WEBVTT
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Thank youaint.
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This is a brief, the equation, and we got the characteristic equation from last time.
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The general topic for today is going to be oscillations, which are extremely important
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in the applications and everyday life.
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But the oscillations we know are associated with a complex root.
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So they correspond to complex roots of that characteristic equation, r squared plus
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br plus k equals 0.
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I'd like to begin.
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Most of the lecture will be about discussing the relations between these numbers, these
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constants and the various properties that the oscillatory solutions have.
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But before that, I'd like to begin by clearing up a couple of questions.
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Everybody has at some point or other when they study the case of complex roots.
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Complex roots are the case which produce oscillations in the solutions.
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That's the relation and that's why I'm talking about this for the first few minutes.
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So you get the, what is the problem?
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The complex roots, of course, there will be two roots and they occur,
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they're complex conjugates of each other.
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So they'll be of the form a plus or minus bi.
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Last time I showed you, I took the root r equals a plus bi, which leads to the solution.
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The corresponding solution is a complex solution, which is e to the a t a plus i b t.
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And what we did was, the problem was to get real solutions out of that.
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We needed two real solutions.
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And the way I got them was by separating this into its real part and its imaginary part.
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And I proved a little theorem for you that said both of those give solutions.
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So the real part was e to the a t times cosine b t.
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And the imaginary part was e to the a t sine b t.
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And those were the two solutions.
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So here was y one.
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And the point was, those out of the complex solutions, we got real solutions.
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We have to have real solutions because we live in the real world.
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The equation is real.
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The coefficients are real. They represent real quantities.
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That's the way the solutions therefore have to be.
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So these are now real solutions.
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These two guys, y one and y two.
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Now, the first question almost everybody has.
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And I was pleased to see that at the end of the lecture, a few people came up and asked me,
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yeah, well, you took a plus bi.
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But there was another root a minus bi.
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We didn't use that one.
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That would give two more solutions, right?
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Of course, they didn't say that.
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They were too smart.
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They just said, what about that other root?
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Well, what about it?
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The reason I don't have to talk about the other root is, because although it does give two solutions,
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it doesn't give two new ones.
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Maybe I can indicate that most clearly here, even though you won't be able to take notes
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by just using colored chalk.
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Suppose instead of a plus bi, I used a minus bi.
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What would have changed?
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Well, this would now become minus here.
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Would this change?
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No.
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Because e to the minus i bt is the cosine of minus b.
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But that's the same as the cosine of b.
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How about here?
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This would have become the sine of minus bt.
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But that's simply the negative of the sine of bt.
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So the only answer change would have been to put a minus sign there.
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Now, I don't care if I get y2 or negative y2, because what am I going to do with it?
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When I get it, I'm going to write y, the general solution, as c1, y1, plus c2, y2.
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So if I get negative y2, that just changes that arbitrary constant from c2 to minus c2,
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which is just as arbitrary a constant.
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So in other words, there's no reason to use the other root because it doesn't give anything new.
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Now, there the story could stop, and I would like it to stop, frankly.
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But I don't dare because there's a second question.
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And I'm visiting recitations, not this semester, but in previous semesters in 1803.
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So many recitations do this.
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I have to partly inoculate you against it, and partly tell you that some of the engineering
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courses do do it, and therefore you probably should learn it.
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Also.
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So there's another way of proceeding, which is what you might have thought.
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Hey, look, we got two complex roots.
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That gives us two solutions, which are different.
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They're not neither one is a constant, multiple of the other.
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Why not?
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So the other approach is use as the general solution, y equals, now I'm going to put a capital
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c here.
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You'll see y in just a second, times e to the a plus b i times t.
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And then I'll use the other solution.
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c2 times e to the a minus b i t.
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These are two independent solutions, and therefore can't I get the general solution in that form.
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Now in a sense, you can, the whole problem is the following, of course, that I'm only
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interested in real solutions.
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This is a complex function.
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This is another complex function.
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It's got an i in it, in other words, when I write it out as u plus i v.
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If I expect to be able to get a real solution out of that, that means I have to make allow
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these coefficients to be complex numbers and not real numbers.
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So in other words, what I'm saying is that an expression like this, where the a plus
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b i and a minus b i are the roots of that, our complex roots of that characteristic equation,
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is formally a very general complex solution to the equation.
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And therefore the problem becomes how from this expression do I get the real solutions?
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So the problem is, I accept these as the complex solutions.
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My problem is to find among all these guys where c1 and c2 are allowed to be complex.
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The problem is which of the green solutions are real.
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Now there are many ways of getting the answer.
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There's a super hack way.
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The super hack way is to say, well, this one is c1 plus i d1.
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This is c2 plus i d2.
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Now write all this out in terms of what it is, cosine plus sine plus i sine and don't
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forget the e to the a t.
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And write all out and it'll take a whole entire board.
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And then I'll just see what the condition is.
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I'll write it's real part and it's an imaginary part.
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And then I'll say the imaginary part has got to be zero.
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And then I'll see what it's like.
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That works fine.
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It just takes too much space.
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And also it doesn't teach you a few things that I think you should know.
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I'm going to give another.
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So let's say we can answer this two ways by hack.
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In other words, multiply everything out.
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Multiply all out.
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Make the imaginary part equal zero.
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Now here's a better way, in my opinion.
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What I'm trying to do is this is some complex function u plus i v.
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How do I know when a complex function is real?
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I want this to be real.
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Well, the hack method corresponds to saying v must be equal to zero.
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It's real if v is zero.
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So expand it out and see what v is zero.
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There's a slightly more subtle method, which
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is to change i to minus i.
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And what?
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And see if it stays the same.
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And see if it stays the same.
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Because if I change i to minus i, and it turns out the expression doesn't
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change, then it must have been real.
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v must have been zero if that's expression doesn't change when I change i to minus i.
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Well, sure, but you'll see it work.
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All right, now that's what I'm going to apply to this.
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If I want this to be real, I rephrase the question for the green equation,
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the green solution as change.
