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. $50\, m,\, 50\, ms^{-1}$
• B. $100\, m, \,100 \,ms^{-1}$
• C. $50 \,m, \,100\, ms^{-1}$
• D. $100\, m,\, 50\, ms^{-1}$

Answer 1

Answer: (B)

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