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: 14.8

Answer

$k = 0.292~N/m$

Work Step by Step

The time to complete half a cycle is 2.60 seconds. Therefore, the period $T$ is 5.20 seconds. We know that $\omega = \frac{2\pi}{T}$ and $k = m\omega^2$. We can find the force constant $k$. $k = m\omega^2$ $k = (m)(\frac{2\pi}{T})^2$ $k = (0.200~kg)(\frac{2\pi}{5.20~s})^2$ $k = 0.292~N/m$
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