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Chapter 16: Q. 30 (page 452)

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

Expert verified

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

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

Step by step solution

01

Given information

The wavelength of the wave is $λ = 2.0 m$

The phase at $r = 4.0 m$ is $ϕ_{1} = π rad$

02

The phase at r=3.5 m.

The phase difference is given by $ϕ_{2} - ϕ_{1} = \frac{2 π}{λ} (r_{2} - r_{1})$ .

Substitute the given values

$ϕ_{2} - π = \frac{2 π}{2} (3.5 - 4.0) ϕ_{2} = - 0.5 π + π ϕ_{2} = \frac{1}{2} π rad$

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

03

The phase at r=4.5 m

Substitute the given values into $ϕ_{2} - ϕ_{1} = \frac{2 π}{λ} (r_{2} - r_{1})$ .

$ϕ_{2} - π = \frac{2 π}{2} (4.5 - 4.0) ϕ_{2} = 0.5 π + π ϕ_{2} = \frac{3}{2} π rad$

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

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