17. Simple Harmonic Motion
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Fundamentals of Physics I17 / 24
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01:13:58
Computer animation
Transcript: English(auto-generated)
00:00
equilibrium, then if you disturb them, they rock back and forth. And there's two simple examples. The standard textbook example is this mass and spring. This spring has a certain natural length, which I'm showing you here. It likes to sit there,
00:23
but if you pull it so the mass comes here, you move it from x equal to 0 to a new location x. Then, there is a restoring force F, which is minus kx, and that force will be equal to
00:42
mass times acceleration by Newton's Law. And so, you're trying to solve the equation d 2x over dt squared equals minus k over m times x. And we use the symbol ω square equals k over m,
01:05
or ω square root of k over m. And what's the solution to this equation? I think we did a lot of talking and said, look, we are looking for a function which, when differentiated twice, looks like the same function up to some constants. And we know they are trigonometric functions. Then, we finally found the
01:23
answer, the most general answer looks like ωt times a cos ωt minus φ. A is the amplitude, φ is called the phase that tells you what your clock is doing when x is a maximum. And we can always choose φ to be 0. That means when the time is 0, x will have a biggest value
01:42
a. This is an example of simple harmonic motion, but it is a very generic situation. So, I give you a second example. Now, we don't have a mass, but we have a mass. We have a bar, let's say, suspended by a cable hanging from the ceiling. And it's happy the way it is, but if you come and twist it
02:03
by an angle θ, give it a little twist, then it'll try to untwist itself. So now, we don't have a restoring force, but we have a restoring torque. What can be the expression for the restoring torque? When you don't do anything, it doesn't do anything. So, it's a function of
02:21
θ that vanishes when θ is 0. If θ is not 0, it'll begin as some function of θ, but the leading term will be just θ. But now, you can put a constant here. It'll still be proportional to θ. And you put a minus sign for the same reason you put a minus sign here to tell you it's a restoring torque. That means if you make
02:42
θ positive, the torque will try to twist you the other way. If you make θ negative, it'll try to bring you back. So, you have to find this κ. If you find this κ, then you can say minus κ times θ is I times d two θ over dt squared. Mathematically, that equation is identical to that.
03:01
And θ then will look like some constant. You can call it A cos ωt minus φ. And ω now will be the ratio of this κ to the moment of inertia. Because mathematically, that's the role played by κ and I. They're like k and m. Now, if someone tells you this
03:24
is κ, then you're done. You just stick it in and mindlessly calculate all the formulas. You can find θ, you can take derivatives, and so on. Sometimes they may not tell you what κ is. In the case of a spring, they would have to give you k. Or they may tell you indirectly, if I pull the spring by nine inches,
03:43
it exerts a certain force. They're giving you f and they're giving you x, and you can find k. So, that's the way. But in rotation problems, the typical situation is the following. Let us take the easiest problem of a pendulum. If a pendulum is hanging, say, a massless rod with some
04:01
mass m at the bottom, so it's very happy to stay this way. If you leave it like this, it'll stay this way forever. No torque, no motion. Now, you come along and you disturb that by turning it by an angle θ. If you do, then the force is like this. The separation vector R from
04:22
the pivot point is this, and R cross F will be non-zero because the angle between them is not zero. And you remember the torque is equal to Rf sin θ. And I told you for small angles, this can be Rf θ.
04:43
And R happens to be, in this problem, the length of the pendulum. So, it's really Rf θ, minus mgl θ. So, you have to do some work to find θ. It wasn't just given to you in a plate. You did this piece of thinking.
05:00
And you, namely, you disturb the system from equilibrium and found the restoring torque. Then, you stare at the formula and say, hey, this guy must be the κ I'm looking for, because that number times θ is the restoring torque. Then, the ω here will be, this κ has a generic κ that goes into whatever it is for that problem. For our problem,
05:22
κ happens to be mgl. That's something you should understand. It's not a universal number that's known to everybody like the spring constant of anything. In the case of mass and this pendulum, it depends on how long the pendulum is, how big the mass is. But you can work it out and extract κ. Downstairs,
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you need the moment of inertia for a point mass m, which is the distance l from the point of rotation, which is ml squared. So, you can cancel the l, you can cancel the m, and you get ω is the root of g over l. And I remind you guys that
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ω is connected to what you and I would call the frequency by this 2Πf, or what you and I would call the time period by 2Π over T. Okay, so this is the situation. Let me give one other example. This is a homework problem,
06:23
but I want to give you a hint on how to do this. Don't take a pendulum with all the mass concentrated there. Take some irregularly shaped object, a flat, planar object. You drive a nail through it there, hang it on the wall. It'll come to rest in a certain configuration and you
06:43
should think about where will the center of mass be if I hang it like this on the wall. The center of mass is somewhere in the body here. So, I want you to think about it. I'll tell you in a second. The center of mass, I claim, will lie somewhere on this vertical line going through that point.
