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Suppose that an object is attached to a coiled spring. It is pulled down a distance of 8 cm from is equilibrium position and then released. It takes 1/3 second to complete one oscillation. Find the position of the object after 1/2 second.

User NSNoob
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1 Answer

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Suppose that an object is attached to a coiled spring. It is pulled down a distance of 8 cm from is equilibrium position and then released.

Therefore, a=8 cm.

It takes 1/3 second to complete one oscillation.

Therefore, t=1/3

we know that,


S(t)=-a\cos wt

So, we have


S(t)=-8\cos wt\ldots\ldots(1)

first we need to find w :

The formula to find w is,


\begin{gathered} w=(2\pi)/(t) \\ =(2\pi)/((1)/(3)) \\ =(6\pi)/(3) \end{gathered}

Therefore, the equation (1) becomes,


S(t)=-8\cos (6\pi)/(3)t

Next to find the position of the object after 1/2 second.

substituting t=1/2 in the above equation we get,


\begin{gathered} S(t)=-8\cos ((6\pi)/(3).(1)/(2)) \\ =-8\cos ((3\pi)/(3)) \\ =-8\cos \pi \\ =-8(-1)_{} \\ =8 \end{gathered}

Hence, the position of the object after 1/2 second is 8 cm.

User Diego Puente
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