Essential University Physics: Volume 1 (4th Edition) Clone

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

Chapter 14 - Exercises and Problems - Page 272: 56

Answer

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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