Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 16 - Traveling Waves - Exercises and Problems: 63

Answer

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

Work Step by Step

We can write an expression for the maximum speed. $v_{max} = A~\omega$ We can write an expression for the maximum acceleration. $a_{max} = A~\omega^2$ We can 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 can find the frequency. $f = \frac{\omega}{2\pi}$ $f = \frac{100~rad/s}{2\pi}$ $f = 15.9~Hz$ We can 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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