College Physics (4th Edition)

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

Chapter 10 - Problems - Page 399: 40


The maximum speed of vibration is $2.5\times 10^{-6}~m/s$

Work Step by Step

We can find the angular frequency $\omega$: $\omega = 2\pi~f = (2\pi)(4000~Hz) = 8000\pi~rad/s$ We can find the maximum speed: $v_m = A~\omega$ $v_m = (0.10\times 10^{-9}~m)~(8000\pi~rad/s)$ $v_m = 2.5\times 10^{-6}~m/s$ The maximum speed of vibration is $2.5\times 10^{-6}~m/s$.
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