Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 20 - Traveling Waves - Exercises and Problems - Page 589: 62


$f = 15.9~Hz$ $A = 2.0~cm$

Work Step by Step

We write an expression for the maximum speed. $v_{max} = A~\omega$ We write an expression for the maximum acceleration. $a_{max} = A~\omega^2$ We divide the second equation by the first equation. $\frac{a_{max}}{v_{max}} = \frac{A~\omega^2}{A~\omega}$ $\omega = \frac{a_{max}}{v_{max}}$ $\omega = \frac{200~m/s^2}{2.0~m/s}$ $\omega = 100~rad/s$ We find the frequency. $f = \frac{\omega}{2\pi}$ $f = \frac{100~rad/s}{2\pi}$ $f = 15.9~Hz$ We find the amplitude. $v_{max} = A~\omega$ $A = \frac{v_{max}}{\omega}$ $A = \frac{2.0~m/s}{100~rad/s}$ $A = 0.020~m = 2.0~cm$
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