College Physics (4th Edition)

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

Chapter 10 - Problems - Page 399: 33


$a = 7.9~m/s^2$

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 point of maximum extension of the spring, the magnitude of acceleration is at its maximum. We can find the magnitude of the acceleration: $a = A~\omega^2$ $a = (0.050~m)~(4\pi~rad/s)^2$ $a = 7.9~m/s^2$
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