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Anderson Video - Cosine Solution to Simple Harmonic Motion

Professor Anderson
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>> Hello class. Professor Anderson here. Let's look at the general solution for simple harmonic motion. When we had a block on a spring and it was going back and forth, what we solved for was the position as a function of time. And it looks like this. It goes like a cosine, okay? Let's understand what these different things mean. A is the amplitude of the oscillation. That's how far you stretch out that spring. Okay? Omega is our angular frequency. That's going to have units of radians per second. But remember, radians is really unitless, okay? Angular frequency. It's just how fast that thing's going back and forth. T is, of course, time. And phi is what we call a phase constant. And that's really just where do you start? Did you start at maximum amplitude? Did you start at x = 0? Did you start somewhere in between? Okay? It has to do with where that oscillation started. All right. We know what this looks like. Right? This looks like a cosine. And if phi equals 0, it's exactly a cosine. It looks like this. This is x as a function of time. So that guess that we had about the solution, that it should be sort of curvy and smooth is exactly right. It looks like a cosine in this case. So, any time you're dealing with differential equations, one thing that you need to do is put the equation -- or put the solution back into the differential equation to solve for some of these constants. And let's do that where we're going to try to solve for this omega. To do that we have to figure out what the derivatives look like. So here's our function x. But we're going to need the first derivative, which is of course v. What's the derivative of cosine? Well, we said it was minus sine. And then there's a complicated argument there. And so you have to take the derivative of the inside as well. And so you pick up an extra omega. And then we're going to want the acceleration, which is the second derivative of that. Which is the first derivative of this. And so we get minus A -- I'm going to pull out another omega. And then sine goes back to cosine. And now I have all the terms that I need to put into our force equation. So let's try it. Okay, our differential equation looked like the following: d squared x dt squared was negative k over m times x. And now we know those terms. Let's put them back in. So d squared x dt squared is negative A omega squared cosine of omega T plus phi. And that's going to equal to negative k over m times x, which is this guy right here, A cosine omega T plus phi. And now a bunch of stuff cancels out. The A cosine omega T, we can cross that out. A capital A, we can cross that out. And we get omega squared equals k over m. The minus signs went away. And we end up with omega as the square root of k over m. And this is what we call our angular frequency. Okay, omega is a square root of k over m. This is how fast that thing goes back and forth. Okay? If I have a stiffer spring, k gets bigger, omega gets bigger. And that should make sense to you, right? If I have a stiff spring, that box is going to back and forth really fast. If I have, instead, a large mass attached to the spring, that spring can't pull on it and move it quite as easily. And so m, big in the denominator, means that omega is small. And so that would be a big sluggish mass going back and forth. Okay, this region from there to there has a very characteristic name. It is called the period, t. And that's where the function is going to start repeating itself, all right? We went down through 0. We came back up. And then we went down through 0 again. And so that amount of time is the period. And the period is, of course, related to things like the frequency. What is the frequency? Well, the frequency is omega over 2 pi. And the period, t, is 1 over the frequency. And so in our case with the block on the spring, we can plug all this in. We've got 1 over omega over 2 pi. And that becomes 2 pi over omega. But we know exactly what omega is, so this is 2 pi divided by the square root k over m. Or 2 pi square root m over k. So when you have a block on a spring, what's the period of that oscillation? It's that. It's 2 pi the square root of m over k. if I have a bigger block, it takes longer for it to go back and forth. If I have a stiffer spring, it takes shorter to go back and forth. Okay? So that's the period. This function that we drew right here looks a lot like a cosine. And so we will call it exactly a cosine. Which means that we let that phase constant equal 0. Okay. If you start up here at a maximum of A and you go down to a minimum of minus A, then that phase constant of 0, it looks exactly like a cosine.