Alright, my name is Brad Langdale and it's the dead of winter, but I want to talk to you about spring. And not the kind of spring that we're all hoping is going to come soon. We're going to talk about springs that have masses attached to them, which are probably less fun and less interesting to look at, but that's what we're going to look at anyways. Under the context of simple harmonic motion, which is motion, which is a repetitive back and forth movement, fueled by a restoring force. And we're going to look at three things here, actually four, I guess, of the system. We're going to calculate acceleration. We're going to calculate force. We're going to calculate velocity. And why not? When we're at it, let's figure out what period is two. So here's an example of a question from this unit of study in the physics 20 curriculum. A mass of three kilograms is attached to a spring. Spring constants 1.4 Newton per meter. The spring is stretched to 0.5 meters past equilibrium and undergo simple harmonic motion is going to vibrate back and forth back and forth. Back and forth kind of like this. I got my little ghost block there to show that. Determine the maximum acceleration, maximum force, maximum velocity, and the period of the mass spring system. All right, so why don't we start off here by just talking about when we're going to get these different things. To find maximum acceleration, that is going to occur when that mass is not at the equilibrium point where it would normally be if you just left this thing alone for a day and came back to check it the next day. This is going to be when you pull this back to the maximum displacement, to the amplitude. That's when you get maximum acceleration. It would also be once it swings back to this side, the acceleration is at a maximum also here where the amplitude is at a maximum on the other side. So I'm going to start from here. Let's think about the forces taking place at that point. Well, the only force really acting at this moment is the force of the spring. And that force of the spring is going to be what we're going to use to find acceleration. Any acceleration or any force I should say can be equal to mass times acceleration according to Newton's laws of motion. So the force of the spring, which is negative kx, is just equal to ma. And we can go through and solve it from there. That's all you really have to do. From there, it's just thrown in the numbers in the right spot and being careful with things like significant digits. So we've got our k, we've got our x, 0.5 meters. We've got our mass, three kilograms, and we're going to solve for a. So we'll grab the handy dandy calculator and see what we're going to get here. Now, one thing that's interesting to note is we're going to get a negative acceleration. And hopefully that makes sense according to the diagram because we pulled this thing back to the right. You pull something back to the right, the acceleration is in the opposite direction. It's going to be to the left. So the terms out, the acceleration we get here is negative 0.23 repeating. 0.23 is a negative meters per second squared. So there's my acceleration. It's not bad to do actually pretty simple. Now let's find the force. Well, you know what, we've got a formula here for force. We can just use that. Force in the spring is the same as the force that the object will experience, the mass will experience. And again, this is at this point when you pull the mass back, the farthest it'll go. Negative kx. Alright, so negative 1.4 Newton's per meter times the by 0.5 meters. I think I can do that on my head, negative 0.70 Newton's. Alright, so there's our force and our acceleration done. Okay, now we're going to find the maximum velocity in the period. So first of all, our maximum velocity. It's not going to occur at the point where we had our maximum acceleration and our maximum displacement. Put our line back. Our maximum velocity is going to occur here at the equilibrium point. At that moment, when all of the energy stored in the spring has been converted to kinetic energy of the block. So this is where our maximum velocity is going to be. Now there's two ways you can find this. The easiest way you can probably look at doing it is using the conservation of energy. So we're going to go through and do that. We're going to say that the energy stored in the spring, the potential spring energy or elastic potential energy, will be equal at that point to the kinetic energy. So the 1.5 kx squared equals 1.5 mv squared. At that point, you can cancel the halves. And that's all you need to do. You just need to plug and chug from there. Again, not too bad with the theory. And then the formulas don't appear in your form, the sheet, made equal to each other. But that's okay. I mean, hopefully you understand where the equality comes from. It comes from the fact that all that energy that was stored in the spring is now been transferred to the mass. And of course, don't forget to square root in this equation. Because people love to forget to square root things. It's just one of those things that are kind of tricky to do until you practice a lot. And even after you practice, you'll probably forget sometimes. So I don't want to go through. Remember to square there. That's good. Divide it by 3. And then I'm going to square root in the end to find what that velocity is. So the velocity is going to be 0.34 m per second. And the last thing you might want to do because you did square root that velocity. It could be a positive or a negative number. Well, it depends on which way you're pulling it. My diagram, because I pulled it back to the right, the first velocity would be going to the left. So maybe I'll stick that negative sign in there. Now for period. I think period's the most interesting one to do. There's one way of finding period that's really straightforward. The formula sheet has a formula for period of a mass spring system. Okay, that's pretty simple. Throw the numbers in the right spot. There's your mass. There's your spring constant. Plug it through. Now that's easy enough. There's another way for finding period two. Because a mass spring system is so darn similar, mathematically, anyways, to a circle in uniform circular motion, we could also do it with uniform circular motion. Did you know that? Look at this. We already have a lot of this info. We know what the speed is. 0.34 m per second. Now this, of course, doesn't make a circle. It goes back and forth, back and forth, back and forth. So it doesn't have a radius. But if we let the radius equal the displacement, the amplitude, if we put in place of that 0.5 m and solve for period, we're going to get the same thing we would get out of the period formula. Which is kind of neat that the two are related, even though they probably don't seem all that related. Alright, so let's just see what we get here. 2 pi times 0.5 divided by 0.34. I think that looks pretty good. It's going to give us a period of 9.2 seconds. And just to prove it to you, we'll run that other calculation as well. If I went 2 pi times the square root of, we're going to go m, which is 3 kilograms, divided by k, which is 1.4. Let's see what we get. Now, isn't that cool? 9.2 seconds to get the same thing. So two ways to find period there at the end. Either one is fine. One just using the formula for the formula sheet. One using the principles of uniform circular motion. So I hope that helps. For more information about mass spring systems, check out my website, www.ldindustries.ca.