Essential University Physics: Volume 1 (4th Edition)

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 14 - Exercises and Problems - Page 272: 56


a) 1.75 b) 29.78 c) .0134 d) 2209 e) 53

Work Step by Step

a) From the wave function, we see that the amplitude is 1.75. b) We see that k is equal to .211. Thus, it follows that the wavelength is: $\lambda = \frac{2\pi}{k}=\frac{2\pi}{.211}=29.78$ c) We see that $466 \ rads/s$ is the wave speed. Thus, it follows: $T=\frac{2\pi}{\omega}=\frac{2\pi}{466}=.0134\ s$ d) We find: $v = f\lambda = \frac{\lambda}{T} = \frac{29.78}{.0134} =2208.5 \ cm/s$ e) We find: $P = \frac{1}{2}\mu v \omega^2 A^2$ This becomes: $P = \frac{2\pi^2 A^2 F }{vT^2}$ Plugging in the known values gives: $53 \ Watts$
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