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)

$100m,100m{s}^{-1}$

The equation of given transverse wave is $y=0.05\sin \pi (2t-0.02x)$ .......(1)

On comparing (1) with $y=A\sin (wt-kx)$

We get, $k=0.02\pi$ and $w=2\pi$

Wave velocity $v=\frac{w}{k}=\frac{2\pi }{0.02\pi }=100$ $m/s$

Minimum distance of separation between two particles which are in phase is equal to the wavelength of the wave.

Wavelength $\lambda =\frac{2\pi }{k}=\frac{2\pi }{0.02\pi }=100$ $m$