## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson

# Chapter 14 - Oscillations - Exercises and Problems: 47

#### Answer

$f = 1.7~Hz$

#### Work Step by Step

When the spring is stretched a distance of $x = 9.0~cm$, the force from the spring is equal in magnitude to the weight. We can write an expression for $\frac{k}{m}$: $kx = mg$ $\frac{k}{m} = \frac{g}{x}$ We can find the frequency of the oscillations as: $f = \frac{1}{2\pi}~\sqrt{\frac{k}{m}}$ $f = \frac{1}{2\pi}~\sqrt{\frac{g}{x}}$ $f = \frac{1}{2\pi}~\sqrt{\frac{9.80~m/s^2}{0.090~m}}$ $f = 1.7~Hz$

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