University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 14 - Periodic Motion - Problems - Exercises - Page 460: 14.12


$f = 0.692~Hz$

Work Step by Step

We can find the angular frequency. $-\omega^2~x = a$ $\omega = \sqrt{\frac{-a}{x}}$ $\omega = \sqrt{\frac{-(-5.30~m/s^2)}{0.280~m}}$ $\omega = 4.35~rad/s$ We can find the frequency of the motion. $f = \frac{\omega}{2\pi}$ $f = \frac{4.35~rad/s}{2\pi}$ $f = 0.692~Hz$
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