Question

The equation of a transverse wave is given by $y=0.05\sin \pi (2t-0.02x)$, where $x,y$ are in metre and $t$ is in second. The minimum distance of separation between two particles which are in phase and the wave velocity are respectively

• A. $50m,50m{s}^{-1}$
• B. $100m,100m{s}^{-1}$
• C. $50m,100m{s}^{-1}$
• D. $100m,50m{s}^{-1}$

Answer 1

Answer: (B)

Given, $y=0.05\sin \pi (2t-0.02x)$
Or $y=0.05\sin (2\pi t-0.02\pi x)$
On comparing this equation with standard equation
$y=a\sin (\omega t-kx)$
where $\omega =\frac{2\pi }{T}$ and $K=\frac{2\pi }{\lambda }$, we have
$\frac{2\pi }{\lambda }=0.02\pi$
$\Rightarrow \lambda -100m$
Also, $\omega =\frac{2\pi }{T}=2\pi$
or $\frac{1}{T}=1$
$\Rightarrow v=1$
So, $v=r\lambda =1\times 100$
$=100m/s$