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So I'm going to change i to minus i in the green.
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And that's going to give me what conditions and that will give conditions on the seas.
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Well, let's do it.
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In fact, it's easier done than talked about.
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Let's change, take the green solution and change.
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So I don't want to re, well, I'd better recopy it.
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See one.
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So these are complex numbers.
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That's why I wrote them as capital letters, because little letters you tend to
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interpret as real numbers.
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So c1 e to the a plus b i t, I'll recopy it quickly plus c2 e to the a minus b i t.
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OK, we're going to change i to negative i.
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Now, here's a complex number.
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What happens when you change i to negative i?
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You change it into its class?
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What do we change it into?
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It's complex conjugate, right?
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And the notation for complex conjugate is you put a bar over it.
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So in other words, when I do that, the c1 changes to c1 bar complex conjugate,
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the complex conjugate of c1.
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What happens to this guy?
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This guy changes to e.
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Sorry.
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This guy changes to e to the a minus b i t.
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This changes to the complex conjugate of c2, now times e to the a plus b i t.
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Well, I want these two to be the same.
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I want the two expressions the same.
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Why do I want them the same?
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Because if there's no change, that will mean that it's real.
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Now, when is that going to happen?
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That happens if, well, here is this that if c2 should be equal to c1 bar.
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That's only one condition.
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There's another condition.
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c2 bar should equal c1.
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So I get two conditions, but there's really only one condition there.
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Because if this is true, that's true too.
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I simply put bars over both things and two bars cancel each other out.
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If you take the complex conjugate and do it again, you get back where you started with.
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Change y to minus i.
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And then y to minus i again, the, well, nevermind.
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Anyway, these are the same.
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This equation doesn't say anything that the first one didn't say already.
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So this one is redundant.
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And that conclusion is that the real solution to the equation
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are in their entirety.
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I now don't need both c2 and c1.
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One of them will do, and since I'm going to write it out as a complex number,
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I'll write it out in terms of its coefficient.
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So it's c1, oh, let's just simply write it.
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c plus i times d.
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That's the coefficient.
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That's what I called c1 before.
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And that's times e to the a plus b i t.
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There's no reason why I put b i here and i d there in case you're wondering.
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Sheer caprice.
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And what's the other term?
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Now the other term is completely determined.
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Its coefficients must be c minus i d times e to the a minus b i t.
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In other words, this thing is perfectly general.
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Any complex number times that first
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proved you used, exponentiated.
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And the second term can be described as the complex conjugate of the first.
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The coefficient is the complex conjugate, and this part is the complex conjugate of that.
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Now, it's in this form, in this form,
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some many engineers write the solution this way.
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And physicists too.
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So scientists and engineers will include,
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write the solution this way.
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Write the real solutions this way.
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In that complex form, well, why do they do something so perverse?
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You'll have to ask them.
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But in fact, when we study Fourier series,
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we'll probably have to do something, have to do that at one point.
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If you work a lot with complex numbers,
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it turns out to be in some ways a more convenient representation
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than the one I've given you in terms of signs and cosines.
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Well, from this, how would I get?
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Suppose I assisted, well, if someone gave it to me in that form,
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I don't see how I would convert it back into signs and cosines.
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And I'd like to show you how to do that efficiently too,
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because again, it's one of the fundamental techniques
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that I think you should know.
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And I didn't get a chance to say it when we studied complex numbers,
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that first lecture.
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It's in the notes, but that doesn't prove anything since I
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don't think I made you use it in an example.
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So the problem is now, by way of finishing this up,
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to how do you change this to the old form?
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I mean the form involving signs and cosines.
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Now again, there are two ways of doing it.
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The hack way is, you write it all out.
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Well, the e to the a t, a e to the a plus b i t,
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turns into e to the a t times cosine, cosine this plus i
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sine that.
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And the other term does too.
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And then you've got stuff out front.
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And you know, they'll think stretches over two boards,
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but you group all the terms together, you finally get it.
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By the way, when you do it, you'll
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find that the imaginary part disappears completely.
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It has to, because that's the way we chose the coefficients.
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So there's the hack method.
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Write it all out.
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Blah, blah, blah, blah, blah, blah, blah, blah.
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And nicer.
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And teach you something you're supposed to know.
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Write it this way.
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First of all, you notice that both terms have
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an e to the a t factor.
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Let's get rid of that right away.
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I'm pulling it out front, because that's automatically real.
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And therefore, it isn't going to affect the rest of the answer at all.
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So let's pull out that.
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00:18:11.540 --> 00:18:14.540
And what's left?
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00:18:14.540 --> 00:18:15.780
What's left?
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00:18:15.780 --> 00:18:19.340
You see involves just the two parameters c and d.
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00:18:19.340 --> 00:18:24.380
So I'm going to have a c term, and I'm going to have a d term.
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What multiplies the arbitrary constant c?
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00:18:29.580 --> 00:18:30.380
Answer.
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00:18:30.380 --> 00:18:33.060
After I remove d to the a t, what multiplies it
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00:18:33.060 --> 00:18:41.700
is e to the b i t plus e to the e to the b i t.
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00:18:41.700 --> 00:18:45.340
Let's write it i b t.
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00:18:45.340 --> 00:18:52.740
And the other term is plus e to the negative i b t.
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00:18:52.740 --> 00:18:53.780
See how I got that?
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00:18:53.780 --> 00:18:55.420
Hold it out.
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00:18:55.420 --> 00:18:57.140
And how about the d?
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What goes with d?
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00:18:59.180 --> 00:19:02.580
d goes with, well, first of all, there's an i in front
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that I better not forget.
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And then the rest of it is i.
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00:19:09.900 --> 00:19:11.700
So it's i d times.
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00:19:11.700 --> 00:19:23.420
It's e to the b i t, e to the i b t minus now e to the minus i b t.
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So that's the way the solution looks.
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00:19:29.140 --> 00:19:32.620
It doesn't look a lot better, but now you must use
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00:19:32.620 --> 00:19:35.780
the magic formulas, which I want you to know, as well as
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you know, oil is formula.
