 I need to show you, I'm going to show you how to do the numerical calculation for something cool like the spring. I told you you could figure it out, but it's a little complicated. I'm going to show you exactly what to do. So these are the important ideas we have here. This is my, I'm going to call this my update position formula. So this says if I know where the mass is, and I know the velocity and the change in time, I can find the new position. This is just from the definition of average velocity. And you did this in the, in throwing the ball up. This is exactly the same. And again, with this, this is my update velocity formula. This is my new velocity, my old velocity plus the acceleration times change in time. Oops, sorry. And in that ball case, this was a negative 9.8 meters per second for the acceleration. Now with the spring, though, there's something different. As I move the force on the, on the mass changes because the spring stretches more. So since the force is this negative K times Y, then if I divide that by the mass, I get the acceleration. So as it moves, as Y changes, the acceleration is going to change too. This is just a equals F over M. That's, that's all this says. And so if I know the position, now here, I'm saying a to the second acceleration is the position before it. And it doesn't really matter too much which position you use, as long as this delta T is small. Okay, so here's, I started to set this up already. Here's my spreadsheet. I put the mass and this, this doesn't do anything. These are just units just so I know a mass of 100 grams K 15 Newton's per meter. And DT is 0.01 seconds. So what's the first time? It's just zero. What's the next time? It's going to be equal to the time before it plus this DT. Now, and up here, I'm going to add in a dollar sign. So that says that as I copy this formula down, don't change that. Okay, so I can just test it out here right there. Okay, that's it. That looks good. Okay, now what about the position? Now, I'm assuming that Y equals zero is at equilibrium, and I'm not going to put gravity in there. That's okay. It is. Just trust me. Okay, so let me just say it starts at, let's say I pull it down 10 centimeters and let go. So the initial Y position is going to be negative 0.1. And I'm going to say I release it from rest. So this is zero. Now, what's the acceleration? Well, I need the force. Okay, and so that depends on the position. I'm hearing use a one equals Y one. I did that wrong before. So I'm going to say this is going to be equal to negative K. So that's I need to put a dollar sign in there because I don't want that to change. Negative K times Y times this. I don't need a dollar sign there because I want that to change divided by the math. And the math was up here. I need a dollar sign. Okay, so that's my acceleration is 15 meters per second squared. Does that seem okay? Maybe this is a high, that's fine. Okay, let's just change the spring constant to five. Make it a little bit smaller. Okay, so now my next row, what do I do for Y? Well, the new Y is going to be equal to the old Y plus the velocity, which I haven't calculated yet. So I use this velocity up here times DT, which is up here and I need a dollar sign. So I want to always use that time. It's the same position. It hasn't moved yet. That's okay. Now for the velocity, I'm going to say it's equal to the old velocity plus the acceleration times DT. And again, dollar sign. Now it's moving up, right? And that makes sense. If I pull it down and let it go, it's going to speed up going up. Now the acceleration, everything's fine. I can just copy that formula down. So, but it has still five. Oh, that's right because it's the same position. So that doesn't look right. But look, as I drag this down, then things start to change. So let's look, the position starts getting smaller, closer, moves up, and the velocity gets faster and faster up too. Okay. So let's just copy all these down some more. Oops, you can't see that. There you go. And so let me go down even further. And then see up here, it gets back to the equilibrium position. It gets back to Y equals zero in between here and here. And then it starts to move up. And when it does that, the force is now pushing down. So I'll say back and forth. Now, I'm not going to graph this. I'll let you do it because this is done. But I do want to point out that here I have a DT of 0.01. That still may not be small enough. You may want to make that 0.001 or something like that. But if you have a smooth graph of position versus time, then that's good. When you do a position versus time graph, you can look at how long it takes to get back to where it started and that's the period. And you can compare that to your experimental result. Okay. So I know that was quick, but you can rewatch it if you want. So it should be fine.