## Essential University Physics: Volume 1 (3rd Edition)

$\omega = \sqrt{\frac{2k}{3m}}$
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: $\omega = \sqrt{\frac{2k}{3m}}$