, density, etc. (b) Under typical conditions, the thunder is heard at f = 100Hz, the speed of sound in air is v = 343 m/s, and the length of the lightning is + 1 km. Estimate the total energy produced by such a lightning and compare it with the energy release of one ton of TNT explosive, 4.6, 109 J. SOLUTION: (a) A sudden release of a large energy along the track of the lightning creates an initially rapidly expanding cylinder of a superhot gas. Remembering that the thermal velocity of molecules coincides with the speed of sound v up to a coe$cient, we conclude that the expansion of a very hot gas is necessarily supersonic. Hence, it creates a shock wave. Since the energy is delivered as a short burst rather than continuously, the expanding gas cools down, slows down, and the shock eventually becomes subsonic. At that moment the sound waves can run ahead of it and be heard as thunder. This description suggests that the most important here are the inertial and sound propagation characteristics of air, i.e., mass density " and the speed of sound v. Temperature, di!usion coe$cient, viscosity, etc, are irrelevant because the process is far from equilibrium. Ambient air pressure P does not add anything either because P + "v2. Hence, we expect W = W (f, ", v). We try the scaling form W = cf#"$v% , where c, !, ,, and - are some dimensionless numbers. The requirement of W to have correct units fixes the last three as follows: W = c"v4/f2 ! W [J/m] = c " [kg/m 3], (v [m/s])4 (f [Hz])2 . Hence, the lower the frequency of the thunder, the higher must be the en- ergy of the lightning. Although it may seem counterintuitive, it can be understood based on the argument that the wavelength of the thunder is set by the radius of the supersonic core around the lightning. Obviously, this radius increases with W . CODE NUMBER: SCORE: 1 #11 : GRADUATE CLASSICAL MECHANICS PROBLEM: Three identical strings are connected to a ring of mass m that can slide frictionlessly along a vertical pole. Each string has tension ! and the linear mass density ". In equilibrium, all strings are in the same horizontal plane. The motion of the strings is in the vertical z-direction. We can choose coordinates x1, x2, and x3 for the three strings, with !" < xi # 0 and the ring position being xi = 0. When a plane wave of a given momentum k is incident on the ring from the first string, it creates trans- mitted waves down the other two strings and a reflected wave on the first string: z1 = f̂(k) exp(ikx1 ! i#t) + ĝ(k) exp(!ikx1 ! i#t) , z2 = ĥA(k) exp(!ikx2 ! i#t) , z3 = ĥB(k) exp(!ikx3 ! i#t) . (a) Write the set of equations of motion for the problem. Define all coe!- cients, e.g., c $ #/k. (b) Find the reflection coe!cient ĝ(k)/f̂(k). (c) Test the correctness of your formula by considering two limits, k % 0 (a very long and slow pulse) and k %" (a very short and fast one). CODE NUMBER: SCORE: 2 SOLUTION: (a) The wave equations read $2zi $x2i = 1 c2 $2zi $t2 , c = ! ! " . Next, if Z exp(!i#t) is the vertical coordinate of the ring, then Newton’s second law implies F = !m#2Z exp(!i#t). Here the force F on the ring is the sum of the vertical components of the tension in the three strings at xi = 0: F = !! 3" i=1 $zi $xi #### xi=0 = !i!ke!i!t(f̂ ! ĝ ! ĥA ! ĥB) . (b) The continuity at the ring demands Z = f̂ + ĝ = ĥA = ĥB . Eliminating ĥA and ĥB from the Newton’s law for the ring, we readily obtain ĝ(k) = ! $ k + iQ k + 3iQ % f̂(k) , where Q $ !