Traveling Wave Basics - Example 4
Wave Equation - Overview
Wave Equation - Example 1
Wave Equation - Example 2
Wave Equation - Example 3
University of North Carolina at Chapel Hill
Traveling Wave Basics - Example 3


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Video Transcript

welcome to our third example video. Considering the basics of traveling waves in this video, we're going to look at plots of traveling waves, an attempt to determine properties based on those plots. So, as before, we're going to need to different plots one versus T in one vs X. So this will be at T equals zero seconds, and this will be at I'm just gonna say, 0 m so we could have put it anywhere, and we could have put it at any time that we like. I'm going to say that we'll have Max and men of three centimeters and negative three centimeters. We're going to start it about 70% of the maximum height. And then as we also late, we follow a pattern like this. Now, I need to put in a couple more numbers here. We'll say that right here. We're at two centimeters. Here were four centimeters, and here we're at five centimeters. Oh, sorry. Six centimeters. And here we're at eight centimeters. And meanwhile, down here, we're going to say we're at one half second and then we're at one second and then we're at 1.5 seconds and then two seconds and then 2.5 seconds. Okay, So, looking at this, we see that we have a new amplitude equal to three centimeters. We see that we have a lambda equal to two centimeters a period equal to one half second, and we're going to have if we're starting at 70% of the height, that's approximately 45 degrees. Sign of 45 is about 450.7. So fi is going to be equal to pi over to. All right. Pardon me? Not not pi over two, but pi over four. There we go. That's better. All right. So we've got our different values here and now. Suppose I want to write the function D of X T. Given all these numbers, Well, I know that I need an amplitude first, so my amplitude is going to be three centimeters, so I have three centimeters, and then I'm gonna have a sign of I know that I have a lambda equal to two centimeters, so that's going to be equal to two pi over two centimeters, X minus. Assuming it's going to the right, it would probably have to say that in a problem statement we have a frequent or rather, a period of 0.5 seconds. That's two pi over 20.5 seconds times time plus pi over four. So here we have been able to write out the function based on the plots.