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Even better than you know, oil is formula.
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00:19:41.620 --> 00:19:43.580
And there are consequences of oil is formula.
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00:19:43.580 --> 00:19:46.260
The oil is formula read backwards.
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00:19:46.260 --> 00:19:49.980
Oil is formula says, you got a complex exponential here.
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This is how to write it in terms of signs and cosines.
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00:19:52.980 --> 00:19:56.580
The backwards thing says, you got a sign or a cosine.
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00:19:56.580 --> 00:19:59.980
Here's the way to write it in terms of complex exponentials.
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00:19:59.980 --> 00:20:07.580
And remember, the way to do it is cosine a is equal to e
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00:20:07.580 --> 00:20:17.220
to the i a t i a plus e to the negative i a divided by 2.
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00:20:17.220 --> 00:20:21.940
And sine of a is almost the same thing
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00:20:21.940 --> 00:20:23.580
except you use a minus sign.
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00:20:23.580 --> 00:20:30.580
And what everybody forgets, you have to divide by i.
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00:20:30.580 --> 00:20:37.020
So this is the backward version of oil is formula.
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00:20:37.020 --> 00:20:40.700
They're equivalent to oil is formula.
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00:20:40.700 --> 00:20:45.180
If I took cosine a, multiply this through by i and added them
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00:20:45.180 --> 00:20:48.820
up on the right hand side, I get exactly e to the i a.
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00:20:48.820 --> 00:20:51.740
I get oil is formula in other words.
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00:20:51.740 --> 00:20:56.780
All right, so what does this come out to be finally?
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00:20:56.780 --> 00:21:00.780
This is, this particular sum of exponentials,
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00:21:00.780 --> 00:21:03.220
you should always recognize as real.
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00:21:03.220 --> 00:21:07.300
You know it's real because when I change i to minus i,
293
00:21:07.300 --> 00:21:10.620
the two terms switch and therefore it's real.
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00:21:10.620 --> 00:21:12.380
The expression doesn't change.
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00:21:12.380 --> 00:21:13.300
What is it?
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00:21:13.300 --> 00:21:19.140
This part is twice the cosine of dt.
297
00:21:19.140 --> 00:21:20.500
What's this part?
298
00:21:20.500 --> 00:21:26.980
This part is 2i to i times the sine of dt.
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00:21:26.980 --> 00:21:29.100
And so what does the whole thing come out to be?
300
00:21:29.100 --> 00:21:40.660
It is e to the a t times 2c cosine dt plus i times,
301
00:21:40.660 --> 00:21:44.180
did I lose a possibly a?
302
00:21:44.180 --> 00:21:45.540
No, it's okay.
303
00:21:45.540 --> 00:21:54.900
Minus i times i is minus, so minus 2d times the sine of dt.
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00:21:54.900 --> 00:21:55.660
Shall I write that out?
305
00:21:55.660 --> 00:22:06.900
So in other words, it's e to the a t times 2c cosine dt minus 2d
306
00:22:06.900 --> 00:22:12.300
times the sine of dt, which is since 2c and negative 2d
307
00:22:12.300 --> 00:22:15.820
are just arbitrary constants, just as arbitrary the constants
308
00:22:15.820 --> 00:22:17.940
as c and z themselves are.
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00:22:17.940 --> 00:22:22.060
This is our old form of writing the solution, real solution.
310
00:22:22.060 --> 00:22:24.900
So here's the way using sine and cosines.
311
00:22:24.900 --> 00:22:28.460
And here's there's the way that uses complex numbers
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00:22:28.460 --> 00:22:30.660
and complex functions throughout.
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00:22:30.660 --> 00:22:34.220
Notice they both have two arbitrary constants in them,
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00:22:34.220 --> 00:22:38.980
c and d, two arbitrary constants that you expect.
315
00:22:38.980 --> 00:22:41.340
But that has two arbitrary constants in it, too.
316
00:22:41.340 --> 00:22:45.820
Just the real and imaginary parts of that complex coefficient
317
00:22:45.820 --> 00:22:49.420
c plus i d.
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00:22:49.420 --> 00:22:52.780
Well, that took half the period, and it was a long.
319
00:22:52.780 --> 00:22:55.980
I don't consider it a digression because learning
320
00:22:55.980 --> 00:22:59.300
those ways of dealing with complex numbers and complex
321
00:22:59.300 --> 00:23:04.180
functions is a fairly important goal of this course, actually.
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00:23:04.180 --> 00:23:07.940
But let's get back now to studying what the oscillations
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00:23:07.940 --> 00:23:11.340
actually look like.
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00:23:11.340 --> 00:23:28.340
OK.
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00:23:28.340 --> 00:23:35.300
Well, I'd like to say it a little time, but I'll very quickly.
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00:23:35.300 --> 00:23:37.460
You don't have to reproduce this sketch.
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00:23:37.460 --> 00:23:40.260
I just want you to re.
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00:23:40.260 --> 00:23:42.500
I remember very well from Friday to Monday,
329
00:23:42.500 --> 00:23:45.980
but I can't expect you to do for a variety of reasons.
330
00:23:51.020 --> 00:23:52.540
Many, I have to think about this stuff
331
00:23:52.540 --> 00:23:56.780
all weekend, and you I got for a bit.
332
00:24:01.700 --> 00:24:02.820
So here's the picture.
333
00:24:02.820 --> 00:24:04.740
And I won't any more explain what's in it,
334
00:24:04.740 --> 00:24:06.780
but except there's the mass.
335
00:24:06.780 --> 00:24:09.060
Here is the spring constant.
336
00:24:09.060 --> 00:24:11.380
The spring with its constant here is the dash
337
00:24:11.380 --> 00:24:13.620
bar with its constant.
338
00:24:13.620 --> 00:24:18.860
The equation is from Newton's law, m x double.
339
00:24:18.860 --> 00:24:20.740
So this will be x.
340
00:24:20.740 --> 00:24:23.500
And here's let's say his equilibrium point is over here.
