Question

A plane wave $y=a \sin (w t-k x)$ propagates through a stretched string. The particle velocity versus $x$ graph at $t=0$ is

• A.
• B.
• C.
• D.

Answer 1

Answer: (D)

Plane wave is given as
$y=a \sin (\omega t-k x)$
at $t=0$
$y=a \sin (\omega \times 0-k x)$
$\Rightarrow =a \sin (-k x) \Rightarrow -a \sin k x$
Particle velocity,
$v_{p a}=\frac{d y}{d t}=\frac{d}{d t}(-a \sin k x)$
$=-a k \cos k x=-a k \cos \frac{2 \pi}{\lambda} x\left[\because k=\frac{2 \pi}{\lambda}\right]$
$=-\frac{2 \pi a}{\lambda} \cos \frac{2 \pi x}{\lambda}$ ...(i)
From Eq. (i),
When $x=0, v_{p a}=\frac{-2 \pi a}{\lambda} \cdot \cos 0=\frac{-2 \pi a}{\lambda}$
When $x=\frac{\lambda}{4}, v_{p a}=\frac{-2 \pi a}{\lambda} \cdot \cos \frac{\pi}{2}=0$
When $x=\frac{\lambda}{2}, v_{p a}=\frac{-2 \pi a}{\lambda} \cdot \cos \pi=\frac{2 \pi a}{\lambda}$
When $x=\frac{3 \lambda}{4}, v_{p a}=\frac{-2 \pi a}{\lambda} \cos \left(\frac{-\pi}{2}\right)=0$
When $x=\lambda, v_{p a}=\frac{-2 \pi a}{\lambda} \cdot \cos 2 \pi=\frac{-2 \pi a}{\lambda}$
Hence, correct graph is