## Essential University Physics: Volume 1 (4th Edition)

Published by Pearson

# Chapter 13 - Exercises and Problems - Page 247: 63

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

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