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3. Driven Harmonic Oscillators

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3. Driven Harmonic Oscillators MITOCW | 3. Driven Harmonic Oscillators The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: Welcome back, and today I'm going to do for you three problems all related to driven harmonic oscillators. Now one of the things you'll notice as I'm doing these problems, that it seems I'm doing the same thing over and over again. You may have also noticed that in the other problems I did earlier for you in harmonic oscillators, et cetera. So at this stage, I have to remind you about something. This is no accident that it is like that. It's to do with the scientific method, and I would want to briefly remind you about how we are solving problems. Here is a diagram of a generic problem. What does it consists of? A typical problem is we are told of a physical situation in words, ordinary language. And in ordinary language, we are asked questions about given that physical situation, what will happen as a function of time? What kind of emotion it has, questions of that kind. What problem solving is, doing the scientific method, is the following. The first thing we have to do is, by studying this situation and using the laws of nature, describing the physical situation in terms of mathematical equations. This is always possible. So starting with as I say this description, using laws of nature, we end up with a description in terms of mathematical equations. The next thing we do is, we now forget that this has anything to do with physics. We have a set of mathematical equations, which we have to solve. This is mathematics. We use mathematics to solve this problem. All right? Once we have found the solution to this mathematical equation, there's only one thing left, we identify the answer with the questions posed in the problem. So understanding of physics comes in this step, which I call step 1, and in the last 1 one. The middle one, step 2, is pure mathematics. Now, it so happens that if you take all driven harmonic oscillators, the mathematics that you end up with is identical. In other words, if I look at the description of the problem in mathematical terms, I cannot tell you which problem that description belongs. That's a generic description of many, many, many situations. Today I'll take three of them, but all three situations, all three problems-- although to you, they may look completely different-- after I've translated them in terms of mathematical equations, are identical. And so it's not surprising that from here on when we then try to solve them, et cetera, we are doing the same problem, or at least it appears so. Having said that, let's immediately dive in and take one of the problems. So the first problem I'm going to consider is an idealized-driven RLC circuit. Why do I see idealized? Because as I've mentioned before, in real life situations, things extremely complicated, and if you take everything into account, it will take you essentially forever to get an exact solution. But we're often not most of the time not interested in the very exact one. So we simplify a physical situation to the bear essential parts and solve that problem. So this is a problem where you have a circuit consisting all a capacitance, inductance, and a resistance, in series, which is driven by an alternating voltage source V0 cosine omega t. Before time t equals 0, I'm assuming the circuit is open. In other words, there is no charges anywhere, no current flowing. At t equals 0, we close the circuit so the charge on the capacitor t equals 0, is 0. The current flowing in the circuit is 0. All right? And the question we want to answer is, what will the current be in the circuit at some time t a specific one, in 17 seconds or whatever, some specific time. Since this is completely specified, I can predict what will happen. That's what it means in this case solving this problem. So let's do it. So we are now entering step 1. We're going to take the physical situation and convert it to a description in terms of mathematics. This is the conceptually hardest step of any problem. It is the one 2 where you need to understand the physics. It may not necessarily be the hardest or the longest. Often the solution of the equations are incredibly tedious and long and take all your time. But they are routine mathematics. Here you've got to conceptualize what's going on and translate it as I say to the language of mathematics. So in the case of I've had have to define coordinates for this problem. So what I have is, I'll assume that at some instant of time that in this circuit, it's closed because it's past equal to 0, there is a current I flowing that on this capacitor on the left there is plus charge Q and minus charge Q on the right side that is this inductance. And over here, we have an alternating potential applied. Now I have to be very specific at any instant of time which side is positive and negative. I'll assume that t equals 0. It's 0 over here, and at this point here, it's plus V0 cosine omega t. So these are the coordinates of this problem. And now what drives the current? What decides what happens here? What are the law of natures applicable? In this case, are the Maxwell's equations or to be specific, the Faraday equation. All right? One more thing I should add that I is a current here flowing charge by conservation of charge. It's law of physics. The current at the instant of time, will be equal to the rate of change of the charge. So at the instant time, I is, in my language, Q with a dot on top of it. This is plus because of the way in which I define my coordinates. Beware if you by mistake say suppose I define my coordinates this being minus and this plus, then this equation would be a minus here, conservation of charge. But with the way I've defined these coordinates, this is plus. So that's one. Next I said what the law of nature which governs the motion of the charges is the Faraday's law, which states that if I take the line integral of the electric field around a closed loop, that's this, it equals to minus the rate of change of total magnetic flux linking this circuit, going through this circuit. 3 Now, you know that the magnetic flux through a circuit is equal to the current times the self-inductance of that circuit or the total inductance in that circuit. So taking this law of nature, this law of nature, I can now translate this into mathematics, and that's what I'm going to do. So first I'm going to take the line integral of E dl around this circuit. If I go around from here to here, this is positive relative to this point, so the electric field from here to here will be in this direction, and so the line integral of E dl along here will be minus, and by definition of a voltage, it will be just minus the voltage drop across here, which is V0 cosine omega t. So this is minus V0 cosine omega t. I continue going around. From here to here, the electric field is in this direction. And so that will be the integral V dl is Q over C, that's from the definition of a capacitance, of course. Then as we go around in these idealized diagrams, all wires have 0 resistance and therefore there cannot be field in them. That's why there's no contribution to this from the field inside the wires. I continue. This inductance is just coiled wire, so there's still no contribution to this in the L. I continue. Here I know there is a current flowing like this means the reason the electric field in this direction and once again knowing the definition of our Ohms law, I get that the integral of the electric fields from here to here is I times R. So this is this quantity, the integral of the electric field around this closed loop. That must equal by Faraday's law to minus the rate of change of magnetic flux, which is minus L dI dt. We're almost finished. Now we do algebra. We're playing around. I can rewrite this like this knowing that the current is the rate of change of Q, and the rate of change of current is the second derivative. So this is just straightforward algebra. Now, I'm going to rewrite just moving terms around to make it look in the way I prefer. And so I've rewritten this equation now in this form. And now, I will just redefine some constants so that I recognize the equation better. So I will define R over L by a constant gamma.
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