Question

The phase difference between two points separated by $ 0.8\,m $ , in a wave of frequency $ 120\,Hz $ , is $ \frac{\pi }{2} $ . The velocity of wave is

• A. $ 720\,ms^{-1} $
• B. $ 384\,ms^{-1}$
• C. $ 250\,ms^{-1}$
• D. $ 1\,ms^{-1} $

Answer 1

Answer: (B)

It is given that $v = 120\, Hz,$
phase difference $ \Delta \phi = \frac{\pi}{2} $ and path difference $ \Delta x = 0.8 \, m. $
As $ \Delta \phi = \frac{2 \pi}{\lambda}.\Delta x $
$ \Rightarrow \, \lambda = \frac{2 \pi}{\Delta \phi} . \Delta x = \frac{2 \pi \times 0.8}{\pi / 2} = 3.2 \, m $
Hence, wave velocity
$ v = v \lambda= 120 \times 3.2 = 384 \, ms^{-1} $ .