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Question: A wave equation is represented as \( r=A\sin \alpha \dfrac{\left( x-y \right)}{2}\cos \omega t+\alph...

A wave equation is represented as r=Asinα(xy)2cosωt+α(x+y)2r=A\sin \alpha \dfrac{\left( x-y \right)}{2}\cos \omega t+\alpha \dfrac{\left( x+y \right)}{2} , where xx and yy are in meters and tt is in seconds . Which of the following options is correct?
(A)The wave is a stationary wave
(B)The wave is a progressive wave propagating along positive xx - axis
(C)The wave is progressive propagating at right angles to the positive x-axis
(D)All point lying on line y=x+4παy=x+ \dfrac{4\pi }{\alpha } are always at rest

Explanation

Solution

Hint : In order to solve this question, first of all we must determine the nature of the given wave, whether it is a standing wave or a progressive wave , that can be determined by comparing the given equation to the general equation for standing and the progressive waves.
General Equation for standing wave: y=2Asinωtcoskxy=2A\sin \omega t\cos kx
General Equation for progressive wave: y=Asin(ωtkx)y=A\sin \left( \omega t-kx \right)
sin(2n+1)π=0\sin \left( 2n+1 \right)\pi =0
v=drdtv=\dfrac{dr}{dt}
Where, vv is velocity.

Complete Step By Step Answer:
Firstly, the given wave equation is:
r=Asinα(xy)2cosωt+α(x+y)2r=A\sin \alpha \dfrac{\left( x-y \right)}{2}\cos \omega t+\alpha \dfrac{\left( x+y \right)}{2}
The general equations for the standing and the progressive waves are:
For standing wave: y=2Asinωtcoskxy=2A\sin \omega t\cos kx
For progressive wave: y=Asin(ωtkx)y=A\sin \left( \omega t-kx \right)
Since the given wave equation does not match with any of the above equations, it is neither a standing nor a progressive wave
Replace yy by y=x+4παy=x+\dfrac{4\pi }{\alpha }
r=Asin[α(xx4πα2)]cos[ωtα(x+x+4πα2)] r=Asin(2π)cos(ωtαx2π) r=0 \begin{aligned} & r=A\sin \left[ \alpha \left( \dfrac{x-x-\dfrac{4\pi }{\alpha }}{2} \right) \right]\cos \left[ \omega t-\alpha \left( \dfrac{x+x+\dfrac{4\pi }{\alpha }}{2} \right) \right] \\\ & \Rightarrow r=A\sin \left( -2\pi \right)\cos \left( \omega t-\alpha x-2\pi \right) \\\ & \Rightarrow r=0 \\\ \end{aligned}
And finding the velocity, we get
drdt=0v=0\dfrac{dr}{dt}=0\Rightarrow v=0
Therefore, we can say that all points lying at the line with the equation y=x+4παy=x+\dfrac{4\pi }{\alpha } are always at rest because the velocity is zero.
Therefore, option (D) is correct.

Note :
According to the wave equation, this wave is neither a standing nor a progressive wave . This rules out the first three options and then yy is replaced by y=x+4παy=x+\dfrac{4\pi }{\alpha } to get a result matching with fourth. But to rule out other options is equally important.