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When two equal and opposite waves meet at a point, destructive interference occurs. But afterwards, waves go about their own ways. But for waves on a string for example, two opposite waves have momentums of the same magnitude and opposite (in both horizontal and vertical) directions at the instance and point of encounter so shouldn't the waves simply cancel?

Is it because the at the point of contact happens a sort of elastic collision (where incoming wave from +x bounces off to equal and opposite -x direction)? If so, what are the condition that enforces it? That is, what property that arises from the geometry of wave conserved this property despite the physical nature of its medium (air, water, string)?

If the conservation of momentum of waves at the point of encounter of propagating medium cannot be seen in terms of elastic collision, what would be an intuitive way to view it?

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  • $\begingroup$ The continued motion of waves in either direction is the result of the geometry of the resultant wave formed during the time they pass the same spot. $\endgroup$ Commented Feb 3, 2020 at 10:19
  • $\begingroup$ I don't quite understand the question, in particular what you are trying to say about elastic collisions. The amplitude of the waves cancel, but not the first derivative. But more to the point, perhaps, is the wave energy and momentum is defined as an average over a cycle. It's not defined at a single point. $\endgroup$
    – garyp
    Commented Feb 3, 2020 at 11:47
  • $\begingroup$ Can you please tell me how you determine the momentum of a wave on a string? $\endgroup$ Commented Feb 3, 2020 at 13:02
  • $\begingroup$ Force acted on for some short period of time gives change in momentum, so portion of the string where wave is present experiences that. But at the point where two opposite waves meet on a small portion of a string, wouldn't that be same as applying two equal and opposite force for a short period of time on a box? In the case of the box, the box is stationary after but for waves some how the momentum seems to pass on, and I want to know why that is. Thanks for the help $\endgroup$
    – VVC
    Commented Feb 4, 2020 at 3:17

2 Answers 2

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I am using two gaussian bell curves as pulses to make the pictures enter image description here The green pulse travels towards right and red one towards left ,note that the points on the front part of wave have a tendency to move up and the back parts have tendency to go down(when the pulse points in positive y axis,you can work out the other conventions) enter image description here When the pulses collide in my diagram ,the portion of string towards left of $x=0$ have a tendency to go up while the portion towards right of $x=0$ wants to go down ,this creates a pull in the actual string ,the left of $x=0$ is pulled down while right is pulled up,note that the back portions of the pulses are still trying to maintain their original state. enter image description here The amplitude of colliding pulses is becoming smaller due to the pull. enter image description here After a certain time pulses flip because the back portion was pulling in the origianl direction (downwards for green and upwards for red)and the pull created in the collision created a pull opposite to the direction of original pull (When the pulses were travelling) enter image description here The back portions still maintain directions but left of $x=0$ is now being pulled up while right is being pulled down ,this creates the scenario for travelling pulses again i.e front and back portion of waves have opposite tendencies of motion ,the pulses grow bigger and leave unaffected. enter image description here

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  • $\begingroup$ Thanks this gives me a better understanding because I see how movement due to restoring force meets with that of displacing force of the other leads to flipping of the wave. But I still cannot understand why two pulls that exist in opposite direction is conserved until the restoring and displacing forces align. That is, I get how restoring tail meet with the head of the other, but shouldn't heads of both wave disappear before that happens? In your second picture, there is no net displacement where heads meet so no net incurred restoring force. $\endgroup$
    – VVC
    Commented Feb 4, 2020 at 4:11
  • $\begingroup$ so shouldn't the displacing heads of two waves cancel then disappear if zero net displacement incurs no motion in the next moment? $\endgroup$
    – VVC
    Commented Feb 4, 2020 at 4:11
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    $\begingroup$ No displacement does not imply no restoring force,remember ,potential energy is minimum at mean position during SHM but in case of an actual real string ,potential energy is maximum at the mean position because string is stretched to max. at that location. $\endgroup$ Commented Feb 4, 2020 at 8:03
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    $\begingroup$ There are forces at the meeting point of the pulses in the second picture ,both pulses are pulling the other in their direction ,the forces being equal and opposite cancel and there is no displacement of that point but deformation does occur ,just think would a soda can be unaffected if you crush it with equal and opposite forces ? 0 net force doesn't mean no activity. $\endgroup$ Commented Feb 4, 2020 at 8:07
  • $\begingroup$ This is more like two identical masses approaching each other under gravity ,because there is 0 net force the COM stays at rest(analogous to our point here) but the masses will move under the influenece of the other's gravity (pulse flliping) $\endgroup$ Commented Feb 4, 2020 at 8:09
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I think what you have done very well is highlight a term which is confusing. Destructive is probably a poor choice of words because it implies that the two travelling waves are interacting with each other in a way that would cause say a loss of energy when in fact that is not really the case, they are passing each other and temporarily creating zero amplitude in the medium. Destructive interference is counter intuitive to our sense of the world, just like extra dimensions are. Destructive interference can create points of silence where we would otherwise expect a cumulative loud sound, when considering for example interfering sound waves

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