14_P34InstructorSolution

14_P34InstructorSolution - 14.34. Solve: The objects...

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14.34. Solve: The object’s position as a function of time is ( ) ( ) 0 cos . xt A t ω φ =+ Letting 0 m x = at 0 s, t = gives 1 00 2 0c o s A φφ π =⇒ = ± Since the object is traveling to the right, it is in the lower half of the circular motion diagram, giving a phase constant between and 0 radians. Thus, 1 0 2 =− and () ( ) ( ) ( ) 11 22 cos sin 0.10 m sin x tA t x tA t t πω = = where we have used 0.10 m
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