WEBVTT chemistry/physical-chemistry/hovasapian
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Hello and welcome back to www.educator.com and welcome back to Physical Chemistry.
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Today, we are going to start our discussion of the harmonic oscillator.
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We are going to spent several lessons on it so let us go ahead and get started.
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The harmonic oscillators are pretty much exactly what you remember from classical physics.
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Basically, it is some system that is moving back and forth.
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It is oscillating just like this, back and forth, back and forth.
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We wanted to discuss the quantum mechanical equivalent of that.
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Let us start with, when radiation interacts with matter,
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it causes changes in the energy levels of the particles making up that matter.
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The whole field spectroscopy is a study of that, the interaction of electromagnetic radiation with matter.
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At different frequencies we get different responses from the molecules or from the atoms,
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Whatever it is that we happen to be irradiating.
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Let us take a molecule like, take a diatomic molecules like hydrogen fluoride.
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We have that, the bond of this.
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We have the H + over here, we have the F - over here, and this molecule vibrates back and forth.
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This goes this way, and this way, this way and this way.
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This idea of a harmonic oscillator is a really nice model for the vibration of a diatomic molecule and the vibration in a bond in general.
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L so imagine a spring with two masses going like this was pretty much the same thing.
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We can study this harmonic oscillator, extract some information from it,
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and come up with a quantum mechanical version of it and it is going to give us a lot of information.
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Irradiation of this molecule by infrared radiation causes the molecule to vibrate at different frequencies.
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Vibrate and oscillate is going to be synonymous.
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In other words, it is going to change the energy level of the molecule.
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This oscillation back and forth is harmonic motion.
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Harmonic motion just refers basically to sin and cos.
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Sin and cos repeat themselves, they are harmonic, periodic, repeating just back and forth.
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That is what harmonic means.
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We want to examine this harmonic behavior and we want to examine the quantum mechanical harmonic oscillation.
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We are going to begin with a classical harmonic oscillator and then move on to the quantum mechanical harmonic oscillation.
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We will get a lot of information by actually studying the classical harmonic oscillation.
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We wish to examine the Q in harmonic oscillation and we begin with the classical harmonic oscillator.
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Let us go ahead and start on a new page.
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Consider the following system.
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We have basically a setup like this and this is a wall.
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We are going to fix one hand and we are going to have a spring with a mass on it and it is going to be a just a spring in its normal unstretched,
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uncompressed string, what we called the equilibrium length of the spring.
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We are going to call that S of 0.
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There is a mass, it has mass M.
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S of 0 is the distance of the undisturbed spring.
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Again, we are making it simple for ourselves at first by holding one end fixed and just having one mass oscillating back and forth.
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It is going to oscillate around some equilibrium position.
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That is how it for all science, we start with the simplest case and
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then we move up in degrees of complication until we arrive at something which matches what we want, our real world.
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It is always like that, start with the simplest case and move forward from there.
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We can stretch the spring or compress the spring at distance S.
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We can stretch or compress the spring at distance S.
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Let us go ahead and take a look at some of those.
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If we stretch the spring, here we stretched it and now this distance is S.
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Or we can go ahead and compress it, push the spring in.
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We have taken it and we have actually pushed it in.
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This is the distance S.
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No gravitation is acting on it, there is no gravitational force.
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The only force that is acting on this mass is the spring.
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I will pull the spring this way it will want pull it back that way.
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I push the spring in, this spring is the one that is going to push back the other direction.
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There is no gravitational force acting on the mass.
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So far so good.
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This spring force is the only force acting on the mass.
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Hooke’s law, you remember hopefully.
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Remember having heard from your physics course.
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The force acting on a mass attached to a restoring medium, in this particular case the restoring medium is the spring.
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It could be something else, it could be a rubber band.
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Anything that will, as you pull it, tries to pull it back.
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As you push it, it tries to push it back, restores it back to its equilibrium position.
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The force acting on a mass attached to a restoring medium is directly proportional to the displacement.
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The displacement is the extent to which it actually pulled away from its equilibrium position.
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From here to here, that is the displacement right there.
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That is the displacement from its equilibrium position.
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It is the S – S0.
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It is directly proportional to the displacement.
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That is Hooke’s law and mathematical statement is the force is actually equal to - K × S - S0.
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If I stretched it, S is bigger than S0.
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The stretched is bigger than the equilibrium position so S – S0 is positive.
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The force we put a negative sin in front of it because now the force is actually pulling the mass back that way.
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If I push it in, our S is smaller than S0.
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This ends up being negative and this negative sin in front actually makes the force go in the positive direction.
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It is opposite.
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I will pull this way, the force wants to pull it back that way.
