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 - Exercises and Problems: 16

Answer

$\lambda = 40~cm$

Work Step by Step

Let $\phi_1 = 0$ be the phase at $r_1 = 20~cm$ and let $\phi_2 = 3\pi~rad$ be the phase at $r_2 = 80~cm$. We can find the wavelength of the wave as: $\lambda = \frac{2\pi~\Delta r}{\Delta \phi }$ $\lambda = \frac{2\pi~(r_2-r_1)}{\phi_2-\phi_1}$ $\lambda = \frac{2\pi~(80~cm-20~cm)}{3\pi~rad-0}$ $\lambda = 40~cm$
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