Question

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

Answer 1

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

Given,
The equation of incident wave,
$y_1=a\cos (kx-\omega t)$
When another wave is superposed with incident wave such that the point $x=0$ is a node,
wave equation of the wave for which $x=0$ is a node,
$y_2=-a\cos (kx+\omega t)$
On superposition of two waves,
$y=y_1+y_2=a\cos (kx-\omega t)-a\cos (kx+\omega t)$
$=a[\cos kx\cos \omega t+\sin kx\sin \omega t-\cos kx\cos \omega t+\sin kx\sin \omega t]$
$y=2a\sin kx\sin \omega t$
At $x=0$
We get $\sin kx=0$
$\Rightarrow y=0$ $\therefore x=0$ is a node.
Therefore, $y_2=-a\cos (kx+\omega t)$ is our other wave equation.

Final answer: (𝑏)