Answer
$\omega = \sqrt{\frac{2k}{3m}}$
Work Step by Step
We use conservation of energy to find:
$E = \frac{1}{2}mv^2 + \frac{1}{2}kx^2+\frac{1}{2}I\theta^2 $
Taking the derivative of this function and setting it equal to zero, we find:
$Ma = \frac{-2kx}{3}$
Equation 13.3 states that $ma = -kx$, so it follows that there is a multiple of $\frac{2}{3}$. Using the equation for angular velocity, this means:
$\omega = \sqrt{\frac{2k}{3m}}$
