## University Physics with Modern Physics (14th Edition)

Published by Pearson

# Chapter 15 - Mechanical Waves - Problems - Exercises - Page 502: 15.66

#### Answer

(a) $A = 1.28~cm$ (b) $a_{max} = 6320~m/s^2$

#### Work Step by Step

(a) We can find the speed of the wave. $v = \sqrt{\frac{F}{\mu}}$ $v = \sqrt{\frac{F}{(m/L)}}$ $v = \sqrt{\frac{330~N}{(0.00300~kg/2.20~m)}}$ $v = 491.9~m/s$ We can find the frequency. $f = \frac{v}{2L}$ $f = \frac{491.9~m/s}{(2)(2.20~m)}$ $f = 111.8~Hz$ We can find the angular frequency. $\omega = 2\pi~f$ $\omega = (2\pi)(111.8~Hz)$ $\omega = 702.5~rad/s$ The expression for the maximum transverse speed at the anti-node is $\omega~A$. We can find the amplitude. $\omega~A = v_{max}$ $A = \frac{v_{max}}{\omega}$ $A = \frac{9.00~m/s}{(702.5~rad/s)}$ $A = 0.0128~m = 1.28~cm$ (b) The expression for the maximum transverse acceleration at the anti-node is $\omega^2~A$. We can find the maximum acceleration. $a_{max} = \omega^2~A$ $a_{max} = (702.5~rad/s)^2(0.0128~m)$ $a_{max} = 6320~m/s^2$

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