Hey, guys. So now that we've seen the forces of mass spring systems, we want to look at what the energy looks like for mass spring systems. So at any point of simple harmonic motion, you've got this mass that's moving back and forth, and so it's exchanging energy. Right? The energy that's associated with its motion is the kinetic energy, and the other type of energy they might have has to do with the compression and stretching of the spring. So it's got some elastic potential energy. Now in all the cases that we've considered, we've considered no friction. So there's no work done by non-conservative forces. And what that allows us to do is use energy conservation. The mechanical energy is conserved. And so what we're going to do is compare the energies at 2 special points of mass spring systems. The 2 special points are going to be the amplitude and the equilibrium. So we've got minus a and plus a, and then we've got the equilibrium in which x equals 0. So let's take a look at what the energy looks like at all of these points. Well, we've got elastic potential energy here when x is equal to positive or negative amplitude. So that means the elastic potential energy is maximized here, 12kaa2. But the velocity at the specific point is equal to 0 at the endpoints, and so there's no kinetic energy. So that contribution actually goes away, so there's no kinetic energy. Which means that the only mechanical energy that we have is just due to the potential energy, the elastic energy at that endpoint. At the equilibrium here, this situation is reversed. Because now what happens is when x is equal to 0, the elastic energy goes away. And what does the kinetic energy look like? Well, at the equilibrium position, we've got this object either flying to the left or to the right and it's got its maximum velocity here. So that means that the kinetic energy, well the elastic energy is equal to 0 and the kinetic energy is maximized. It's got 12mv_max2. So that means that the total mechanical energy is only made up of the kinetic energy, k zero. Well, what about at any other points? At any other points, so for instance, if I just label this point right here and I call that p. So at x equals p, the elastic energy has to do with what the deformation looks like. So how much it's stretched or compressed. But it also has some non-zero velocity. It's also going either to the right or to the left with some velocity. So it's got some kinetic energy. The elastic energy is just going to be 12kx_p2, whereas the kinetic energy is going to be 12mv_p2, and those are both not 0. So that means that the total mechanical energy is actually the sum of all of those energies. So now, if we've compared all of these energies at these different points, and we remember that the total mechanical energy is conserved, it means that these things just get exchanged. They never get lost or destroyed. And so we can actually just plug in our expressions here, for our energies. 12ka2, and we've got 12mv_max2, and that's equal to 12kx_p2 plus 12mv_p2. So this is the energy conservation equation for springs. It's really powerful. And what questions are going to ask you is to relate all of these energies to each other. So they'll give you one and ask you for another, and so you need to know how to use this equation right here. So what we can do is we can use this equation and actually solve for the last kinematic quantity that we need. That's the velocity as a function of position. So to do that, I'm just going to cancel out some halves. I'm going to manipulate some numbers around, and then I get that the velocity as a function of position is equal to the square root of k over m times the square root of a squared minus x_p squared. So that's basically it. Let's go ahead and take a look at an example. So in here, in this example, we've got a 5 kilogram mass, we've got the k constant, and we've got the amplitude and we're supposed to find the maximum speed. So in this first part here, if I'm asked to find what v_max is, I'm just going to write out my conservation of energy equation. Now this is the conservation of energy equation. So I got all these things that are equal to each other. So I'm looking for specifically v_max. So I'm looking for this guy right here. So let's take inventory of all my variables. I know what the mass is. I'm given what the amplitude and the k constant is. So I can actually just take a look at these two relationships in order to figure out what v_max is. So v_max, if I go ahead and solve for that, I'm going to cancel out the half terms and I'm going to move the m to the other side and then take the square roots. So I'm going to get k over m, and then I've got the amplitude that's squared. So, yeah, the amplitude. So you might recognize this equation, v_max is equal to a times omega, and that's because these two things are the same. So when I got the square root of k over m, you might recognize that as that omega symbol. So if I've got all I need, I've got all of these variables I need, so I just take the square root of 30 over 5 which is k over m, multiply it by the amplitude, and what I get is a v_max of 0.98, and that's meters per second, not meter per second squared. So what does part b ask? Part b is asking us for a speed at a specific position. So for part b, we've got x equals negative 0.2 meters, and we're supposed to find the velocity. So we're going to use our velocity as a function of position equation. So we've got v when x is equal to minus 0.2 is going to be what? Well, we've got the square root of 30 over 5, that's k over m. And then we've got 0.4 squared minus 0.2 squared, and we've got this negative sign right here. So if you go ahead and take the square root of that, what you're going to find is that the velocity is equal to 0.85 meters per second. So that is the velocity at that specific position. So now this last part here asks us to figure out what the total mechanical energy of the system is, which means that we're going to use that conservation of energy equation. So that conservation of energy equation is this guy right here. So now, which one is the one that we can use? Well, the answer is you can actually use any of them. So notice here how I have the k, I have a, I also have m and v_max, so I can use all of those. So if I use this guy right here, so if I use the first terms, what I'm going to use is 12k30a0.42, what I get is a total mechanical energy of 2.40 joules. Right? So if I use those other two variables, so if I go down here, I've got 125v_max 0.982, so I get a total mechanical energy of 2.40 joules, which is the same exact answer. So the idea is that you can actually use any one of these to relate the energy at different positions. Alright, guys. So that's it for this one. Let's keep.
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Energy in Simple Harmonic Motion
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