Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 16 - Traveling Waves - Conceptual Questions - Page 450: 7


$\lambda = 40~cm$

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: $\lambda = \frac{2\pi~\Delta r}{\Delta \phi }$ $\lambda = \frac{2\pi~(r_2-r_1)}{\phi_2-\phi_1}$ $\lambda = \frac{2\pi~(80~cm-20~cm)}{3\pi-0}$ $\lambda = 40~cm$
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