Question

A wave represented by the equation $y_1$ = $a\, cos (kx - \omega \,t$) is superimposed with another wave to form a stationary wave such that the point $x = 0$ is node. The equation for the other wave is

• A. $a\,cos\left(kx-\omega t+\pi\right)$
• B. $a\,cos\left(kx+\omega t+\pi\right)$
• C. $a\,cos\left(kx+\omega t+\frac{\pi}{2}\right)$
• D. $a\,cos\left(kx-\omega t+\frac{\pi}{2}\right)$

Answer 1

Answer: (B)

Since the point x = 0 is a node and reflection is taking place from point x = 0. This means that reflection must be taking place from the fixed end and hence the reflected ray must suffer an additional phase change of $\pi$ or a
path change of $\frac{\lambda}{2}.$
So, if $y_{incident} = a\, COS \left( kx - \omega t \right)$
$\Rightarrow \quad y_{incident} = a \,COS \left(-kx - \omega t + \pi\right)$
$= - a\, cos \left(\omega t + kx\right)$
Hence equation for the other wave
$y = a \,cos\left(kx+\omega t + \pi\right)$