Question

Derive the formula for plane progressive wave moving in positive direction of x axis

Answer 1

The displacement due to the wave at position xx at a time tt is y=f(kx±ωt)y=f(kx±ωt)where f(kx±ωt)f(kx±ωt) is a function which satisfies the wave equation.
It might help your visualisation if you think of that displacement corresponding to a peak.

Since the wave and hence the "information" is travelling to the right at some later time t+Δtt+Δt a particle at x+Δxx+Δx will have the same displacement yy.
In other words the peak has travelled a distance ΔxΔx in a time ΔtΔt with both ΔxΔxand ΔtΔt both positive.

So y=f(kx±ωt)=f(k(x+Δx)±ω(t+Δt)y=f(kx±ωt)=f(k(x+Δx)±ω(t+Δt)

which implies kx±ωt=k(x+Δx)±ω(t+Δt)⇒0=kΔx±ωΔtkx±ωt=k(x+Δx)±ω(t+Δt)⇒0=kΔx±ωΔt

Since in the case of a right travelling wave k,Δx,ωk,Δx,ω and ΔtΔt are all positive the only way to satisfy this equation is to have 0=kΔx−ωΔt0=kΔx−ωΔt which implies that the original function was y=f(kx−ωt)y=f(kx−ωt).

Out of this analysis you also get that ΔxΔt=ωkΔxΔt=ωk which is the speed of the wave.

When the information is travelling to the left then either −Δt−Δt and +Δx+Δx or +Δt+Δtand −Δx−Δx.

To satisfy 0=k(−Δx)±ωΔt0=k(−Δx)±ωΔt or 0=kΔx±ω(−Δt)0=kΔx±ω(−Δt) it must be that y=f(kx+ωt)y=f(kx+ωt).

So if you want to watch the progress of a right travelling wave, ie follow a peak as time tt and position of peak xx increases you need to keep kx−ωtkx−ωt constant.

Answer 2

$\textbf{Answer is in Attachment !!}$