>> When we have the horizontal picture and we stretch it out to a distance A. Here we were at X equals zero. We're going to stretch it out to A. We know it's going to go all the way to negative A, and then, go back and forth. When we let that thing go, it, of course, going to head that way. It's eventually going to come back. Where is it going its maximum speed? Well, the spring is pulling on it, right. The spring is trying to pull it back towards equilibrium. And the spring is pulling that whole way, until it gets to X equals zero. And once it gets past X equals zero, the spring is starting to compress. And so, the spring is pushing back the other way. So, clearly, Vmax has to be when X equals zero. Because soon as you pass it it's going to switch, pushing back the other way. Okay. How do we figure out what that Vmax is? It's not too bad. What's the energy in the system, here? It is kinetic plus potential. But there's no kinetic when it's out at maximum stretch. The box is at rest. And so, it's all potential energy. When it comes back to X equals zero, what do we have? We have some kinetic plus potential. But now, the potential is zero, because it's at X equals zero. And we just have 1/2m Vmax squared, plus potential energy at zero. All right. And now, we can set them equal. So, 1/2mVmax squared is equal to 1/2ka squared. And you can quickly solve this for V. I multiplied both sides by two. I divide by m. And I get Vmax is equal to k over m square root times A. But we already knew this. because this is just omega. And when we wrote down the equation for V, we said it was negative omega A times the sine of omega T plus V. and the biggest that sine can be is one. So, what's the maximum speed? Omega times A. Everything that's multiplying that sine. But let's say you're at some arbitrary point. Let's say you're not at the maximum. Okay. So, let's redraw this picture slightly at some other location. Let's say you're there at some position X, and you want to calculate the speed. Okay. What do we do? Well, conservation of energy tells us that energy has to go somewhere. This is 1/2kA squared. At this location, we have kinetic energy. One half mV squared. But we still have a little bit of stretch of that spring, and so, we have to include that in our equation. And now, if we set these equal, we can solve this thing for V. Multiply both sides by two. That gets rid of the half. Divide everything by m. And move this over to the other side, and you get V squared is equal to k over M, times A squared minus X squared. And now, you can take the square root. V is square root k over M, A squared minus X squared. And that is omega A squared minus X squared. And then, we have a little half up there for square root. And let's convince ourselves that this is the old solution in the case where X equals zero. Right. If I put X equals zero in there, that thing goes away, and I just get omega times A. And so, it does work out to be our Vmax in the case of X equals zero. It also tells us that if X equals A, if it goes back to full stretch, then this whole thing goes to zero, and we know that. The speed of the box is zero at either extreme. Okay. So, this is a good one. This is the general equation for V at any position X.