(I) The two pulses shown in Fig. 15–37 are moving toward each ...

Question

(I) The two pulses shown in Fig. 15–37 are moving toward each other. In Fig. 15–37, at the moment the pulses pass each other, the string is straight. What has happened to the energy at this moment?

Accepted Answer

Everyone. Let's take a look at this practice problem dealing with wave interference. So in this problem, we have two transverse pulses are traveling towards each other through a slinky pulse A is moving to the right and pulse B is moving to the left when the pulses intersect, the slinky is completely straight. Question wants to discuss what happens to the energy of the pulses at this point below this, we have a diagram showing the two pulses. Pulse A is on the left and pulse B is on the right, the two pulses have the same width and the same amplitude. However, pulse A is below the equilibrium point on the string. And pulse B is above the equilibrium line on the string. We're also given four choices as our answers. Choice A is the energy is completely lost. Choice B the energy remains the same but is entirely elastic potential. Choice C is the energy is partially kinetic and partially potential. In choice D, the energy remains the same but is entirely kinetic. Now, the first bit of information I want to focus on is the fact that when the pulses intersect the blinky is completely straight this means we have completely, completely destructive interference. So I'm actually gonna draw this situation on the screen. So I'm just gonna have a horizontal line for my string. What I'm going to do is superimpose what the waves would have looked like, had they not interfered with each other. So I have pulse V on top and pulse A on the bottom. So pulse A is going to be moving to the right and pulse B is going to be moving to the left. Now, if I look at these two pulses, every point um has basically the opposite with the other pulse. So the maximum of B will actually cancel out with the maximum of A since one is above the equilibrium point. Um And the other is below the equilibrium point for the strength. And so if you think of the upper half as being positive and lower half being negative, you have basically B minus A and they cancel each other out to give you zero along at every point. However, the moment after this, the incident in time after this um diagram that we have here on the screen occurs, we're gonna have B re emerging on the left and pulse A re emerging on the right. So eventually those pulses will look just like they do in the original diagram that we were shown underneath the question, except the B would be on the left and A would be on the right with B still traveling to the left and a still traveling to the right. So that means that the energy is not completely lost, the energy remains actually the same. But we don't know what type it is yet. We need to figure that out. And to figure that out, we're going to look at the direction that the points move for each of the pulses. If I look at pulse B here and look at the points to the left or the maximum, those points are going to be moving upward. So as the pulse moves to the left, those points are getting pulled upwards up until it reaches a maximum height, then we for the left points, the points left to the maximum of A, those points are also moving upward. Those points um A is already past its points. And so those points are going back towards the equilibrium position. But since A is below the equilibrium line, those points are now moving upward. So on the left hand side of both pulse A and pulse B, we have the points moving upward. And so even though their displacement is going to be zero, their displacements have canceled out, their motion is going to be upwards on the left hand side there. So all those points there on the left hand side are gonna have some sort of velocity moving upwards. Now, the magnitude of that velocity is going to vary from point to point, but you will have a velocity moving upwards. Similarly, if we look at the right side of those pulses on the right side of B, those points are going to be moving downward. And for the points on the right side of the maximum of pulse A, those are also going to be moving downward. So again, we have um a downward velocity for the points even though the displacement is zero. So we will have kinetic energy for those points as they're moving, they have a velocity. Therefore, they have a kinetic energy. So we've at least established that we have kinetic energy. Now, we need to look at the potential energy. And here, if we noticed that at all these points that are underneath these two wave um pulses, we have zero displacement. So when I add these two waves together to get the true wave, everything is completely straight. That's what we were given the problem. So that means that none of those points are displaced from equilibrium. And since they're not displaced from equilibrium, the string is not exerting a restoring force on any of those points. So it's very similar to a spring. If you don't stretch the spring, there's gonna be no potential energy um being stored in that spring. Likewise, here we have no elastic potential energy being stored in the string to the displacement of the string points from his equilibrium position. So that means we have no elastic potential energy. And so now if we look at our answers. We're gonna have the energy remaining the same as the um the wave will reemerge the two pulses, pulse B and pulse A will reemerge after the intersection. But at the point where they're intersected, we have just completely kinetic energy, there's no potential energy. So I hope that the explanation has been helpful and I will see you in the next video.

(I) The two pulses shown in Fig. 15–37 are moving toward each other. In Fig. 15–37, at the moment the pulses pass each other, the string is straight. What has happened to the energy at this moment?

Key Concepts

Wave Interference

Energy Conservation in Waves

Amplitude and Energy Relationship

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