Question

Two waves $y_1=A\sin (\omega t-kx)$ and $y_2=A\sin (\omega t+kx)$are superimpose to produce a stationary wave, then

• A. $x=0$ is a node
• B. $x=0$ is an antinode
• C. $x=\frac{\pi }{k}$ is a node
• D. $x=\frac{2\pi }{k}$ is an antinode

Answer 1

Answer: (B), (D)

$x=\frac{2\pi }{k}$ is an antinode

By the principle of super position
$y=y_1+y_2=\left(2A\cos kx\right)\sin \omega t=A_x\sin \omega t$
Here, $A_x=2A\cos kx$
At $x=0,A_x$ is maximum or $2A$.
So, it is an antinode.

Next antinode will occur at
$\cos kx=\pm 1$
$kx=2\pi$
$x=0,\frac{\pi }{k},\frac{2\pi }{k},....$