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Unformatted text preview: x =16 . 1 1 / 3 . 11 + A cos0 . 567 t + B sin0 . 567 t x = 50 + A cos0 . 567 t + B sin0 . 567 t (1) The initial conditions for the problem are 2 @ t = , x = @ t = , x = Using these on equation (1) we get A =50 B = x = 50(1cos0 . 567 t ) When x = 25 f t , the weight is at the top, 25 = 50(1cos0 . 567 t TOP ) t TOP = 1 . 567 cos1 1 2 = 1 . 85 sec At top, the velocity is determined as x = 50(0 . 567)sin(0 . 567)1 . 85 = 24 . 6 f t / sec 3...
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 Spring '10
 richard
 Quadratic equation, Harshad number, xc, Introductory Dynamics, S. Balachandar

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