/mc2 has dimensions of inverse length. Substituting this into formulas for ĥA and ĥB, we have ĥA(k) = ĥB(k) = $ 2iQ k + 3iQ % f̂(k) . (c) For a very long wavelength pulse, composed of plane waves for which |k| & Q, we have ĝ(k) ' !13 f̂(k). Thus, the reflected pulse is inverted, and is reduced by a factor of 3 in amplitude. The other outgoing pulses have amplitudes (2/3)f̂ . This is consistent with the energy conservation: 12 = (1/3)2 + (2/3)2 + (2/3)2 (for # % 0, the ring oscillates very slowly, and so it has no appreciable kinetic energy). Conversely, for a very short wavelength pulse, k ( Q, we have perfect reflection with inversion, and no transmission. This is due to the inertia of the ring. CODE NUMBER: SCORE: 3 #12 : GRADUATE CLASSICAL MECHANICS PROBLEM: The pivot of an inverted simple pendulum is rapidly oscillated vertically with amplitude a and frequency # (see diagram). Find a condition on # such that % = 0 is a point of stable equilibrium. Hint: Separate the equation of motion into “fast” and “slow” parts. Elim- inate the former. The remaining equation for the slow part determines whether the system is stable. SOLUTION: In the oscillating frame of the pivot, ge! = g + d2y dt2 = g ! #2a cos #t . Therefore, %̈ = g & (1! # 2a g cos #t) % . Let us decompose % into a “slow” % and a “fast” %1 parts: % = %(t) + %1(t) , then d2%1 dt2 + d2% dt2 = g & (1! # 2a g cos #t)(% + %1) . The “fast” equation is d2%1 dt2 = !g & #2a g cos #t % ! %1 = a & % cos #t . CODE NUMBER: SCORE: 6 #14 : GRADUATE ELECTROMAGNETISM PROBLEM: A small amount of water is being warmed in a microwave oven. (a) Derive the formula for the amplitude E of the electric field in terms of the power P dissipated in a unit volume of water, microwave frequency f (in Hz), and the complex dielectric function of water ( = (1 + i(2. Assume that the field penetrates the water uniformly. (b) Compute the voltage drop V (assuming uniform field) across the longest dimension of the oven, L = 0.3 m. Use the following information. Typically, it takes about a minute to heat a cup of water by 10 "C, so that P , 106 W/m3. The microwave frequency is # = 2)f , where f = 2.45 GHz. At such frequency the dielectric function of water is ( , 80 + 10i. SOLUTION: (a) Let " be the real part of the conductivity. In the Gaussian units " = #(2/4) = f(2/2. The ac Joule heating is P = 1 2 "E2 = 1 4 f(2E 2 . Therefore, E = 2 & P/f(2 . (b) Substituting the numbers, we get E [statV/cm] , 2 & 107 [erg/cm3]/ 2.45+ 109 [Hz]/ 10 = 0.04 [statV/cm] = 1.2 kV/m . Accordingly, the voltage drop across the oven is V = EL , 360 V. One more consideration is in order to get the correct estimate for V . (How- ever, failure to acknowledge it was not penalized by taking o" points in grading the exam). Strictly speaking, the field E we computed is actually the field inside the water. The magnitude of the field Eout in the rest of the oven is larger: Eout = E[1 + ((! 1)N ] where N is referred to as the depolarization factor. If water forms a very shallow puddle and the microwave field is parallel to its surface, i.e., hori- zontal, N is very small and Eout , E, so the above estimate of V stands. CODE NUMBER: SCORE: 7 However, if the dimensions of the volume occupied by the water are compa- rable (e.g, water filling a common mug), then f - 1/3, so that |Eout| - |(E|/3 , 30|E| . Therefore, the actual total voltage across the microwave oven is - 360 V + 30 - 10 kV. This explains why safety features are necessary. CODE NUMBER: SCORE: 8 #15 : GRADUATE QUANTUM MECHANICS PROBLEM: A particle of mass m moves in a spherically symmetric potential well V (r) = !V0 < 0 at r < a and V (r) = 0 at r > a. Find the smallest V0 at which a bound state exists at zero angular momentum. SOLUTION: The Schrödinger equation for zero angular momentum reads ! ! 2 2m ' *##(r) + 2 r *#(r) ( = E* , r > a , ! ! 2 2m ' *##(r) + 2 r *#(r) ( = (V0 + E)* , r < a , The sought solution for E < 0 is *(r) = A exp(!+r)/r , + = . !2mE/! , r > a , *(r) = B sin(,r)/r , , = & 2m(V0 + E)/! , r < a , The continuity of *#(r)/*(r) at r = a demands , cot ,a = !+ , As E % 0!, + % 0+, and so ,a% )/2. Thus, minV0 = )2!2 8ma2 . CODE NUMBER: SCORE: 11 #17 : GRADUATE STATISTICAL MECHANICS PROBLEM: A system is composed of N identical classical oscillators, each of mass m, defined on a one-dimensional lattice. The potential for the oscillators has the form U(x) = ( |x/a|n , ( > 0, n > 0 . (Thus, the oscillators are harmonic for n = 2 and anharmonic otherwise). Find the average thermal energy at temperature T . Hint : An integral that appears in the course of evaluating the partition function cannot be computed in terms of elementary functions. Fortunately, it amounts only to an unimportant overall coe!cient. SOLUTION: Classical partition function for a single oscillator is . = %+ !