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

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