This question arises from analysis of standing waves.The incident wave has the equation $$y=Asin(kx-\omega t)$$The reflected wave has the equation$$y=Asin(kx+\omega t +\pi)$$When determining the standing wave equation , we add these two , but that should be correct only if ONE wavepulse from left proceeds exactly when ONE wavepulse from right does , simultaneously , It should imply that the incident pulse take one time period $T$ to change its phase by $\pi$, but I am unable to prove it.

enter image description here

Here is an animation.To express my question better, I have a diagram too. enter image description here

In the diagram the $RED$ wave is a case where reflection takes $T$ time to occur so the by the time an incident pulse turns around completely , the source finishes generating an identical and complete pulse simultaneously.

In the $ORANGE$ wave the pulse takes less than the time period $T$ to turn around so the source is unable to generate a complete pulse.This case cannot be handled by adding the equations I wrote formerly (I believe).

So how much time does the pulse take?

  • $\begingroup$ What are you defining to be the "beginning" and "end" of the phase change? When the peak is not at its maximum? $\endgroup$ – Aaron Stevens Mar 27 at 14:31
  • $\begingroup$ As seen in the animation-When the pulse ( I mean the whole bump thing and not only the peak) just reaches the boundary, I take this to be the "beginning" of phase change and when the whole bump turns around completely and resembles the original pulse completely ( except the upside-down orientation ) to be the "end" of phase change. $\endgroup$ – ADITYA PRAKASH Mar 27 at 14:45
  • $\begingroup$ Shouldn't it be half the time period? It is pretty obvious I guess. $\endgroup$ – Akash Roy Mar 27 at 14:53
  • $\begingroup$ The $T$ I was referring to is not for one crest and a trough combined , the pulses I am sending along the string are "one bump" each. $\endgroup$ – ADITYA PRAKASH Mar 27 at 14:59
  • 1
    $\begingroup$ @ADITYAPRAKASH I am not saying the entire process is instantaneous. I am saying the direction change is instantaneous. Just like when throwing a ball in the air, you can determine the point in time where the velocity is $0$, and hence at that instant the direction changes. The entire process of the ball moving up and down is of course not instantaneous, but the direction change is. The ball doesn't hit the top of its trajectory, levitate for an extended period of time, and then fall down. In your case the instantaneous direction change is at some time $t_0$ when $y(x,t_0)=0$ for all $x$ $\endgroup$ – Aaron Stevens Mar 27 at 16:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.