Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 20 - Traveling Waves - Conceptual Questions - Page 585: 7

Answer

The amplitude is 4 cm The frequency is 2 Hz The phase constant is $\frac{\pi}{6}$

Work Step by Step

The wave oscillates vertically between -4 cm and 4 cm. Therefore, the amplitude is 4 cm. From the graph, we can see that the horizontal length of one complete oscillation is 12 cm. That is, the wavelength is 12 m. We can find the frequency as: $f = \frac{v}{\lambda}$ $f = \frac{24~m/s}{12~m}$ $f = 2~Hz$ At t = 0, $D = \frac{A}{2}$ at the point x = 0. Therefore; $D(t) = A~cos(\omega~t+\phi)$ $cos(\phi) = \frac{1}{2}$ $\phi = arccos(\frac{1}{2})$ $\phi = \frac{\pi}{6}, \frac{5\pi}{6}$ From the graph, we can see that the point x = 0 is moving downward at t = 0. Therefore, $\phi = \frac{\pi}{6}$.
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.