Question

A plane progressive wave of frequency $25Hz$, amplitude $2.5\times {10}^{-5}m$ and initial phase zero moves along the negative x-direction with a velocity of $300m{s}^{-1}$ . A and B are two points $6m$ apart on the line of propagation of the wave. At any instant the phase difference between A and B is $\Theta$ . The maximum difference the displacement of the particles at A and B is $\Delta$ , then

• A. $\Theta =\pi$
• B. $\Theta =0$
• C. $\Delta =0$
• D. $\Delta =5\times {10}^{-5}m$

Answer 1

Answer: (A), (D)

The correct options are
A $\Theta =\pi$
D $\Delta =5\times {10}^{-5}m$

Given : $f=25Hz,A=2.5\times {10}^{-5}m,v=300m/s$,

from , $v=f\lambda$,

or $\lambda =v/f=300/25=12m$,

As points A and B are 6cm apart hence path difference between A and B will be , $\Delta x=6cm$,

by $\Delta \varphi =\frac{2\pi }{\lambda }\Delta x$,

$\Delta \varphi =\frac{2\pi }{12}\times 6=\pi$ (phase difference between A and B),

Now, displacement for particle A,

$y_A=A\sin \omega t$,

or $y_A=2.5\times {10}^{-5}\sin \omega t$,

and , displacement for particle B,

$y_B=A\sin (\omega t+\pi )$, (as wave is travelling in -ive x-direction)

or $y_B=2.5\times {10}^{-5}\sin (\omega t+\pi )$,

or $y_B=-2.5\times {10}^{-5}\sin \omega t$,

therefore , difference in displacements of A and B,

$y_A-y_B=[2.5\times {10}^{-5}-(-2.5\times {10}^{-5}\left)\right]\sin \omega t$,

this difference will be maximum when , $\sin \omega t=1\left(maximum\right)$,

therefore , $\Delta$=$[2.5\times {10}^{-5}-(-2.5\times {10}^{-5}\left)\right]=5\times {10}^{-5}m$