University of North Carolina at Chapel Hill
Wave Equation - Example 4


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Video Transcript

welcome to our fourth example video considering waves on strings and the wave equation In this video, we're looking at other options that we could use for our displacement. So say we have, ah, displacement function that looks like some constant a times X squared minus some constant Be times t squared, plus some constancy. Let's see if this satisfies the wave equation. Remember, the wave equation looks like this. And in fact, we could do more than just see if it satisfies the wave equation. We could say, Hey, here's a function. See if it satisfies the wave equation and tell me the way of speed. All right, well, first things first. If I take the second derivative of this with respect to time, what I'm going to get is this isn't have time and it's so goes away. This doesn't have time again. Go. That goes away. So we have a negative to be t for the first derivative with respect to time and then for the second derivative with respect to time, we get negative to be now Similarly, when we take the derivative with respect to X, we're gonna end up with is to a clearly these are constants. And so you have to ask yourself, Are these constants the same? Well, unless a equals b uh, then no. And also a cannot equal be because notice the units up here. If displacement is in meters, see, you would have to be in meters B would have to be in meters per second squared and they would have to be in, uh, one over meters. So clearly not thes two cannot be the same. So this function does not satisfy the wave equation. What if on the other hand, we can pick the simplest one here we have dfx t is equal to eight times cosine of K X minus omega t plus Fine. Now this one should be clear to you that it is going to satisfy the wave equation because we're taking to derivatives and it's gonna have the exact same result as we had when we when we did sign. So the second derivative of this, with respect to time, is going to be negative. Omega squared a cosign of K x minus omega T plus five and then the second derivative with respect to X is gonna be equal to K squared a sorry negative k squared a co sign, have a K X minus omega T plus fi. So it is going to satisfy it because one we see that we have a cosign k X minus omega T on both sides, which will cancel. And then we have Constant's left over which we can use to satisfy, or to find what wave speed is eso. You could do this with all sorts of equations. It's a kind of good practice to go and try it. Uh, it's simple problem that can show up easily on an exam or homework set.

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