sln4 - 7224 Homework Solution 4 Problem 1 (a) m + + C sin =...

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7224 Homework Solution 4 Problem 1 (a) m ¨ θ + α ˙ θ + C sin θ = A sin ω 0 t Let t = c t ¯ t , the equation changes into d 2 θ d ¯ t 2 + α m c t d ¯ t + C c 2 t m sinθ = A m c 2 t sin ( ω 0 c t ¯ t ) (b) To eliminate a second coefficient, we have three possible ways i) Let αc t /m = 1 and c t = m/α , which is the time scale for exponential decay. The equation changes into d 2 θ d ¯ t 2 + d ¯ t + C m α 2 sin θ = Am α 2 sin( ω 0 m α ¯ t ) ii) Let Cc 2 t /m = 1 and c t = q m/C , which is the time scale for the oscillation.The equation changes into d 2 θ d ¯ t 2 + α Cm d ¯ t + sin θ = A C sin( ω 0 r m C ¯ t ) iii) Let Ac 2 t /m = 1 and c t = q m/A , which is the time scale for the driving.The equation changes into d 2 θ d ¯ t 2 + α Am d ¯ t + C A sin θ = sin( ω 0 r m A ¯ t ) (c) From the equations above, Am α 2 is the Rayleigh-like parameter with the meaning of “ratio of strength of drinving to strength of dissipation.” and c t = m/α is the only choice. (d) For a damped pendulum system X 00 + βX 0 + ω 2 X = 0 The general solution is approximately X ( t ) C 0 e - βt/ 2 sinωt 1
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and the quality factor can be define as Q = ω β where in this system
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sln4 - 7224 Homework Solution 4 Problem 1 (a) m + + C sin =...

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