Midterm%20Solutions - Copy (6)

Midterm%20Solutions - Copy (6) - Prof. Raghuveer...

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Prof. Raghuveer Parthasarathy University of Oregon; Fall 2007 NAME : SOLUTIONS Physics 351 – Vibrations and Waves MIDTERM EXAMINATION Format: Closed book; no calculators. Four problems. Time: 50 minutes. (Suggested times: # 1 , 5 min; # 2 , 10 min., # 3 & # 4 , each 15-20 min.) Write clearly! You may make use of the following if needed: For the steady state response of a damped oscillator driven at angular frequency ω : The amplitude ( A ) and phase offset ( δ ) are given by: () 0 1 22 2 2 2 0 0 / ); t a n ( ) ( Fm A γω ωδ ω ωω γ == ⎡⎤ −+ ⎢⎥ ⎣⎦ The power absorbed (averaged over one cycle): 1 2 2 00 0 2 0 ( 1 ) 2 F kQ Q P ⎛⎞ =− + ⎜⎟ ⎝⎠ As usual: m is the mass; k is the spring constant.; γ is the damping factor (“= b / m ”) ω 0 is the angular frequency of the undamped system. F 0 is the amplitude of the driving force
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(1, 7 pts. total ) A driven oscillator. The power-absorption versus angular frequency, ) ( P ω , for a driven oscillator is measured and plotted (see graph, right). (a, 2 pts. ) If the Q-factor of the system is increased, will the peak become narrower or wider? [ Just provide a one-word answer .] Answer: Narrower. (b, 5 pts. ) The driving is turned off and the oscillations freely decay from an initial value x 0 . I’ve plotted x (t) – see below but due to an accident with scissors I chopped the numbers off the time scale of the graph. Which of the time scales (A, B, C) shown below the plot is the correct one for this system? Very briefly explain your answer . Answer: A. The width (FWHM) of the ) ( P curve is about 0.2 rad/sec. We know that the FWHM is equal to γ , the damping factor, and that the amplitude of free oscillations decays as /2 t e . Therefore the 1/e decay time is 2/ 10 seconds, corresponding to scale A.
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(2, 7 pts. ) Oscillation in a sinusoidal potential. Consider a particle of mass m subject to a potential energy 0 () c o s2 x Ux U π λ ⎛⎞ = ⎜⎟ ⎝⎠ , where U 0 and λ are constants and x is position. Determine the angular frequency ( ω ) of small oscillations about any equilibrium position. Answer: The equilibrium positions are the minima of U(x), which occur at x 0 = 2 , 3 2 , 5 2 , . .. If this isn’t obvious, note that 0 2 sin 2 0 d U x x U d λλ =− = at x=integer* 2 , and 2 2 2 0 2 cos 2 0 U U dx x d = −> only at the odd-integer multiples. All the equilibrium points are equivalent. Taylor expanding U(x) around x 0 , 2 2 00 0 2 0 0 1 ( ) ( ) ... 2 x x Ux dU d U xx dx d x +− + + , which for x 0 being an equilibrium point becomes 2 2 2 0 1 ) ... 2 ( x dU U d x x ++ 2 2 0 12 )( ) ( 2 U x x ≈+ , limiting ourselves to small (x-x 0 ) so that we can neglect higher-order terms. As we are well aware, having done similar exercises before, this looks just like the potential energy function for a spring: 2 1 " " 2 kx = , with x being deviation from equilibrium (i.e. like (x-x 0 ) above) and spring constant k , except for an irrelevant offset (U(x 0 )). Here, 2 0 2 "" kU = . Therefore the oscillation frequency 0 2 U k m m ω ==
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(3, 8 pts. ) A damped, oscillating sphere. Consider a sphere of radius r and density ρ attached to a spring of spring constant k . Damping is due to a drag force FC r v = − ,where v is the (1-
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This note was uploaded on 11/08/2009 for the course PHY 245 taught by Professor Barzda during the Spring '09 term at Adelphi.

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Midterm%20Solutions - Copy (6) - Prof. Raghuveer...

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