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I push this way, the force wants to push it back that way.
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That is why the negative sin is there.
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This is Hooke’s law.
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In the case that the force is constant that is called the force constant or the spring constant.
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A large K means you have a very stiff spring and a small k means you have a very loose spring, very easy to push and pull.
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We have F is equal to - K × S - S0.
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Let us go ahead and draw this out.
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This is the equilibrium position of the spring.
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If I had it stretched out, this distance would be my displacement.
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This is my S - S0.
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The force, if I stretch it, the force is going to want to pull it back that way.
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Of course, the compressed, I have pushed it in from its equilibrium position.
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This is the equilibrium position.
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We have right there, that is the S, that is the S0, S - S0, that is this displacement.
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The force is going to be pushing that way.
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Trying to push it back to its equilibrium, the restoring force, the restoring medium.
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We have this, we also know from Newton’s second law that force = the mass × the acceleration.
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It is equal the mass × the acceleration is the second derivative of the displacement.
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The displacement is S.
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The first derivative of that is going to be the velocity.
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The second derivative is going to be the acceleration.
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We can write D² S DT².
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Let us let X be the net displacement.
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Net displacement means this distance, the actual distance I have pulled or push away from the equilibrium position.
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This is X and this is X, net displacement.
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In other words, it is S - S0.
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Mathematically, we are going to let X equal S – S of 0 because we are ultimately concerned with.
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S0 is the initial, S is the final, the difference between them is the displacement.
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S = S - S0 means that S = X + S0.
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Now F and F, I'm going to set them equal to each other.
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I have MD² S DT² = -K × S - S0.
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We said the X is equal to S - S0.
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I’m going to put X in for here and the D² of S DT², I’m going to take the derivative of this.
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DS DT = DX DT, this is a constant.
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It is just the equilibrium position so derivative is just 0.
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Of course, if I differentiate twice, it is just going to be D² S/ DT² = D² X/ DT².
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That is it, nothing strange going on here.
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When I put this in for this, I get M D² X DT² = -KX.
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Or more appropriately M D² X DT² + KX is equal to 0.
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This is the differential equation for a harmonic oscillator.
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This is the equation that we solved in order to find the function X of T.
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When we find X of T, what that tells us that any time T is going to tell us exactly how far this mass is from the equilibrium position, where is it.
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That is what we are doing.
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This is the differential equation we solved.
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The general solution of this equation, I’m not going to go through the actual solution of it.
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I will just go ahead and give it to you here.
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The general solution of this differential equation.
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I need a little more room here.
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X of T = C1 cos of ω T + C2 × the sin of ω T.
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Where ω is actually equal to the force constant ÷ the square root mass or to the ½ power.
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Instead of putting this, we just use the ω.
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Now X of T, as we just said, X of T tells me how far the mass is from the equilibrium position at any time T.
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You have this mass equilibrium position, if I pull it and let it go, it is going to just bounce back and forth, oscillate.
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This equation tells me where, how far from the equilibrium position it is, X at time T.
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That is all I have done.
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Let us go ahead and take this mass and that is the equilibrium position.
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What I have done is I have actually hold it, I have stretched it out, and I’m going to let it go.
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When I let it go, it is going to oscillate back and forth, back and forth, back and forth.
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Let us actually work this out.
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This is the general solution.
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We want to find some specific cases.
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The initial position, let me go to blue here.
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The initial position, in other words X at time 0 is equal to A.
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A is the distance that I have pulled.
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I have pulled it to distance A.
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Let us go ahead and call that A.
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The initial velocity, velocity is the first derivative.
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X prime at T = 0.
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At 0, I have to let it go.
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When I let it go, that is when it starts.
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The initial velocity is actually 0.
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We have C1 cos ω T.
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We have our equation which is X of T = C1 × cos of ω T + C2 × the sin of ω T.
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The X of 0 is equal to C1 × the cos of 0, when I put 0 in for T here and here, + C2 × the sin of 0.
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And we know that X of 0 is equal to A.
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Sin of 0 is 0 so that term goes to 0.
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Cos of 0 is 1, therefore, C1 is equal to A.
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We found the C1, we found that coefficient.
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Let us take X prime.
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Let me rewrite the equation up here.
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X of T = C1 × the cos of ω T + T2 sin ω T.
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X prime of T that is going to equal -ω C1 × the sin of ω T + ω C2 × the cos of ω T.
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X prime of 0 = - WC 1 × the sin of 0 + ω C2 × the cos of 0.
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We said that that actually = 0.
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This is 0, this becomes C2 W = 0.
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W is not equal 0 which means that C2 = 0.
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Now we went ahead and found C2.
