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Unformatted text preview: Problem 13.50 Solution This is a physical pendulum. Since it takes 120s to complete 100 swings. Period is T = 120s / 100 = 1 . 2s Use the formula (1414), replace h by d , the distance from center of gravity to pivot point, T = 2 s I mgd We can solve for moment of inertia I with respect to the pivot, I = MgdT 2 4 2 Plug in numbers I = . 1286 kg m 2 . A Wobbling Bridge Solution This is a resonating, driven, damped oscillator. Part A From (1416) and (1417), x = Ae b 2 m t cos( t ) Amplitude of damped oscillator decays as e b 2 m t , plug in t = 6 T , 1 /e = e b 2 m 6 T , or b 2 m 6 T = 1 By T = 2 / , where = p k/m is the angular frequency of undriven, undamped oscillator, we have b/m = 1 6 (1) The given formula of driven amplitude can be rewriten as A = F/m q ( 2 2 d ) 2 + ( b/m ) 2 2 d (2) where m = 2000kg / m * 144m = 2 . 88 10 5 kg is the mass of the oscillator. Maximize this amplitude with respect to d , which is equivalent as minimizing the expression inside of square root, by expanding the expression and completing square, ( 2 2 d ) 2 + ( b/m ) 2 2 d = ( 2 d ) 2 2 2 ( b/m ) 2 / 2 d + ( 2 ) 2 = 2 d 2 1 2 ( b/m ) 2 2 + ( b/m ) 2 ] 2 1 4 ( b/m ) 4 Therefore, resonating angular frequency is 2 d = 2 1 2 ( b/m ) 2 (3)...
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 Spring '08
 Lanzara
 Physics, Gravity

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