>> I want you guys to come up with questions, such as questions that you asked right here, often. And just the act of sitting there and thinking about a question gets you thinking about physics, gets your creative juices flowing in terms of what do I understand? What do I not understand? What was unclear in his lecture? What am I really struggling with? And it allows me to sort of peer into your mind. Okay. So we'll take a look at some of these questions as we go here just to sort of break up the lecture a bit. All right. We have a good idea what the model is that we're starting from. It's this. Tie a spring to the wall, put a block on the spring, stretch it out a bit, and let it go. And if it's a frictionless surface that it's on, it will just slide back and forth forever. So what can we say about the energy in this system, the energy in this block right here? Well, we know that energy is, of course, kinetic plus potential. And the potential energy in this case is the potential energy in the spring. Okay. But we know little bit about that stuff. Right. We know that kinetic energy is 1/2 mv squared. And we know that spring potential energy is 1/2 kx squared where x is how far you stretch the spring from its equilibrium position. Since it's squared, you get the same answer if you compress the spring by the same amount. It doesn't matter that it's a negative. Once you square it, it becomes positive. Okay. That's good. But we know little bit more about those for simple harmonic motion. Right. We wrote down the differential equation for simple harmonic motion. And what we discovered was x has a particular solution. We can write it like this: A cosine omega t plus phi. And if that is x, then we can get v very easily by taking the derivative of x. And when we do that, cosine becomes negative sine. I have to pull out an omega from inside the argument and so v is this negative omega A sine of omega t plus phi . And now I can put all that back into this equation. So what does energy become? 1/2 m times v squared. Here's my v right here. Negative omega A sine of omega t plus phi. We're going to square that. And then we're going to add 1/2k times x squared but there's my x. A cosine omega t plus phi quantity squared. And now we can write out this stuff. Right. We've got 1/2m omega squared A squared times sine squared of omega t plus phi. And then the next one is 1/2kA squared cosine squared omega t plus phi. And it looks like maybe we can't combine those unless we remember that omega is, in fact, related to k. That was one of those things that we derived before. Omega is, in fact, the square root k over m. And if omega is that then omega squared just becomes k over m. And if omega squared is k over m, that m cancels out and I get 1/2kA squared on both of those terms. That term becomes 1/2kA squared and that term was already 1/2kA squared. And now everything that's left is just sine squared omega t plus phi plus cosine squared omega t plus phi. And now we go back to our joke. Right. Anybody remember the joke? What's the geek pickup line? All right. Geek approaches the girl at the bar and he says I wish I was sine squared and you were cosine squared so together we could be one. You are allowed to boo in here if you want. That's fine. That is equal to 1. And so look. The total energy, what is it simplified to? 1/2kA squared. That's it. That's the total energy in the system. However far you stretch this thing out from, x equals 0 to x equals A and you let it go, that determines how much energy is in the system. At full stretch, it's all that. Okay. And that is, of course, a constant. k is a constant. A is a constant. And so it doesn't change as a function of time. The total energy of the system is fixed as a function of time.