College Physics (4th Edition)

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

Chapter 10 - Problems - Page 399: 34


The maximum speed of the needle is $7.0~cm/s$

Work Step by Step

We can find the period: $T = \frac{9.0~s}{24~cycles} = 0.375~s$ We can find $\omega$: $T = \frac{2\pi}{\omega}$ $\omega = \frac{2\pi}{T}$ $\omega = \frac{2\pi}{0.375~s}$ $\omega = 16.755~rad/s$ The amplitude is half the distance between the lowest and highest point of the motion. We can find the amplitude: $A = \frac{8.4~mm}{2} = 4.2~mm$ We can find the maximum speed of the needle: $v_m = A~\omega$ $v_m = (4.2~mm)~(16.755~rad/s)$ $v_m = 70~mm/s$ $v_m = 7.0~cm/s$ The maximum speed of the needle is $7.0~cm/s$
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