Answer
$T=2\pi \sqrt{\frac{7}{10ga}}$
Work Step by Step
We first find the kinetic energy:
$ E_k = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 $
$ E_k = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)\frac{v^2}{r^2} $
$ E = \frac{7}{10}mv^2 $
$E = \frac{7}{10}m(\frac{dx}{dt})v^2 $
Thus, we can find omega. Using the equation for omega, we obtain:
$\omega = \frac{x}{\frac{d^2x}{dt^2}}$
$\omega= \sqrt{\frac{7}{10ga}}$
The period is equal to the angular velocity times 2 pi, so we find:
$T=2\pi \sqrt{\frac{7}{10ga}}$