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We have X of T is actually equal to A × the cos of ω T.
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We found a specific solution under the circumstance of me pulling it and letting it go.
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This is harmonic motion, you know what a cos function looks like.
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In this particular case, A is the amplitude and it goes like this, it oscillates back and fourth.
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This is harmonic motion, it is always going to be in terms of sin and cos.
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The period of this oscillation is equal to 2 π ÷ ω.
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The frequency of the oscillation is equal to ω/ 2 π.
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It is the reciprocal of the period and it is in cycles per second.
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Cycles per second, otherwise known as a Hertz.
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We often do not include the cycles, we just put second.
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It is usually like this, inverse second.
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This is frequency.
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Let us take ω / 2 π cycles per second.
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Let us multiply by how many radians are there in a cycle.
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If you do one cycle circle it was 2 π radians.
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It is 2 π radians per cycle.
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The 2 π and the 2 π cancel and I'm left with radiance per second.
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Ω, the unit is radians per second, this is the angular velocity.
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We have our period, our frequency, we have our angular velocity which is ω and A is called the amplitude.
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A is the amplitude of the oscillation.
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It is the largest value of the amplitude can actually have.
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If a stretch something A, when it goes the other direction it can only compress the spring at distance A maximum.
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It will just go back and forth, AAA –AA – AA, something like that.
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A is the amplitude of the oscillation.
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All this information is extractable from this amplitude angular velocity ÷ 2 π gives you the frequency.
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Take the reciprocal of the frequency will give you the period.
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Or you can just take 2 π over the angular velocity and that will give you the period.
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All of this information comes from that.
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Recall from physics, I will go back to blue here.
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Recall from physics that a force is equal to - the derivative of the potential energy.
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DV = - F of X DX.
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If I want the potential energy it equals- the integral of DF of X DX.
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Just integrate that equation.
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Here our F of X is equal to – KX, that is Hooke’s law.
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The force acting on a mass is directly proportional to the extent to which I actually pull that mass away from its equilibrium position.
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Our potential energy is equal to - the integral of - KX DX, which means it is equal to ½ KX² + some constant.
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We can set C equal to 0, it is just a 0. energy.
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We can go ahead and set it to 0 so what we get is that the potential energy stored
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When I stretch the spring or compress the spring is equal to ½ KX².
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This is very important, that is a potential energy of a mass spring system.
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We have our V of X is equal to ½ KX², that is nice.
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Our kinetic energy, we know the kinetic energy is equal to ½ the mass × the velocity².
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Velocity is just a derivative of DX is DX DT.
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I’m just going to go ahead and call it X prime².
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X of T, we know that X of T is A × the cos of ω T.
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When we take the first derivative which is going to be the velocity, it is going to end up being -ω A sin of ω T.
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Let us go ahead and put those back in.
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This is X of T X prime of T, we are going to put these in to here.
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Our potential energy is actually equal to ½ K × A cos of ω T² which is going to equal ½ K A² cos² ω T.
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This is a potential energy of our classical harmonic oscillator.
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Let us go ahead and calculate the kinetic energy.
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Kinetic energy is equal to ½ the mass × the derivative².
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The derivative² was - A ω sin because we take the derivative sin ω T².
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We end up with, the -² cancels out so we end up with ½ MA² O² sin² ω T.
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This is our kinetic energy of the harmonic oscillator.
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The total energy of the system we know is equal to the kinetic energy + the potential energy.
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Our total energy we just add them up.
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It is just ½ MA² cos² ω T + ½ M ω².
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I just switched the ω² and A².
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Sin² ω T.
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That is the expression for the total energy of a harmonic oscillator.
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When I plot these, this is what I get.
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Equilibrium position that is A, this is –A.
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A equilibrium –A, it is going to oscillate back and forth like this.
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Here is what it looks like.
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We have this up here, let me do this one in red.
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Let me go ahead and go to black.
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This right here, this is the potential energy curve.
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This right here, the red, that is the kinetic energy curve.
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Notice what is going on.
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I pull the thing to A, all the energy is potential and there is no velocity.
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I have not let go, 0 kinetic energy.
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I release it, all that potential energy starts to go to kinetic.
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At some point they meet, where the kinetic and the potential are equal.
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As it passes through the equilibrium position because it is actually at the equilibrium position X is 0.
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Therefore, the potential energy is 0.
00:29:25.400 --> 00:29:28.400
However, the kinetic energy is at its maximum.
00:29:28.400 --> 00:29:33.800
It will start to slow down as it passes through the equilibrium position, it starts to go toward the negative.