07:04
I know that by default, because the center of mass is here, for example. We know all the force of gravity can be imagined to be acting here. Then, if you do that separation and do the torque, then you will find this is able to exert a torque around this point. But that cannot be in equilibrium. So, the center of mass will
07:23
always lie, will align itself. The body will swing a little bit and settle down the center of mass somewhere there. If you now disturb the body, I don't want to draw another picture, but I'll probably fail. This is a rotated body, off from its ideal position. Then, there will be a torque. What will the torque be?
07:42
It'll be the same thing, minus mgL sin θ, which I'm going to replace by θ. L is now the distance between the pivot point and the center of mass. That's your tau. So, you can read the κ. In fact, it's just like
08:02
the pendulum. In other words, as far as the torque is concerned, it's as if all the mass were sitting here. The difference will be in moment of inertia. The moment of inertia of this is not mL squared. So, don't make the mistake. You're going to take all the masses not sitting here, all the masses sitting all over the place. So, you should know that if you want the moment of
08:21
inertia, it is I with respect to center of mass, plus this mL squared. That's the old parallel axis theorem. So, take that moment of inertia, put that into this formula here, take this κ and put it there, you'll get some frequency. That's the frequency with which it'll oscillate. So, every problem that you
08:41
will ever get will look like one of these two. Either it's something moving linearly with a coordinate that you can call x, or it's rotating or twisting by an angle you can call θ. And if you want to find out the frequency of vibration, you have to disturb it from equilibrium, either by pulling the mass or by twisting the cable or
09:01
displacing this pendulum from its ideal position here, then finding the restoring torque. Yes? Student response to this question is, if I give you a cable and I twist the cable by some angle, how are we going to calculate
09:23
restoring torque? Here is the good news. This problem is so hard, we will give it to you. In other words, that is of course an underlying answer, given a cable made of some material and its torsional properties, how much of a torque you will get if you twist it by some amount. But it's not something you can
09:42
calculate from first principle, so they will simply have to calculate the torque and give it to you, okay, and such problems. The only time you will have to find the torque on your own is in a problem like this, where I believe you know enough to figure out the torque. Okay? What you will find is, if you leave it alone, it'll go to a position where there is no torque. If you move it off that
10:02
position, there will be a torque, and the torque will always be proportional to the angle by which you displaced it. You read off the proportionality constant, and that's your κ. By the way, this is not a problem. You should look at the forces too, which is pretty interesting. This body, when it's hanging in its rest position,
10:20
has two forces on it. The nail, which is pushing up, and the weight of the body, which is pushing down, and they cancel each other. The nail will keep it from falling. The nail will not keep it from swinging, because the force of the nail acting as it does at the pivot point is unable to exert a torque. Whereas, the minute you
10:40
rotate the body, mg is able to exert a torque. That's why if you twist it, it'll start rattling back and forth. All right, so what I'm going to do now is to go over more complicated oscillations using some of the techniques we learned last time.