341
00:24:23.500 --> 00:24:25.060
It looks like m x double prime.
342
00:24:25.060 --> 00:24:31.100
We derived this last time plus c x prime plus k x equals 0.
343
00:24:31.100 --> 00:24:33.260
And now if I put that in standard form,
344
00:24:33.260 --> 00:24:38.420
it's going to look like x double prime plus c over m x prime
345
00:24:38.420 --> 00:24:43.900
plus k over m times x equals 0.
346
00:24:43.900 --> 00:24:47.580
And finally, the standard form in which your book writes it,
347
00:24:47.580 --> 00:24:51.100
which is good, it's a standard form in general that
348
00:24:51.100 --> 00:24:55.260
is used in the science and engineering courses.
349
00:24:55.260 --> 00:25:08.580
One writes this as a just to be perverse.
350
00:25:08.580 --> 00:25:10.140
I'm going to change x back to y.
351
00:25:13.780 --> 00:25:16.900
So mostly just to be eclectic to get you
352
00:25:16.900 --> 00:25:19.020
used to every conceivable notation.
353
00:25:19.020 --> 00:25:21.780
So I'm going to write this as change x to y.
354
00:25:21.780 --> 00:25:23.900
So that's going to be one y double prime.
355
00:25:23.900 --> 00:25:27.220
And now this is given a new name p,
356
00:25:27.220 --> 00:25:31.180
except to get rid of lots of twos, which would really
357
00:25:31.180 --> 00:25:36.140
screw up the formulas, make it 2p.
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00:25:36.140 --> 00:25:38.100
You'll see y in a minute.
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00:25:38.100 --> 00:25:41.420
So there's 2p times a y prime.
360
00:25:41.420 --> 00:25:46.100
And this thing we're going to call omega naught squared.
361
00:25:46.100 --> 00:25:46.780
Now that's OK.
362
00:25:46.780 --> 00:25:49.660
It's a positive number.
363
00:25:49.660 --> 00:25:52.460
Any positive number is the square of some other positive
364
00:25:52.460 --> 00:25:54.460
number.
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00:25:54.460 --> 00:25:58.300
You'll see why it just makes the formula as much pretty
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00:25:58.300 --> 00:25:59.100
to call it that.
367
00:26:02.380 --> 00:26:07.420
And it makes it also a lot of things much easier to remember.
368
00:26:07.420 --> 00:26:10.140
So all I'm doing is changing the names of the constants
369
00:26:10.140 --> 00:26:14.340
in that way in order to get better formulas,
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00:26:14.340 --> 00:26:17.020
easier to remember formulas at the end.
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00:26:17.020 --> 00:26:19.580
Now we're interested in the case where there's
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00:26:19.580 --> 00:26:22.820
oscillations.
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00:26:22.820 --> 00:26:25.540
In other words, I only care about the case
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00:26:25.540 --> 00:26:28.220
in which this has complex roots.
375
00:26:28.220 --> 00:26:30.500
Because if it has just real roots,
376
00:26:30.500 --> 00:26:32.060
that's the over-damped case.
377
00:26:32.060 --> 00:26:34.500
I don't get any oscillations.
378
00:26:34.500 --> 00:26:37.580
By far, oscillations is by far the more important
379
00:26:37.580 --> 00:26:39.740
of the cases, two cases.
380
00:26:39.740 --> 00:26:42.100
I mean, just because I don't like,
381
00:26:42.100 --> 00:26:44.300
I could go on for five minutes listing things that
382
00:26:44.300 --> 00:26:51.420
oscillate oscillations like this.
383
00:26:51.420 --> 00:26:54.460
So they could oscillate by going to sleep and waking up,
384
00:26:54.460 --> 00:26:56.380
and going to sleep and waking up.
385
00:26:56.380 --> 00:26:57.380
They could oscillate.
386
00:27:01.100 --> 00:27:01.700
OK.
387
00:27:01.700 --> 00:27:04.820
So that means we're going to get complex roots.
388
00:27:04.820 --> 00:27:08.940
The characteristic equation is going to be r squared plus 2p.
389
00:27:08.940 --> 00:27:11.420
So p is a constant now, right?
390
00:27:11.420 --> 00:27:16.100
I often p I use in this position to indicate a function of t,
391
00:27:16.100 --> 00:27:18.820
but here p is a constant.
392
00:27:18.820 --> 00:27:23.500
So r squared plus 2p times r plus omega naught squared
393
00:27:23.500 --> 00:27:25.060
is equal to 0.
394
00:27:25.060 --> 00:27:26.260
Now what are its roots?
395
00:27:26.260 --> 00:27:28.340
Well, you see right away the first advantage
396
00:27:28.340 --> 00:27:31.780
in putting in the 2 there.
397
00:27:31.780 --> 00:27:36.580
When I use the quadratic formula, it's negative 2p over 2.
398
00:27:36.580 --> 00:27:38.380
Remember that 2 in the denominator?
399
00:27:38.380 --> 00:27:40.740
So that's simply negative p.
400
00:27:40.740 --> 00:27:42.460
And how about the rest?
401
00:27:42.460 --> 00:27:44.500
Plus or minus the square root of.
402
00:27:44.500 --> 00:27:45.860
Now do it in your head.
403
00:27:45.860 --> 00:27:52.500
b squared, 4p squared, minus 4 omega naught squared.
404
00:27:52.500 --> 00:27:55.660
So there's a 4 in both of those terms.
405
00:27:55.660 --> 00:27:58.740
When I pull it outside, it becomes a 2.
406
00:27:58.740 --> 00:28:00.420
And then the 2 in the denominator
407
00:28:00.420 --> 00:28:03.620
is lurking, waiting to annihilate it.
408
00:28:03.620 --> 00:28:05.940
So that 2 disappears entirely.
409
00:28:05.940 --> 00:28:10.180
And what we're left with is simply p squared minus
410
00:28:10.180 --> 00:28:13.580
omega naught squared.
