Harmonic motion part 2 (calculus)

So where I left off in the last video, I'd just rewritten the spring equation. And I just wrote force is mass times acceleration. And I was in the process of saying, well if x is a function of t, what's acceleration? Well, velocity is this derivative of x with respect to time, right? Your change in position over change of time. And acceleration is the derivative of velocity, or the second derivative of position. So you take the derivative twice of x of t, right? So let's rewrite this equation in those terms. Let me erase all this--I actually want to keep all of this, just so we remember what we're talking about this whole time. Let me see if I can erase it cleanly. That's pretty good. Let me erase all of this. All of this. I'll even erase this. That's pretty good, all right. Now back to work. So, we know that-- or hopefully we know-- that acceleration is the second derivative of x as a function of t. So we can rewrite this as mass times the second derivative of x. So I'll write that as-- well, I think the easiest notation would just be x prime prime. That's just the second derivative of x as a function of t. I'll write the function notation, just so you remember this is a function of time. Is equal to minus k times x of t. And what you see here, what I've just written, this is actually a differential equation. And so what is a differential equation? Well, it's an equation where, in one expression, or in one equation, on both sides of this, you not only have a function, but you have derivatives of that function. And the solution to a differential equation isn't just a number, right? A solution to equations that we've done in the past are numbers, essentially, or a set of numbers, or maybe a line. But the solution to differential equations is actually going to be a function, or a class of functions, or a set of functions. So it'll take a little time to get your head around it, but this is as good an example as ever to be exposed to it. And we're not going to solve this differential equation analytically. We're going to use our intuition behind what we did earlier in the previous video. We're going to use that to guess at what a solution to this differential equation is. And then, if it works out, then we'll have a little bit more intuition. And then we'll actually know what the position is, at any given time, of this mass attached to the spring. So this is exciting. This is a differential equation. When we drew the position-- our intuition for the position over time-- our intuition tells us that it's a cosine function, with amplitude A. So we said it's A cosine omega t, where this is the angular velocity of-- well, I don't want to go into that just yet, we'll get a little bit more intuition in a second. And now, what we can do is, let's test this expression-- this function-- to see if it satisfies this equation. Right? If we say that x of t is equal to A cosine of wt, what is the derivative of this? x prime of t. And you could review the derivative videos to remember this. Well, it's the derivative of the inside, so it'll be that omega, times the outside scalar. A omega. And then the derivative-- I'm just doing the chain rule-- the derivative of cosine of t is minus sine of whatever's in the inside. I'll put the minus outside. So it's minus sine of wt. And then, if we want the second derivative-- so that's x prime prime of t. Let me do this in a different color, just so it doesn't get monotonous. That's the derivative of this, right? So what's the derivative of-- these are just scalar values, right? These are just constants. So the derivative of the inside is an omega. I multiply the omega times the scalar constant. I get minus A omega squared. And then the derivative of sine is just cosine. But the minus is still there, because I had the minus to begin with. Minus cosine of omega t. Now let's see if this is true. So if this is true, I should be able to say that m times the second derivative of x of t, which is in this case is this, times minus Aw squared cosine wt. That should be equal to minus k times my original function-- times x of t. And x of t is a cosine wt. I'm running out of space. Hopefully you understand what I'm saying. I just substituted x prime prime, the second derivative, into this, and I just substituted x of t, which I guess that's that, in here. And now I got this. And let me see if I can rewrite. Maybe I can get rid of the spring up here. I'm trying to look for space. I don't want to get rid of this, because I think this gives us some intuition of what we're doing. One of those days that I wish I had a larger blackboard. Erase the spring. Hopefully you can remember that image in your mind. And actually, I can erase that. I can erase that. I can erase all of this, just so I have some space, without getting rid of that nice curve I took the time to draw in the last video. Almost there. OK. Back to work. Make sure my pen feels right, OK. So all I did is I took-- we said that by the spring constant, if you rewrite force as mass times acceleration, you get this. Which is essentially a differential equation, I just rewrote acceleration as the second derivative. Then I took a guess, that this is x of t, just based on our intuition of the drawing. I took a guess. And then I took the second derivative of it. Right? This is the first derivative, this is the second derivative. And then I substituted the second derivative here, and I substituted the function here. And this is what I got. And so let me see if I can simplify that a little bit. So if I rewrite there, I get minus mAw squared cosine of wt is equal to minus kA cosine of wt. Well it looks good so far. Let's see, we can get rid of the minus signs on both sides. Get rid of the A's on both sides. Right? We can divide both sides by A. Let me do this in black, just so it really erases it. So if we get rid of A on both sides, we're left with that. And then-- so let's see, we have mw squared cosine of omega t is equal to k cosine of omega t. So this equation holds true if what is true? This equation holds true if mw squared-- or omega squared, I think that's omega. I always forget my-- is equal to k. Or another way of saying it, if omega squared is equal to k over m. Or, omega is equal to the square root of k over m. So there we have it. We have figured out what x of t has to be. We said that this differential equation is true, if this is x of t, and omega is equal to this. So now we've figured out the actual function that describes the position of that spring as a function of time. x of t is going to be equal to-- we were right about the A, and that's just intuition, right, because the amplitude of this cosine function is A-- A cosine-- and instead of writing w, we can now write the square root of k over m. The square root of k over m t. That to me is amazing. We have now, using not too sophisticated calculus, solved a differential equation. And now can-- if you tell me at 5.8 seconds, where is x, I can tell you. And I just realized that I am now running out of time, so I will see you in the next video.