00:29:33.800 --> 00:29:38.400
It will start to slow down so the kinetic energy is going to drop down to 0 and
00:29:38.400 --> 00:29:44.700
the potential energy is going to rise to its maximum, back and fourth.
00:29:44.700 --> 00:29:54.100
Notice, the sum of energy is actually constant so the kinetic and potential energy they switch off.
00:29:54.100 --> 00:29:56.300
The total energy of the system is constant.
00:29:56.300 --> 00:30:00.400
I will go ahead and do an algebraic so you can see.
00:30:00.400 --> 00:30:02.500
Let us go ahead and write this out.
00:30:02.500 --> 00:30:05.200
Let us go ahead and write it in blue.
00:30:05.200 --> 00:30:27.700
At maximum displacement, the mass has stopped which implies that the kinetic energy is equal to 0.
00:30:27.700 --> 00:30:40.400
V is at a maximum, here and here.
00:30:40.400 --> 00:31:02.500
As the mass passes through the equilibrium position, now the potential energy is equal to 0 and K is a max.
00:31:02.500 --> 00:31:07.800
Here and here the potential is 0, kinetic energy is at a max.
00:31:07.800 --> 00:31:12.200
At the ends it is the potential energy that is on the max and the kinetic energy is at 0.
00:31:12.200 --> 00:31:14.200
They switch off.
00:31:14.200 --> 00:31:18.600
In between, we have some of both.
00:31:18.600 --> 00:31:23.100
Let us go ahead.
00:31:23.100 --> 00:31:43.700
We have our energy is equal to ½ M A² cos² ω T.
00:31:43.700 --> 00:31:46.600
I’m checking my equations here.
00:31:46.600 --> 00:32:08.100
I’m messing up the sin and cos.
00:32:08.100 --> 00:32:13.200
I will make sure I have this to maintain the order that I actually used in here.
00:32:13.200 --> 00:32:20.200
Potential is going to be the K and it is going to be square O².
00:32:20.200 --> 00:32:22.400
Sorry about.
00:32:22.400 --> 00:32:42.600
We have got ½ mass A² ω² sin² ω T + ½ K A² cos ω cos² ω T.
00:32:42.600 --> 00:32:45.300
Kinetic energy + potential energy.
00:32:45.300 --> 00:32:57.000
Recall that that is equal to K/ M ^½.
00:32:57.000 --> 00:33:06.400
So ω² is equal to K/ M.
00:33:06.400 --> 00:33:13.500
I can go ahead and put that into there.
00:33:13.500 --> 00:33:35.700
Therefore, I have my total energy is equal to ½ M A² K/ M sin² ω T + 1/2 K A² cos² ω T.
00:33:35.700 --> 00:33:40.800
The M cancels the M, I can go ahead and I combined terms and factor out.
00:33:40.800 --> 00:33:45.000
I have ½ A² K, ½ A² K.
00:33:45.000 --> 00:33:55.100
I have 1/2 A² × K × sin² ω T + cos² ω T.
00:33:55.100 --> 00:33:58.500
You remember from trigonometry, the sin + cos² is equal to 1.
00:33:58.500 --> 00:34:05.200
Therefore, my total energy is 1/2 A² × K.
00:34:05.200 --> 00:34:07.700
The total energy of the harmonic oscillator.
00:34:07.700 --> 00:34:10.800
This is a constant, A is the amplitude as a constant.
00:34:10.800 --> 00:34:13.600
K is a constant.
00:34:13.600 --> 00:34:22.000
This is an algebraic representation of the fact that the energy of the harmonic oscillator is constant.
00:34:22.000 --> 00:34:25.600
It oscillates back and forth between the potential and the kinetic.
00:34:25.600 --> 00:34:28.200
They tradeoff and one of them is maximized, the other is minimized.
00:34:28.200 --> 00:34:35.400
The one is minimized, the other is maximized, but it is a constant.
00:34:35.400 --> 00:34:42.200
Let us see.
00:34:42.200 --> 00:35:08.500
Now a system where the total energy is conserved is called appropriately a conservative system.
00:35:08.500 --> 00:35:12.400
The harmonic oscillator that we describe is a conservative system.
00:35:12.400 --> 00:35:18.800
Energy is a constant, it is transferred between kinetic and potential but it is a constant.
00:35:18.800 --> 00:35:21.900
It does not go away.
00:35:21.900 --> 00:35:26.700
Energy is conserved, it is not lost to anything.
00:35:26.700 --> 00:35:28.900
Thank you so much for joining us here at www.educator.com.
00:35:28.900 --> 00:35:32.500
We will see you next time for a continuation of the discussion of the harmonic oscillator.
00:35:32.500 --> 00:35:33.000
Take care, bye.