Question

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.

Answer 1

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.