% dp exp , !,p2/2m - %+ !% dx exp (!,(|x/a|n) . By change of variables, we bring this to the form . = #(1/2)#(1/n) 2a n $ 2m , %1/2 $ 1 ,( %1/n , where #(z) $ %+ 0 dt tz!1 exp(!t) . A learned reader would recognize this as the Euler Gamma-function. How- ever, knowing this is not necessary. The product #(1/2)#(1/n) is just a numerical coe!cient, which will disappear from the final result. The average energy is given by E = !N $ $, ln . = $ n + 2 2n % NkT . This result resembles the equipartition theorem in the sense that material constants do not enter. CODE NUMBER: SCORE: 12 #18 : GRADUATE STATISTICAL MECHANICS PROBLEM: The Hamiltonian of N noninteracting spin-1/2 particles in mag- netic field H is given by H0 = !HM , M = µ N" i=1 "i , "i = ±1 . (a) Calculate the average magnetization )M*, the averare square of the magnetization )M2*, and the magnetic susceptibility / = (d/dH))M* at temperature T . (b) Verify that your results obey the thermodynamic identity )M2* ! )M*2 = kT/ . (c) Prove that the above identity holds even in the presence of interactions, H0 % H0 +Hint, for arbitrary Hint({"i}). SOLUTION: A shorter derivation can be given if we start with part (c). (c) The partition function is Q = " {#i} exp , ,HM({"i})! ,Eint({"i}) - , whence )M* = 1 ,Q $Q $H , )M2* = 1 ,2Q $2Q $H2 . Now / = $ $H )M* = 1 ,Q $2Q $H2 ! 1 ,Q2 $ $Q $H %2 = ,()M2* ! )M*2) , which proves the identity. (a) We can now apply the above formulas to the problem in hand. We have Q = . i " #i exp(,µH"i) = [2 cosh(,µH)]N . CODE NUMBER: SCORE: 13 Taking the requisite derivatives, we find )M* = Nµ tanh(,µH) , )M2* = µ2[N(N ! 1) tanh2(,µH) + N ] , / = ,µ2Nsech2(,µH) . (b) We have µ2[N(N ! 1) tanh2(,µH) + N ]! [Nµ tanh(,µH)]2 = µ2N [1! tanh2(,µH)] = µ2Nsech2(,µH) = kT/ . CODE NUMBER: SCORE: 16 #20 : GRADUATE GENERAL PROBLEM: In the absence of other forces, surface tension causes a liquid droplet to assume a spherical shape. Lord Rayleigh has shown that this is no longer true for an electrified droplet of a su!ciently large charge Q. (This instability has found a practical application in ink-jet printers.) Compute the corresponding critical charge Qc for a droplet of radius R and surface tension ". Hint: The capacitance C of a nearly spherical object is related to its surface area S by the Aichi-Russel formula C = & S/4) (Gaussian units) SOLUTION: The total energy of the droplet is E(S) = Q2 2C + "S = ! ) S Q2 + "S . Function E(S) has the minimum at S = Sc, Sc = $. ) 2 Q2 " %2/3 . However, since the sphere has the mimimal surface area for a given fixed volume, S cannot be smaller than 4)R2. As a result, for small Q the sphere remains the optimal shape. The critical charge is determined by the condi- tion Sc = 4)R2, which gives Qc = 4 . )"R3 , 7.1 . "R3 , in agreement with Lord Rayleigh (1882). At Q somewhat larger than Qc the droplet deforms into a prolate ellipsoid. This is the answer the student is expected to give for this problem. Actually, an astute reader may realize that this answer may be incomplete. In principle, the droplet can also change its shape discontinuosly, e.g., by splitting into two smaller droplets. Let us examine this “first-order tran- sition” scenario assuming the new droplets are also spherical and equal in size. CODE NUMBER: SCORE: 17 For the droplet of charge q and radius r the energy is E = q2 2r + 4)"r2 . Comparing the energies of one droplet with q = Q and r = R with that of two droplets with q = Q/ 2 and r = R/ 21/3, we conclude that the first-order instability occurs at Q > Qm = / 8) 21/3 ! 1 1! 2!5/3 01/2. "R3 , 3.1 . "R3 . We see that Qm < Qc, and so the first order transiton wins. More precisely, the spherical droplet with charge Qm < Q < Qc is metastable, and so in practice it still may have a long lifetime.