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Anderson Video - Simple Harmonic Motion

Professor Anderson
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>> Hello class. Professor Anderson here, welcome to another edition of the Learning Glass Lectures on physics. We're going to talk about something very key in physics, and in fact all of science, which is simple harmonic motion. Turns out a lot of the universe that you interact with behaves according to simple harmonic motion. And so, the models that we're going to develop here, the mathematical models apply to a great number of problems. Okay, so let's talk about the basic problem. It looks like this. Let's take a spring and let's tie it to the wall. And let's stretch it out from its equilibrium position. So, it was sitting there at X equals 0, we're going to stretch it out to some distance, X, and then we're going to let it go. Okay, when I let this block go, we're ignoring any friction here. What does it do? We know what it's going to do. It's going to start moving that way. And at some later time, it's going to be cruising to the left and the spring will be at equilibrium, and then at some later time, it will compress the spring and it will come back to rest. And then it will do that again, and again, and again. All right and it will just go back and forth between those two points, if we stretch it out to X. This is of course, negative X. This is the basic model for simple harmonic motion. And this is a great point to start to tackle the math. Because we understand a little bit about springs. We understand a little bit about forces. And hopefully now we understand about acceleration. So, let's ask the following question. What does X as a function of time, look like? What is the position of this thing as a function of time? Well we know we start out here at some amplitude. We know it's going to move to the left. It's going to go through X equals 0. It's got to go to some negative amplitude. And then it's going to go back and forth. And so, you might think, oh, this is going to look like this. It started up here at some amplitude. It moved to the left, which means X was going down. It went through X equals 0, and then it went to some negative amplitude. And then it goes the other way and it just does that over and over again. That is a possible solution to this problem. But it looks a little funny. Does anybody not like the graph that I just drew? Carlos, what do you think? Why don't you grab that mic? >> (student speaking) I don't know what's wrong with the graph. >> Okay, but you're immediately regretting sitting the front row, aren't you? Yeah. Okay. So, this looks like a possible solution, right. It started up here at some amplitude, some positive X. It went through zero and then it went to some negative amplitude, and then it went back and forth. But if I think about this, right here. That's a very quick turnaround, all right. I was going to the end and then all of the sudden I started going the other way. And that's sort of a fast, jerky motion. Does that make sense with something like this? >> (student speaking) No, it's probably going to go, it's going to go slow and then stop and then going to start increasing the speed level. >> Yeah, it seems like we should smooth it out, right? A spring acts sort of smoothly, right? So, this doesn't look like a very good solution here. Let's smooth out the ends. And if we do that, maybe our graph looks a little more like this. That looks a little smoother, right? We've smoothed out the ends. We've smoothed out the whole curve. Everything is sort of behaving nicely. This is in fact, the correct curve. And this is a sinusoidal curve. Sinusoidal refers to anything that goes like a sine or a cosine. And this curve right here in fact looks like a cosine. Okay, so this is our guess. We're going to say it's got to look something like this. Let's see if we can understand how to get to that mathematically. Okay, let's see if we can derive simple harmonic motion from Hooke's Law. Okay and we said that we're going to take the problem where we have a mass on a spring. Let's tie it to the wall. And we know what that thing's going to do, as I stretch it out and release it, it's going to go back and forth. Let's understand this now using strict mathematics that we understand. We know that Hooke's Law applies to a spring, right. When you stretch out a spring, there is a restoring force on that object, and that restoring force is negative KX. K is a spring constant for the spring. X is the position of the object. Okay. All right. But we also know that Newton's second told us the following. If I sum the forces in the X direction. I only have on in this case, it's negative KX. They have to equal the mass times the acceleration. Okay, if you have more than one force you would put it in here, but in this case, we're ignoring friction, ignoring air resistance. All we have is spring. All right. We can solve this equation now for acceleration. Acceleration is negative K over M times X. And this is the basis for simple harmonic motion. Simple harmonic motion is when the acceleration A, is proportional to the negative of the displacement. The negative just means restoring, okay. It's trying to restore itself to equilibrium. If that was a positive, it's like the spring would keep pushing out forever. If you stretched the spring it would push harder in that direction, and that box would of course never return to X equals 0. So, the negative sign just means restoring force. It's got to pull it back towards X equals 0. And this is the basis for simple harmonic motion. Any time you have a system where the acceleration is proportional to the negative of the displacement, you get simple harmonic motion. And this is the power of this approach, because it's not just blocks on springs, it's electrons in their orbits. It's molecules when you stretch them away from each other, they tend to rebound. And when they do that, they exhibit simple harmonic motion. All right, let's take a look at this math right here. We have A sub X, but we know what A sub X is. A sub X is d squared X, dt squared. And that is negative K over M times X. And that looks a little bit formidable. Anybody know what this sort of equation is called? Other than Randy? Anybody know what this sort of equation right here is called? Faye what do you think? Can you hand the microphone to Faye, let's have a discussion with Faye. Faye, what sort of an equation are we dealing with here? >> (student speaking) I'm not sure. >> Okay. What do you see on the left side here, what is this thing? >> (student speaking) Acceleration? >> It's a derivative, right? It's a second derivative. And so, any time you have a second derivative and it's equal to something else, the general category is differential equation. And the differential just means there's a derivative in there somewhere. Okay. This happens to be a second order differential equation. If you haven't learned about these things yet, don't worry about it. Okay. It's not that critical that you fully understand how to solve differential equations yet. But it is kind of important that you recognize that this is a differential equation. Now, there's a lot of ways to solve this thing, but let's look at the basic premise. The basic premise is, if I start with some X and I take two derivatives of it, I have to get back to that original function, X. And there's this constant out in front. Okay, so you might think, oh maybe an exponential would work. E to the X, if I take a derivative of E to the X, I get E to the X. If I take a derivative of that, I get another E to the X. But the problem is that minus sign right out in front. That minus sign turns out to be really key, because we need the function to flip its sign. And so, one possible solution is the following. X equals A cosine of omega T plus phi. Let's just think about that for a second. If I take the derivative of a cosine, I get minus sine. And if I take the derivative of a minus sine, I get minus cosine. So, that's my original function, but now I have that minus sign built into it. That looks like it's a good solution to this problem. And that's the one that we're going to use in attacking it.