411
00:28:13.580 --> 00:28:15.900
Now whenever people write quadratic equations
412
00:28:15.900 --> 00:28:19.220
and arbitrarily put a 2 in there, it's
413
00:28:19.220 --> 00:28:22.300
because they're going to want to solve the quadratic equation
414
00:28:22.300 --> 00:28:24.220
using the quadratic formula.
415
00:28:24.220 --> 00:28:26.740
And they don't want all those 2s and 4s
416
00:28:26.740 --> 00:28:29.780
to be cluttering up the formula.
417
00:28:29.780 --> 00:28:30.780
That's what we're doing here.
418
00:28:33.300 --> 00:28:39.100
OK, now the first case is where p is equal to 0.
419
00:28:39.100 --> 00:28:42.060
This is going to explain immediately why I wrote that omega
420
00:28:42.060 --> 00:28:43.020
naught squared.
421
00:28:43.020 --> 00:28:45.220
That should probably already know from physics.
422
00:28:45.220 --> 00:28:48.900
If p is equal to 0, the mass isn't 0.
423
00:28:48.900 --> 00:28:52.580
Otherwise, nothing would be happening here.
424
00:28:52.580 --> 00:28:55.020
It must be that the damping is 0.
425
00:28:55.020 --> 00:28:58.460
So p is equal to 0 corresponds to undamped.
426
00:29:01.740 --> 00:29:04.740
There is no dashpot.
427
00:29:04.740 --> 00:29:07.820
The oscillations are undamped.
428
00:29:07.820 --> 00:29:11.660
And the equation then becomes the solutions then
429
00:29:11.660 --> 00:29:14.540
are, well, the equation becomes the equation
430
00:29:14.540 --> 00:29:18.260
of simple harmonic motion, which I think you already
431
00:29:18.260 --> 00:29:20.460
are used to writing in this form.
432
00:29:20.460 --> 00:29:22.220
And the reason you're writing in this form
433
00:29:22.220 --> 00:29:24.180
because you know when you do that,
434
00:29:24.180 --> 00:29:28.540
this becomes the circular frequency of the oscillations.
435
00:29:28.540 --> 00:29:30.860
The solutions are pure oscillations,
436
00:29:30.860 --> 00:29:33.820
and omega naught is the circular frequency.
437
00:29:33.820 --> 00:29:36.460
So right away from the equation itself,
438
00:29:36.460 --> 00:29:39.500
if you write it in this form, you can read off
439
00:29:39.500 --> 00:29:41.940
what the frequency of the solutions is going to be.
440
00:29:41.940 --> 00:29:44.220
The circular frequency of the solutions.
441
00:29:44.220 --> 00:29:46.340
Now, the solutions themselves, of course,
442
00:29:46.340 --> 00:29:51.420
look like the general solutions look like y equal.
443
00:29:51.420 --> 00:29:56.540
In this particular case, the p part is 0.
444
00:29:56.540 --> 00:29:57.420
This is 0.
445
00:29:57.420 --> 00:30:05.100
It's simply so in this case, r is equal to omega naught
446
00:30:05.100 --> 00:30:07.940
y times omega naught plus or minus.
447
00:30:07.940 --> 00:30:11.460
But as before, we don't bother with the minus sign
448
00:30:11.460 --> 00:30:13.540
since one of those roots is good enough.
449
00:30:13.540 --> 00:30:19.300
And then the solutions are simply c1 cosine omega naught t
450
00:30:19.300 --> 00:30:22.820
plus c2 sine omega naught t.
451
00:30:22.820 --> 00:30:24.820
That's if you write it out in the sign.
452
00:30:24.820 --> 00:30:26.700
And if you write it using the trigonometric
453
00:30:26.700 --> 00:30:28.940
identity, then the other way of writing it
454
00:30:28.940 --> 00:30:33.340
is a times the cosine of omega naught t.
455
00:30:33.340 --> 00:30:37.100
But now you'll have to put it in a phase lag.
456
00:30:37.100 --> 00:30:39.180
So you have those two forms of writing it.
457
00:30:39.180 --> 00:30:41.700
And I assume you remember the little triangle, which
458
00:30:41.700 --> 00:30:45.460
converts one into the other.
459
00:30:45.460 --> 00:30:50.300
OK, so this justifies calling this omega naught squared
460
00:30:50.300 --> 00:30:53.980
rather than k over m.
461
00:30:53.980 --> 00:30:58.340
And now the question is, what does the damp case look like?
462
00:30:58.340 --> 00:31:06.420
That's more requires a somewhat closer analysis
463
00:31:06.420 --> 00:31:10.740
and requires a certain amount of thinking.
464
00:31:10.740 --> 00:31:15.500
So let's begin with an epsilon bit of thinking.
465
00:31:15.500 --> 00:31:17.180
So here is my question.
466
00:31:17.180 --> 00:31:26.500
So in the undamped case, how do I get?
467
00:31:26.500 --> 00:31:29.020
I want to be sure that I'm getting oscillations.
468
00:31:29.020 --> 00:31:35.220
When do I get oscillations if, well, we get oscillations
469
00:31:35.220 --> 00:31:40.780
if those roots are really complex and not masquerading.
470
00:31:40.780 --> 00:31:42.940
Now, when are the roots going to be really complex?
471
00:31:45.380 --> 00:31:51.620
This has to be the inside has to be negative.
472
00:31:51.620 --> 00:31:57.860
P squared minus omega squared must be negative.
473
00:31:57.860 --> 00:32:04.060
P naught, P squared minus omega naught squared must be less
474
00:32:04.060 --> 00:32:07.140
than 0 so that we're taking the square root of a negative
475
00:32:07.140 --> 00:32:09.820
number and we're getting real complex roots.
476
00:32:09.820 --> 00:32:13.180
Really complex roots.
477
00:32:13.180 --> 00:32:15.340
In other words, now this says, remember, these numbers
478
00:32:15.340 --> 00:32:19.140
are all positive, P and omega naught are positive.
