Question

A plane wave $y=a\sin (wt-kx)$ 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
$y=a\sin (\omega t-kx)$
at $t=0$
$y=a\sin (\omega \times 0-kx)$
$\Rightarrow =a\sin (-kx)\Rightarrow -a\sin kx$
Particle velocity,
$v_{pa}=\frac{dy}{dt}=\frac{d}{dt}(-a\sin kx)$
$=-ak\cos kx=-ak\cos \frac{2\pi }{\lambda }x\left[\because k=\frac{2\pi }{\lambda }\right]$
$=-\frac{2\pi a}{\lambda }\cos \frac{2\pi x}{\lambda }$ ...(i)
From Eq. (i),
When $x=0,v_{pa}=\frac{-2\pi a}{\lambda }\cdot \cos 0=\frac{-2\pi a}{\lambda }$
When $x=\frac{\lambda }{4},v_{pa}=\frac{-2\pi a}{\lambda }\cdot \cos \frac{\pi }{2}=0$
When $x=\frac{\lambda }{2},v_{pa}=\frac{-2\pi a}{\lambda }\cdot \cos \pi =\frac{2\pi a}{\lambda }$
When $x=\frac{3\lambda }{4},v_{pa}=\frac{-2\pi a}{\lambda }\cos \left(\frac{-\pi }{2}\right)=0$
When $x=\lambda ,v_{pa}=\frac{-2\pi a}{\lambda }\cdot \cos 2\pi =\frac{-2\pi a}{\lambda }$
Hence, correct graph is