Question

A spherical wave with a wavelength of 2.0 m is emitted from the origin. At one instant of time, the phase at r = 4.0 m is p rad. At that instant, what is the phase at r = 3.5 m and at r = 4.5 m?

Short answer

The phase at $r=3.5m$is $\frac{1}{2}\mathrm{\pi }\mathrm{rad}$.

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

Step-by-step solution

Given information

The wavelength of the wave is $\lambda =2.0m$

The phase at$r=4.0m$is${\mathrm{\varphi }}_1=\mathrm{\pi }\mathrm{rad}$

The phase at r=3.5 m.

The phase difference is given by ${\varphi }_2-{\varphi }_1=\frac{2\mathrm{\pi }}{\lambda }\left(r_2-r_1\right)$.

Substitute the given values

$$\begin{gathered} {\varphi }_2-\mathrm{\pi }=\frac{2\mathrm{\pi }}{2}\left(3.5-4.0\right) \\ {\mathrm{\varphi }}_2=-0.5\mathrm{\pi }+\mathrm{\pi } \\ {\mathrm{\varphi }}_2=\frac{1}{2}\mathrm{\pi }\mathrm{rad} \end{gathered}$$

Therefore, phase difference at$r=3.5m$is$\frac{1}{2}\mathrm{\pi }\mathrm{rad}$.

The phase at r=4.5 m

Substitute the given values into ${\varphi }_2-{\varphi }_1=\frac{2\mathrm{\pi }}{\lambda }\left(r_2-r_1\right)$.

$$\begin{gathered} {\varphi }_2-\mathrm{\pi }=\frac{2\mathrm{\pi }}{2}\left(4.5-4.0\right) \\ {\mathrm{\varphi }}_2=0.5\mathrm{\pi }+\mathrm{\pi } \\ {\mathrm{\varphi }}_2=\frac{3}{2}\mathrm{\pi }\mathrm{rad} \end{gathered}$$

Therefore, the phase at$r=4.5m$is$\frac{3}{2}\mathrm{\pi }\mathrm{rad}$.