479
00:32:19.140 --> 00:32:22.460
So the condition is that P should be less
480
00:32:22.460 --> 00:32:25.820
than omega naught.
481
00:32:25.820 --> 00:32:30.740
In other words, the damping should be less than the frequency,
482
00:32:30.740 --> 00:32:33.820
the circular frequency.
483
00:32:33.820 --> 00:32:35.740
Except that's P is not the damping.
484
00:32:35.740 --> 00:32:37.780
It's half the damping of twice.
485
00:32:37.780 --> 00:32:41.900
It's half the damping and it's not really the damping either
486
00:32:41.900 --> 00:32:44.660
because it involved the M2.
487
00:32:44.660 --> 00:32:47.540
You better just call it P.
488
00:32:47.540 --> 00:32:49.900
Naturally, I could write the condition out in terms
489
00:32:49.900 --> 00:32:53.340
of Cm and K. So your book does that,
490
00:32:53.340 --> 00:32:56.660
but I'm not going to.
491
00:32:56.660 --> 00:33:06.060
Gives it in terms of Cm and K, which somebody might want
492
00:33:06.060 --> 00:33:10.420
to know, but we don't have to do everything here.
493
00:33:10.420 --> 00:33:14.220
OK, so let's assume that this is true.
494
00:33:14.220 --> 00:33:16.100
Then what does the solution look like?
495
00:33:16.100 --> 00:33:18.860
Well, we already experimented with that last time.
496
00:33:18.860 --> 00:33:20.700
I'm going to first draw it.
497
00:33:20.700 --> 00:33:24.420
Remember, there was some guiding thing, which
498
00:33:24.420 --> 00:33:25.940
was an exponential.
499
00:33:25.940 --> 00:33:29.300
And then down here, we wrote the negative.
500
00:33:29.300 --> 00:33:31.420
So this was an exponential.
501
00:33:31.420 --> 00:33:35.780
In fact, it was the exponential e to the negative Pt.
502
00:33:39.820 --> 00:33:44.500
And in between that, the curve tried to do its thing.
503
00:33:44.500 --> 00:33:47.420
So the solution then looks like this.
504
00:33:47.420 --> 00:33:52.580
It oscillated, but it had to use that exponential function
505
00:33:52.580 --> 00:33:56.940
as its guidelines, as its amplitude, in other words.
506
00:33:56.940 --> 00:33:59.500
Now, this is a really truly terrible picture.
507
00:33:59.500 --> 00:34:02.300
It's so terrible it's unusable.
508
00:34:02.300 --> 00:34:04.620
OK, this picture never happened.
509
00:34:04.620 --> 00:34:17.820
Unfortunately, this is not my forte, along with a lot of other things.
510
00:34:17.820 --> 00:34:19.220
All right, let's try it better.
511
00:34:19.220 --> 00:34:21.380
Here's our better picture.
512
00:34:21.380 --> 00:34:22.900
OK, there's the exponential.
513
00:34:27.820 --> 00:34:31.260
At this point, I'm supposed to have a lecture demonstration.
514
00:34:31.260 --> 00:34:34.180
It's supposed to go up on the thing so you can all see it.
515
00:34:34.180 --> 00:34:35.860
But then you wouldn't be able to copy it.
516
00:34:35.860 --> 00:34:39.580
So at least we're on even terms now.
517
00:34:39.580 --> 00:34:41.860
OK, how does the actual curve look?
518
00:34:41.860 --> 00:34:47.060
Well, I'm just trying to be fair, that's all.
519
00:34:47.060 --> 00:34:50.300
OK, after a while, the point is, let's try to.
520
00:34:53.700 --> 00:34:56.140
So just so we have something to aim at, let's say.
521
00:34:56.140 --> 00:34:58.500
OK, here we're going to go.
522
00:34:58.500 --> 00:35:00.460
I'm going to get down through there.
523
00:35:00.460 --> 00:35:03.100
And then this is a good, this is our better curve.
524
00:35:03.100 --> 00:35:12.820
OK, so I am a solution.
525
00:35:12.820 --> 00:35:16.420
A particular solution, satisfying this initial condition.
526
00:35:16.420 --> 00:35:20.180
I started here, and that was my initial velocity.
527
00:35:20.180 --> 00:35:22.540
The slope of that thing gave me the initial velocity.
528
00:35:22.540 --> 00:35:33.700
OK, now the interesting question is, the first, in some ways,
529
00:35:33.700 --> 00:35:36.380
the most interesting question, though there'll be others too,
530
00:35:36.380 --> 00:35:40.900
is what is this spacing?
531
00:35:40.900 --> 00:35:43.020
Well, that's a period.
532
00:35:43.020 --> 00:35:47.980
And now the period would be, it's half a period.
533
00:35:47.980 --> 00:35:51.220
I clearly want to think of this as the whole period.
534
00:35:51.220 --> 00:35:57.540
So let's call that, I'm going to call this pi over,
535
00:35:57.540 --> 00:36:02.620
so this spacing here, from there to there.
536
00:36:02.620 --> 00:36:06.060
I'll call that pi divided by omega 1,
537
00:36:06.060 --> 00:36:10.820
because this, from here to here, should be, I hope twice that,
538
00:36:10.820 --> 00:36:13.020
2 pi over omega 1.
539
00:36:13.020 --> 00:36:20.060
Now, my question is, so this, for a solution,
540
00:36:20.060 --> 00:36:26.580
it's, in fact, is going to cross the axis regularly in that way.
541
00:36:26.580 --> 00:36:31.820
My question is, how does this period, so this is going to be
542
00:36:31.820 --> 00:36:37.780
its half period, I'll put period in quotation marks,
543
00:36:37.780 --> 00:36:40.100
because this isn't really a periodic function,
544
00:36:40.100 --> 00:36:43.380
because it's decreasing all the time in amplitude,
545
00:36:43.380 --> 00:36:45.260
but it's trying to be periodic.
546
00:36:45.260 --> 00:36:47.300
At least it's doing something periodically.
547
00:36:47.300 --> 00:36:51.260
It's crossing the axis periodically.
