test2a_fall07

test2a_fall07 - PHY 206 EXAM II Oct. 24, 2007 NAME:...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: PHY 206 EXAM II Oct. 24, 2007 NAME: SIGNATURE: ID: 1. 2. 3. Some relations you may need: pV = nRT dW = pdV dQ = mCdT dQ = dU + dW, adiabatic process : pV = const. dS = dQ/T, S = kB ln w 2y 1 2y = 2 2 x2 v t fL = v + vL fS v + vS 224 (130) 1. An object of mass m and specific heat C is heated from initial temperature 100 K to final temperature 300 K by being placed in contact with a heat reservoir at temperature 300 K. The object and reservoir are isolated from the surroundings. a) Compute the entropy change of the object. b) Compute the entropy change of the reservoir. c) Find the entropy change of the universe. d) Suppose instead the object is heated in two steps: first it is placed in contact with an intermediate temperature reservoir at temperature 200 K and then it is placed in contact with the reservoir at 300 K. What is the entropy change of the universe in this case? Is this entropy change greater, less than, or the same as the result from c)? [You might find it useful to employ ln A+ln B = ln(AB)] Solution: a) 300K So = 100K dQ T dT = mc ln(3) T 300K dQ = mCdT So = mC 100K b) Tr = const. Sr = Qr Tr . Qo +Qr = 0 Qr = -Qo = -mcT = -mc(300K-100K) = -mc(200K) Sr = -mc c) Suniv. = So + Sr = mc ln 3 - d) 200K 200K 2 = - mc. 300K 3 2 3 . So = 100K dQ + T 300K 200K dQ T So = mc[ln(2) + ln(3/2)] = mc ln(3). Sr = Qr1 Tr1 + Qr2 Tr2 . Qr1 = -Qo1 = -mc(200K - 100K) = -mc(100K) Qr2 = -Qo2 = -mc(300K - 200K) = -mc(100K) Sr = -mc 100K 100K 5 - mc = - mc. 200K 300K 6 5 6 . Suniv. = So + Sr = mc ln 3 - The entropy change is lower with the intermediate reservoir (since 5/6 is greater than 2/3) as it must be because the temperature differences during heat transfer are smaller than with the single higher-T reservoir and hence the processes are "more reversible." 2. A general expression for the wave function of transverse periodic waves moving to the right is, y(x, t) = A cos(kx - t + ), where is a phase angle satisfying 0 2. a) For the case = /2, use the axes given to sketch one full wavelength for the wave function at t = T /2 (T is the period). b) Calculate the transverse particle velocity. With = /2, what is this velocity at x = 0, t = T /2? c) Suppose you don't know . If you are told that a particle at x = is moving with transverse velocity A at t = 0, what is the value of ? y Solution: A a) -A x b) vy = y = A(-)(-) sin(kx - t + ) = A sin(kx - t + ). t vy (0, T /2) = A sin[0 - (2/T )T /2 + /2] = A sin(-/2) = -A. c) vy (, 0) = A sin[(2/) - 0 + ] = A sin(2+) = 1 2+ = (2m+1/2) and -(2m+3/2), m = 0, 1, 2, . . . The only allowed value that satisfies 0 2 is = /2. 3. A military helicopter and submarine are stationary above and below a buoy, respectively, at altitude h and depth y. Each emits identical sound waves (synchronized in time) of frequency f0 directly to the buoy at the ocean surface. The sound velocities in air and water are v and v , respectively (primed symbols will refer to the water throughout). a) Write down expressions for the wavelengths (, ) and wave numbers (k, k ) in air and water. b) Write down the condition for the signals to interfere constructively at the buoy and use this to show that the depths for which this condition is satisfied are, y= h y v (h m), m = 0, 1, 2 . . . v c) Suppose the submarine dives straight down at velocity u = v /100 while continuing to emit signals at f0 . Find the frequency ratio f /f0 , where f is the new frequency received by the buoy from the submarine. Solution: a) = v/f0 , = v /f0 , k = 2/ = 2f0 /v, k = 2/ = 2f0 /v . b) The phase difference between the signals must be a multiple of 2 for constructive interference, = kh - k y = 2m, m = 0, 1, 2, . . . h y - = m y h = m y = h m Now using the relations found in a), w have / = v /v, thus y= v (h m). v c) Here we have a stationary listener (the buoy) and a receding source (the submarine at velocity u = v /100). The sound velocity in the Doppler equation is that of the water, v . The vector from the listener to the source is downward, the same direction in which the source moves, so according to convention u is positive in the denominator of the Doppler frequency relation: f = v f0 v +u f v 100 = = . f0 v + v /100 101 ...
View Full Document

Ask a homework question - tutors are online