Question

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

• A. $y=-a\cos (kx+\omega t)$
• B. $y=-a\cos (kx-\omega t)$
• C. $y=a\sin (kx+\omega t)$
• D. $y=-a\sin (kx-\omega t)$

Answer 1

The correct option is A $y=-a\cos (kx+\omega t)$

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 form the fixed end and hence the reflected ray must suffer an additional phase change of 𝜋 or a path change of $\frac{\lambda }{2}$
So, if
$y_{incident}=a\cos (kx-\omega t)$
$\Rightarrow y_{reflected}=a\cos (-kx-\omega t+\pi )$
$\Rightarrow y_{reflected}=-a\cos (\omega t+kx)$

Final answer: (𝒃)