The two wave shown in the diagram have the same frequency wavelength and initial displacement. If the two waves interfere, which of the other diagrams A, B, C, and D best shows how the resultant wave compares to the two identical waves?
Okay, so in this question, we’ve been told that we’ve got two waves — this orange one and this blue one here — that have the same frequency wavelength and initial displacement. Now, the first two conditions are important: the fact that they have the same frequency and wavelength because this is what allows the two waves to interfere as well as this the fact that they have the same initial displacement means that essentially these two waves are in phase.
In other words, at any point along the wave, let’s say this point here, the two waves are at the same point in their cycle. And we can see this at these points because both of these waves at this point have a zero displacement. And as we move towards the right, the displacement increases in both cases.
Now, we’ve been told that the two waves interfere. And we’ve been asked to find which of the diagrams A, B, C, or D best shows the resultant wave, whether it’s this one or this one or this one or this one. Now, one of the most convenient ways to do this is to think of the two waves in the following way.
Let’s think of labelling the points on the two waves by using axes, as we do with the graphs, for example. Now, let’s start with the orange wave. We can arbitrarily assign values to each point on the wave. So let’s say that this point right at the beginning is point 𝐴, where the wave peaks is point 𝐵, this point here is 𝐶, where the wave peaks in the other direction is 𝐷, and this final point here is 𝐸.
Now, that’s the orange wave. But of course, if the orange wave is to interfere with the blue wave, then they both have to be in the same position in space. And hence, what we actually have is the blue and orange wave both being in the same position in space. The only reason we’ve drawn them separately here is because it makes them easier to see that way.
So essentially, what we have here is a superposition of the blue and orange wave. Now, at each point along the wave, we can assign a value on the vertical axis as well. So once again, very arbitrarily, let’s just say that the peak of the wave has a value of one on the vertical axis and the trough has a value of negative one.
Now, again, we could have chosen any value that we wanted just as long as we were consistent with this positive value and this negative value having the same magnitude because the way it goes up in this direction exactly the same amount as it goes down in this direction. But so as long as we were consistent here and here, we could have picked any value. We could have chosen one and negative one or 100 and negative 100 or 2042 and negative 2042; it doesn’t really matter.
But the reason that we’ve done this is the following. When two waves interfere, essentially we can add together their amplitudes. In other words, let’s say we start at point 𝐴. Well, at point 𝐴, the amplitude of the orange wave is zero and the amplitude of the blue wave is also zero. So the amplitude of the resultant wave which we’ll label in black is going to be zero plus zero; in other words, it’s zero.
Moving on to point 𝐵 then, at point 𝐵, the amplitude of the orange wave is one and the amplitude of the blue wave is also one because remember the orange and the blue waves in this case are identical and so the amplitude of the resultant wave is going to be one plus one and that is two. Moving on to point 𝐶, the amplitude of the orange wave is zero and the amplitude of the blue wave is also zero. So adding those together, the amplitude of the resultant wave is zero once again.
At point 𝐷, the amplitude of the orange wave is negative one and the amplitude of the blue wave is also negative one. So the amplitude of the resultant wave is negative one plus negative one. That’s negative two.
And finally, at point 𝐸, the amplitude of the orange wave is zero and the amplitude of the blue wave is zero. So the amplitude of the resultant wave is zero plus zero once again which is zero. Now, in this case, we’ve only chosen to do this for points 𝐴, 𝐵, 𝐶, 𝐷, and 𝐸 because they should give us a very good idea of what the resultant wave is going to look like. But ideally, we’d do this for every point along the waves. And if we did, we’d see that this is what the resultant wave looks like or something like that anyway if it wasn’t so badly drawn.
But the point is we add the amplitudes of the two waves that are interfering at every point along the horizontal axis. And this gives us the resultant wave which has an amplitude now of two. And of course, the trough is at negative two. In other words, it has twice the amplitude of the orange and blue waves.
So out of these four options, which one gives us the correct resultant wave? Well, starting with A, this resultant wave does look correct because we can see that the two original waves both have half the amplitude of the supposed resultant wave. We can see this by drawing arrows from the zero point all the way up to the amplitude in both cases.
The arrow showing the amplitude of the orange and blue waves are about half the size as the arrow showing the amplitude of the black wave. So option A looks correct. But let’s go for the other ones just to make sure.
Well, option B obviously is not even close to being correct; it’s a flat line. Option C shows the resultant wave as having the same amplitude as the orange and blue waves. We can see that all of the pink arrows are of the same size. And option D actually shows the amplitude of the resultant wave to be less than the amplitudes of the orange and blue waves. We can see that this pink arrow is smaller than these two pink arrows.
And so as we thought initially when we first looked at option A, that is our correct answer. In other words, diagram A best shows how the resultant wave compares to the two identical waves.