Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 14 - Exercises and Problems: 54

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: $53 \ Watts$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.