11:03
So, here is the mega formula. I'm going to give it to you guys. You have to, you're allowed to tattoo it on your face. You can carry it with you. I will allow you to bring it, but you cannot forget this formula. I think you know the formula I'm driving at. Ready? e to the i x or
11:22
θ, I don't care. Let me call it θ here. It's cos θ plus i sin θ. That's the great formula from Euler. From this formula, if you take the complex conjugate of both sides, that means change every i to
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minus i, you will get e to the minus i θ is cos θ minus i sin θ. That also you should know. If you got this in your head, by the way, this is something you're not going to derive on the spot. It takes a lot of work. So, this is something,
12:02
if you tell me what are the few things you really have to cram that you don't want to carry in your head, well, this is one of them. Okay, this is a formula worth memorizing. Unlike formula number 92 of your text, it's not worth memorizing. This is worth memorizing. Once you got this, you should realize that in any expression involving complex numbers, you can get another
12:23
equation where every i is changed to minus i. That will give you this one. That's called complex conjugation. I generally said if you have a complex number z, z star is equal to x minus iy, if z is equal to x plus iy. So, in our simple example,
12:41
this fellow here is z, this is the x, and this is the y. Now, you should be able to invert this formula. To invert the formula, you add the 2 and divide by 2, then you will find cos θ is e to the i θ plus e to the minus i θ over 2.
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If you subtract and divide by 2i, you will find sin θ is e to the i θ minus e to the minus i θ over 2i. In other words, my claim is that this funny sum of two exponentials is the familiar cosine. It'll do everything your cosine will do. It'll oscillate, sine squared plus cosine
13:22
squared will be 1. It'll satisfy all the trigonometric identities like sine 2θ is 2 sine θ, cos θ. Everything comes out of this expression. So, you should know, here is the main point you should learn. We don't need trigonometric functions anymore. Once you've got the exponential function, provided you let the exponent be complex or
13:44
imaginary, you don't need trigonometric functions. This is one example of grand unification. People always say Maxwell unified this and Einstein tried to unify this. Unification means things that you thought were unrelated are in fact related and there are
14:00
different manifestations of the same thing. When we first discovered trigonometric functions, we were drawing right angle triangles and opposite side and adjacent side. Then, we discovered the exponential function, which by the way was computed by bankers who were trying to calculate compound interest every second. The fact that those functions are related is a marvelous result,
14:20
but it happens only if you go to complex numbers. So, this is another thing you should know. Okay, now armed with this, we are ready to do anything we want. Let me just tell you that every complex number z can be written as its absolute value times some phase. In other words, if my complex number z is here, it is x plus iy.
14:42
That's one way to write it. You can also write it as its absolute value, e to the i φ, rr, e to the i φ. r is the radial length, which also happens to be the length of the complex number. That's called the polar form of the complex number. x plus iy is the Cartesian form of the complex number.
15:04
I may have called them r and θ or r and φ. I just don't remember. I'm going to use these symbols back and forth because people do use both the symbols, which is why I'm sometimes calling this angle as θ and sometimes this angle is φ. But the basic idea is the same. This entity here can be
15:22
written either in this form or in this form. This is something you should know. If you don't know or don't understand, you should stop and ask because everything is built on the same axis. So, what I'm going to do now is to go back to this rather simple equation, d two x over dt squared equal to minus ω
15:45
squared x. By the way, today I'm going to call this ω as ωo. It's the same guy. I'm going to call it ωo because we will find as the hour progresses, there are going to be many ωs in the game. Two more ωs are going to
16:04
tell them apart. As long as there's only one ω in town, you just call it ω. If there are many, you call this ωo to mean it's the frequency of vibration of the system left to its own design. When you pull it and let it go, what's the frequency or angle of frequency? That's what we call ωo. Now, how did we solve this
16:22
equation? The way I tried to solve it for you is to say, turn it into a word problem. I'm looking for a function x of t with a property that two derivatives of the function look like the function, and we raked our brains and we remembered, hey, sines and cosines are the property. One derivative is no good. Turn sine into cosine.
16:42
Two derivatives bring back the function you started with, which is why the answer could be sines or cosines. But now, I'm going to solve it in a different way. My thinking is going to be, I know a function that repeats itself even when I differentiate it once, and I'm going to take the derivative of the function.
17:01
Namely, the function has a property. Its first derivative looks like the function itself. It's obvious that its 92nd derivative will also look like the function because taking the derivative leaves the function alone except for pulling out some numbers. So, why not say, I want an answer that looks like this, x equals some a e to the
17:20
αt. That certainly has the property that if you take two derivatives, it's going to look like e to the αt on the left-hand side and e to the αt on the right-hand side, and then you can cancel them and you got yourself a solution. Do you guys remember why I didn't follow this solution for the oscillator?
