14_P41InstructorSolution

14_P41InstructorSolution - 14.41. Model: The block on a...

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14.41. Model: The block on a spring is in simple harmonic motion. Solve: (a) The position of the block is given by ( ) ( ) 0 cos . xt A t ω φ =+ Because ( ) x tA = at 0 s, t = we have 0 0 rad, = and the position equation becomes ( ) cos . x tA t = At 0.685 s, t = () 3.00 cm cos 0.685 A = and at 0.886 s, t = 3.00 cm cos 0.886 . A −= These two equations give ( ) ( )
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This note was uploaded on 12/10/2011 for the course PHYS 1211 taught by Professor Geller during the Fall '09 term at UGA.

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