Wave Equation - Example 4
University of North Carolina at Chapel Hill
Wave Equation - Example 3


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Video Transcript

welcome to our third example video looking at waves on strings and the wave equation. In this video, we're going to go back to the wave equation where we have if you recall d squared displacement et squared equal to V squared D squared displacement respect two X. Okay, So here's our wave equation, and we're going to ask ourselves, how can we look at displacement then and figure out what wave speed is? So we have d of X is equal to a sign of K X minus omega T plus five. And we already did this once in the video. But in case you've forgotten, what we can do then is just take a second derivative here, then the second derivative here with respect to X, and we'll come up with me now, I'm going to do this in slow motion to illustrate. And in the next video, we're actually gonna look at alternative functions that satisfy the wave equation. So here we're going to take a first derivative all by itself. So the first derivative of this, with respect to time, we're gonna pull out a negative omega. So that's negative. Omega A and then we go from sign to co sign of K X minus omega T plus fi. Okay. And then we take a second derivative with respect to time. So we pull out another negative omega. But we go from kind of coastline to sign, which is a third negative value. So we have negative Omega squared, a sign of K X minus omega T plus. By now, this has to be equal to on the other side, some constant V squared multiplied by second derivative of D with respect tax. Now the first derivative of D with respect tax, we're gonna pull out a case. So we're gonna have okay, a cosine of K X minus omega T plus fi taking a second derivative with respect to X. We obtain K squared a sign. Okay, X minus Omega T plus five. But again, we went from co signed a sign, which means we get a negative here. So the negatives cancel. I want to solve for v squared. I c h cancel. I see the sine function cancels and I end up with omega squared over K squared is equal to b squared. Excellent. So in other words, omega over K equals V which we already knew as the wave speed for this particular wave. So again, every displacement is not necessarily going to look like this function. It could look like a different function. So in the next video, we're going to take a look at making sure that any function satisfies the wave equation.