College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 10 - Problems - Page 399: 36


The speed at the equilibrium point is $0.63~m/s$

Work Step by Step

We can find $\omega$: $T = \frac{2\pi}{\omega}$ $\omega = \frac{2\pi}{T}$ $\omega = \frac{2\pi}{0.50~s}$ $\omega = 4\pi~rad/s$ At the equilibrium point, the speed reaches a maximum. We can find the speed at the equilibrium point: $v_m = A~\omega$ $v_m = (0.050~m)~(4\pi~rad/s)$ $v_m = 0.63~m/s$ The speed at the equilibrium point is $0.63~m/s$.
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