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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 π s I mgd We can solve for moment of inertia I with respect to the pivot, I = MgdT 2 4 π 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( ω 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 π/ω , where ω = p k/m is the angular frequency of undriven, undamped oscillator, we have b/m = 1 6 π ω (1) The given formula of driven amplitude can be rewriten as A = F/m q ( ω 2 ω 2 d ) 2 + ( b/m ) 2 ω 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 ω d , which is equivalent as minimizing the expression inside of square root, by expanding the expression and completing square, ( ω 2 ω 2 d ) 2 + ( b/m ) 2 ω 2 d = ( ω 2 d ) 2 £ 2 ω 2 ( b/m ) 2 / ω 2 d + ( ω 2 ) 2 = ‰ ω 2 d • ω 2 1 2 ( b/m ) 2 ‚ 2 + • ( b/m ) 2 ] ω 2 1 4 ( b/m ) 4 ‚ Therefore, resonating angular frequency is ω 2 d = ω 2 1 2 ( b/m ) 2 (3)...
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This homework help was uploaded on 04/02/2008 for the course PHYSICS 7A taught by Professor Lanzara during the Spring '08 term at Berkeley.
 Spring '08
 Lanzara
 Physics, Gravity

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