548
00:36:51.260 --> 00:36:54.180
So this is the half period.
549
00:36:54.180 --> 00:36:57.660
2 pi over omega 1 would be its full period.
550
00:36:57.660 --> 00:37:01.420
What I want to know is, how does that half period,
551
00:37:01.420 --> 00:37:12.180
or how does omega 1 is called its pseudo frequency?
552
00:37:12.180 --> 00:37:14.140
This should really be called its pseudo period.
553
00:37:14.140 --> 00:37:19.020
Everything's pseudo, everything's fake here.
554
00:37:19.020 --> 00:37:21.900
The emeba has its fake foot and stuff like that.
555
00:37:21.900 --> 00:37:27.620
OK, so this is its pseudo period, pseudo frequency.
556
00:37:27.620 --> 00:37:32.140
Sudo circular frequency, but that's hopeless.
557
00:37:32.140 --> 00:37:34.660
I guess it should be circular pseudo frequency,
558
00:37:34.660 --> 00:37:38.700
or pseudo, I don't know how you say that.
559
00:37:38.700 --> 00:37:46.260
I don't think pseudo is a word all by itself, not even in 1803.
560
00:37:46.260 --> 00:37:47.260
Circular.
561
00:37:53.700 --> 00:37:55.900
OK, here's my question.
562
00:37:55.900 --> 00:38:04.060
If the damping goes up, this is the damping term.
563
00:38:04.060 --> 00:38:08.860
If the damping goes up, what happens to the pseudo frequency?
564
00:38:14.620 --> 00:38:18.820
The frequency is how often the curve crosses, you know,
565
00:38:18.820 --> 00:38:24.540
how this is high frequency, and this is low frequency.
566
00:38:24.540 --> 00:38:31.020
OK, so my question is, which way does the frequency go?
567
00:38:31.020 --> 00:38:36.580
If the damping goes up, does the frequency go up or down?
568
00:38:36.580 --> 00:38:37.580
Down.
569
00:38:48.100 --> 00:38:50.460
I mean, I'm just asking you to answer intuitively
570
00:38:50.460 --> 00:38:53.140
on the basis of your intuition about how this thing
571
00:38:53.140 --> 00:38:56.260
explains how this thing goes.
572
00:38:56.260 --> 00:39:01.020
And well, now let's get the formula.
573
00:39:01.020 --> 00:39:04.660
What in fact is omega 1?
574
00:39:04.660 --> 00:39:06.140
What is omega 1?
575
00:39:06.140 --> 00:39:15.380
The answer is, when I solve the equation, so r is now,
576
00:39:15.380 --> 00:39:19.620
so in other words, if omega 1 is, sorry, if I have p,
577
00:39:19.620 --> 00:39:23.700
if p is no longer 0 as it was in the umdamm case,
578
00:39:23.700 --> 00:39:25.060
what is the root now?
579
00:39:25.060 --> 00:39:25.460
OK.
580
00:39:25.460 --> 00:39:30.260
Well, the root is minus p plus or minus the square root
581
00:39:30.260 --> 00:39:31.940
of p squared.
582
00:39:31.940 --> 00:39:33.140
Now I'm going to write it this way.
583
00:39:33.140 --> 00:39:37.580
Minus to indicate that it's really a negative number, omega
584
00:39:37.580 --> 00:39:39.580
squared minus p squared.
585
00:39:45.540 --> 00:39:49.340
Now, I'm going to call this because you see when I change
586
00:39:49.340 --> 00:39:54.020
this to signs and cosines, the square root of this number
587
00:39:54.020 --> 00:39:57.300
is what's going to become that new frequency.
588
00:39:57.300 --> 00:40:02.060
I'm going to call that minus p plus or minus the square root
589
00:40:02.060 --> 00:40:04.900
of minus omega 1 squared.
590
00:40:04.900 --> 00:40:08.060
That's going to be the new frequency.
591
00:40:08.060 --> 00:40:15.580
And therefore, the root is going to change
592
00:40:15.580 --> 00:40:21.860
so that the corresponding solution that is going to look out.
593
00:40:21.860 --> 00:40:27.060
Well, it's going to be e to the negative p t times,
594
00:40:27.060 --> 00:40:29.740
let's write it out first in terms of signs and cosines,
595
00:40:29.740 --> 00:40:35.100
times the cosine of, well, the square root of omega 1 squared
596
00:40:35.100 --> 00:40:36.460
is omega 1.
597
00:40:36.460 --> 00:40:41.540
But there's an i out front because of the negative sign
598
00:40:41.540 --> 00:40:42.380
in front of that.
599
00:40:42.380 --> 00:40:45.700
So it's going to be the cosine of omega 1t
600
00:40:45.700 --> 00:40:48.980
plus c2 times the sine of omega 1t.
601
00:40:48.980 --> 00:40:52.020
Or if you prefer to write it out in the other form,
602
00:40:52.020 --> 00:40:55.380
it's e to the minus p t times some amplitude, which
603
00:40:55.380 --> 00:41:00.860
depends on c1 and c2, times the cosine of omega 1t
604
00:41:00.860 --> 00:41:02.700
minus the phase lag.
605
00:41:06.100 --> 00:41:09.460
Now, when I do that, you see omega 1
606
00:41:09.460 --> 00:41:13.620
is this pseudo frequency.
607
00:41:13.620 --> 00:41:17.180
In other words, this number omega 1 is the same one
608
00:41:17.180 --> 00:41:19.900
as I've identified here.
609
00:41:25.180 --> 00:41:26.060
And why is that?
610
00:41:26.060 --> 00:41:29.180
Well, because what are two successive times?
611
00:41:29.180 --> 00:41:34.380
Suppose it crosses, suppose the solution crosses the x-axis
612
00:41:34.380 --> 00:41:39.300
for the first, sorry, y, the t-axis.
613
00:41:39.300 --> 00:41:44.700
For the first time at the point t1, what's the next time
614
00:41:44.700 --> 00:41:48.540
it crosses at t2?
