1
$\begingroup$

When a wave gets reflected, doesn't its phase change by 180 degrees. So shouldn't a wave and its reflected counterpart get canceled out as their phase differs by 180 degrees?

$\endgroup$
3
  • $\begingroup$ It is due to destructive interference itself that the nodes and internodes are formed $\endgroup$
    – Agrim Arsh
    Mar 12 at 15:36
  • 1
    $\begingroup$ The phase does not (generally) change by 180 degrees. $\endgroup$
    – Prahar
    Mar 12 at 16:09
  • $\begingroup$ I've learnt that the reflection of waves off a rigid boundary changes their phase by π. Am i wrong to assume so? $\endgroup$
    – Gokul L
    Mar 12 at 16:17

5 Answers 5

3
$\begingroup$

At the wall there is indeed a phase change of $\pi$ between the incident and the reflected waves and that is the reason for a node at that position.
Away from the wall there are positions where the incident and reflected waves are again $\pi$ out of phase and that is where there are nodes, but there are also positions where the incident and reflected waves are exactly in phase and that is where there are anti-nodes.
So the "cancellation" only occurs at some positions.

$\endgroup$
2
$\begingroup$

Imagine $2$ identical sinusoidal waves except that one is going to $x+$ and other to $x-$. It is easy to see that they will cancel out for some specific times, but not always. For example, a little while after the state of canceling, the crest of $x+$ wave will be added to the node of the $x-$ wave.

The sum of waves: $Asin(kx - \omega t) + Asin(kx + \omega t)$ can be expanded to:

$$A(sin(kx)cos(\omega t) - sin(\omega t)cos(kx) + sin(kx)cos(\omega t) + sin(\omega t)cos(kx))$$

Two terms cancel out leaving: $$2Asin(kx)cos(\omega t)$$

What in general is different of zero, except for the $x$'s such that $sin(kx) = 0$

$\endgroup$
1
$\begingroup$

Imagine a square wave pulse of length $L$ traveling at speed $v$ on a string reflecting off a fixed boundary, which causes the reflected wave to change phase by 180 degrees. Suppose the square wave pulse reaches the boundary at $t=0$.

  • For $t<0$, no part of the wave has been reflected, and there is no destructive interference.

  • For $0 < t < L/(2 v)$, the length of the pulse that has not hit the boundary is greater than $L/2$, while the length of the pulse that is reflected is less than $L/2$. Therefore there is only partial cancellation of the wave pulse, in the region where the wave has been reflected.

  • At $t=L/(2v)$, the phenomenon you anticipated occurs. The length of the wave pulse that has not yet hit the boundary is $L/2$. The length of the pulse that has reflected off the boundary is $L/2$. These two pieces of the pulse overlap and have opposite signs, so there is complete destructive interference. If you took a snapshot of the string at this instant, you would not see any displacement of the string from equilibrium.

  • For $L/(2v) < t < L/v$, the length of the pulse that has not hit the boundary is less than $L/2$, while the length of the pulse that is reflected is greater than $L/2$. Therefore there is only partial cancellation of the wave pulse, in the region where the wave has not yet been reflected.

  • For $t > L/v$, the entire wave has been reflected, and there is no destructive interference.

As you can see, you cannot use the fact that the phase of the reflected wave is shifted by 180 degrees to conclude that the reflected wave is zero. You need to do a more careful analysis to establish when and where there is destructive interference.

In the case of a standing wave, you don't need to worry about when there is destructive interference, as the pattern (locations of the nodes and antinodes) of a standing wave does not change -- this is why standing waves are special. (Except, of course, there are times when the entire string will have zero amplitude because the amplitude of the pattern is oscillating in time). However, you do need to worry about where there is destructive interference. In particular, there is perfect destructive interference at the nodes of a standing wave.

$\endgroup$
0
$\begingroup$

It is 180 right there, at the reflection point. At various points along the propagation direction the phase difference has different values, in the range 0 to 180 degrees. So you can get maxima, minima and in between values.

$\endgroup$
0
$\begingroup$

Yes the magnitude of oscillations of particles that we see in standing waves is a result of interfernce of the original and reflected wave at all instants . When the crest of original wave interferes with crest of 180• reflected wave the particles in the standing wave are at their max displacment (fully constructive interference ) . Similarly when the crest of one wave interferes with trough of other all the particles in the standing waves Crosses their mean position (since this a fully destructive interference). Remember two interfering waves only produce resulting wave they do not cancel out each other forever. Thus destructive interference is only when crest and valley crosses each other. And then other values of displacement occurs.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.