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

Published by Pearson

# Chapter 20 - Traveling Waves - Exercises and Problems: 15

#### Answer

The phase at r = 3.5 m is $\frac{\pi}{2}$ The phase at r = 4.5 m is $\frac{3\pi}{2}$

#### Work Step by Step

Let $\phi_1 = \pi$ be the phase at $r_1 = 4.0~m$ and let $\phi_2$ be the phase at $r_2 = 3.5~m$. Therefore; $\Delta \phi = \frac{2\pi~\Delta r}{\lambda}$ $\phi_2-\phi_1 = \frac{2\pi~(r_2-r_1)}{\lambda}$ $\phi_2 = \phi_1+\frac{2\pi~(r_2-r_1)}{\lambda}$ $\phi_2 = \pi+\frac{2\pi~(3.5~m-4.0~m)}{2.0~m}$ $\phi_2 = \frac{\pi}{2}$ The phase at r = 3.5 m is $\frac{\pi}{2}$. Let $\phi_1 = \pi$ be the phase at $r_1 = 4.0~m$ and let $\phi_3$ be the phase at $r_3 = 4.5~m$. Therefore; $\Delta \phi = \frac{2\pi~\Delta r}{\lambda}$ $\phi_3-\phi_1 = \frac{2\pi~(r_3-r_1)}{\lambda}$ $\phi_3 = \phi_1+\frac{2\pi~(r_3-r_1)}{\lambda}$ $\phi_3 = \pi+\frac{2\pi~(4.5~m-4.0~m)}{2.0~m}$ $\phi_3 = \frac{3\pi}{2}$ The phase at r = 4.5 m is $\frac{3\pi}{2}$.

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