615
00:41:48.540 --> 00:41:51.100
Let's jump to the two times it crosses.
616
00:41:51.100 --> 00:41:55.380
So I want this to be a whole period, not a half period.
617
00:41:59.700 --> 00:42:01.020
What's t2?
618
00:42:01.020 --> 00:42:07.620
Well, I say the t2 is nothing but 2 pi divided by omega 1.
619
00:42:07.620 --> 00:42:12.820
And you can see that because when I plug in,
620
00:42:12.820 --> 00:42:19.580
if it's 0, if I have a point where it's 0,
621
00:42:19.580 --> 00:42:24.700
so omega 1t minus phi, when will it be 0 for the first time?
622
00:42:24.700 --> 00:42:28.580
Well, that will be when the cosine has to be 0,
623
00:42:28.580 --> 00:42:30.900
so it'll be some multiple of a, it'll
624
00:42:30.900 --> 00:42:33.980
be say pi over 2.
625
00:42:33.980 --> 00:42:41.580
Then the next time this happens will be, if that happens at t1,
626
00:42:41.580 --> 00:42:46.940
then the next time it happens will be at t1 plus 2 pi
627
00:42:46.940 --> 00:42:48.740
divided by omega 1.
628
00:42:51.780 --> 00:42:59.900
That will also be pi over 2 plus how much plus 2 pi,
629
00:42:59.900 --> 00:43:02.180
which is the next time the cosine gets around
630
00:43:02.180 --> 00:43:07.220
and is doing its thing, becoming 0 as it goes down,
631
00:43:07.220 --> 00:43:10.020
not as it's coming up again.
632
00:43:10.020 --> 00:43:12.260
In other words, this is what you should add
633
00:43:12.260 --> 00:43:16.660
to the first time to get this second time that the cosine
634
00:43:16.660 --> 00:43:20.380
becomes 0 coming in that, coming in the direction
635
00:43:20.380 --> 00:43:23.700
from top to the bottom.
636
00:43:23.700 --> 00:43:30.700
So this is in fact the frequency with which it's crossing
637
00:43:30.700 --> 00:43:32.700
the axis.
638
00:43:32.700 --> 00:43:43.620
Now, notice running out of boards, what a disaster.
639
00:43:43.620 --> 00:43:46.220
In that expression, take a look at it.
640
00:43:46.220 --> 00:43:48.900
I want to know what depends on what?
641
00:43:48.900 --> 00:43:54.220
So p, in that we got constants.
642
00:43:54.220 --> 00:43:58.660
We got p, we got phi, we got a, what else we got?
643
00:43:58.660 --> 00:44:01.660
Omega 1.
644
00:44:05.660 --> 00:44:08.180
What are these things depend upon?
645
00:44:08.180 --> 00:44:10.380
You got to keep it firmly in mind.
646
00:44:10.380 --> 00:44:12.780
This depends only on the ODE.
647
00:44:17.220 --> 00:44:18.500
It's basically the damping.
648
00:44:18.500 --> 00:44:22.100
It depends on C and M. It's essentially, it's like,
649
00:44:22.100 --> 00:44:25.180
it's C over 2 M actually.
650
00:44:25.180 --> 00:44:26.260
How about phi?
651
00:44:26.260 --> 00:44:29.380
Well, phi, what else depends only on the ODE?
652
00:44:29.380 --> 00:44:31.700
Omega 1 depends only on the ODE.
653
00:44:37.060 --> 00:44:38.860
What's the formula for omega 1?
654
00:44:38.860 --> 00:44:42.980
Omega 1 squared, where do we have it?
655
00:44:42.980 --> 00:44:45.180
Omega 1 squared, I didn't, I've never
656
00:44:45.180 --> 00:44:47.260
wrote the formula for you.
657
00:44:47.260 --> 00:44:52.740
So we have omega naught squared minus p squared equals
658
00:44:52.740 --> 00:44:53.860
omega 1 squared.
659
00:44:53.860 --> 00:44:55.460
What's the relation between them?
660
00:44:55.460 --> 00:44:57.380
That's the Pythagorean theorem.
661
00:44:57.380 --> 00:45:03.940
If this is omega naught, then this omega 1, this is p.
662
00:45:03.940 --> 00:45:08.300
They make a little right triangle, in other words.
663
00:45:08.300 --> 00:45:11.420
The omega 1 depends on the spring.
664
00:45:11.420 --> 00:45:14.220
So it's equal to, well, it's equal to that thing.
665
00:45:14.220 --> 00:45:16.820
So it depends on the damping.
666
00:45:16.820 --> 00:45:18.620
And it depends upon the damping.
667
00:45:18.620 --> 00:45:21.700
And it depends on the spring, the spring constant.
668
00:45:21.700 --> 00:45:25.980
How about the phi and the a?
669
00:45:25.980 --> 00:45:28.380
What do they depend on?
670
00:45:28.380 --> 00:45:30.780
They depend upon the initial conditions.
671
00:45:43.260 --> 00:45:47.740
So this mess of constants, they have different functions.
672
00:45:47.740 --> 00:45:50.580
What's making this complicated is that there are four
673
00:45:50.580 --> 00:45:51.980
parameters.
674
00:45:51.980 --> 00:45:53.260
Our answer has four parameters.
675
00:45:53.260 --> 00:45:56.300
Needs four parameters to describe it.
676
00:45:56.300 --> 00:45:59.900
This tells you how fast it's coming down.
677
00:45:59.900 --> 00:46:02.220
This tells you the phase lag.
678
00:46:02.220 --> 00:46:07.740
This tells you this amplitude modifies the e to the minus.
679
00:46:07.740 --> 00:46:10.540
It tells you whether the exponential curve starts going
680
00:46:10.540 --> 00:46:13.980
like that, or goes like this.
681
00:46:13.980 --> 00:46:17.660
And finally, the omega 1 is this pseudo frequency, which
682
00:46:17.660 --> 00:46:20.820
tells you how it's bobbing up and down.