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

Published by Pearson

# Chapter 13 - Exercises and Problems - Page 240: 69

#### Answer

$\frac{1}{2\pi}\sqrt{\frac{2a}{m}}$

#### Work Step by Step

We know that the force is given by $F=-\frac{dU}{dx}=-2ax$. Thus, we use Hooke's Law and the equation for frequency to find: $-kx=-2ax \\ k=2a$ Thus, it follows: $f= \frac{1}{2\pi}\sqrt{\frac{k}{m}}=\frac{1}{2\pi}\sqrt{\frac{2a}{m}}$

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