17:42
Why was it rejected? Yes? Ah, that's a very good point. So, maybe I will take e to the minus αt. How about that? Will that give me a second derivative which is negative? You didn't fall for that, right, because first time you will get minus α, but second time you'll get
18:02
plus α squared. That's correct. So, this function is no good. Also, it doesn't look like what I want, even without doing much mathematical physics. I know if you pull the spring, it's going to go back and forth. Whereas these functions are exponentially growing or exponentially falling, they just don't do the trick. But now, we'll find that if you're willing to work with
18:22
complex exponentials, this will do the job. So, we're just going to take this guess and put it in the equation and see if I can get an answer that works, okay? This is called an ansatz. Ansatz is something we use all the time. It's a German word. I'm not sure exactly what it means, but we all use it to mean a tentative guess.
18:43
That's what it is. If you're lucky, it'll work. If you're not lucky, that's fine. You move on and try another solution. So, we're going to say, okay, can a solution of this form be found with A and α completely free? Maybe the equation will tell me what A and α should be. So, let's take it,
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put it here. By the way, I'm going to write this equation in a form that's a little easier. x double dot is minus ω zero squared. Each dot is a derivative. It's just more compact that way, rather than to write all of this stuff. So, let's take that guess and put it here, and see what we get.
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I get, when I take two derivatives, you can say, do I want e to the αt or e to the minus αt? You will find it doesn't matter, so I'm going to take e to the αt and put it in. Now, the beauty of the x function now is if you take one derivative, x dot,
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I hope everybody knows the x dot of this is A e to the αt, and x double dot is another e to the αt. So, I'm going to write it here as Aα square e to the αt. That's the beauty of the exponential. The act of differentiation is trivial. You just multiply it by the exponent α. So, this being the result of
20:03
taking two derivatives, let me write it here now. What do I get? So, I'm going to go to x double dot equal to minus ω zero squared x. Into that, I'm going to put e to the αt. So, I'm going to put in the guess x equals A e to the αt.
20:23
And what do I find? I find that it's a, in fact, let me write the equation in a nicer way. x double dot plus ω zero squared x has to vanish. I just brought everything to one side. Now, put e to the αt,
20:41
A e to the αt as your guess. Then, you find it's α square A e to the αt plus ω zero squared A e to the αt has to be zero. So, we know what to do. Let's group a few terms. So, this means I get A
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times α square plus 1 e to the αt has to vanish. If you can make that happen, you've got a solution because you're going to satisfy the equation. No one's going to say, well, you got it by guess work. Well, it turns out solving differential equation is only
21:21
guess work. There's no other way to solve the equation other than to make a guess, stick it in, fiddle with the parameters, see if it'll work. So, we're not done yet because we want this to vanish. How many ways are there to kill this answer? You can say maybe A is zero. If A is zero, you've got what's called a trivial solution. Yes? Ah, yes, thank you.
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Yep, thank you very much. This one is α squared plus ω zero squared. Does everybody follow that? Yes, from here to here it's very clear what to do. So, we cannot kill A.
22:03
Yep, yeah, yeah, yeah. I'm trying to rule out certain options I don't like. The option I don't like to get this to be zero is to say that is zero. No, that's not happening. e to the αt is not going to
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vanish. Maybe it'll vanish for some negative infinity time, but I want this to be true at all times. The equation has to be obeyed at all times. Well, this guy certainly is not zero at all times. So, the only way to fix that is to have α squared plus ω zero squared equal to zero. That's the only solution
22:41
which is not a trivial solution. Well, that's a simple equation, right? It's a quadratic equation. That means α squared is minus ω zero squared, or α is plus or minus i ω zero. So, when we were young and we
23:06
didn't know about complex numbers, we will come to this stage and we will quit and say, look, exponentials won't work, but now we're not afraid of complex exponentials. So, we embrace the solution. So, now I've got two solutions. By the way, what's the value of A? Think about this.
23:21
If this condition is satisfied, you realize A can be whatever you like. A is completely arbitrary. So, what this equation has told me is the following. Yes, there are solutions of this form. For A, you can pick any number you like.
23:40
In fact, real or complex, it doesn't matter. The equation is satisfied, but your α is not an arbitrary number. It can be only one of two solutions. So, we've got a problem now that I look for one answer, I've got two. So, I will write them both. So, let's say there's a
24:00
solution x one of t, which is some number A, e to the i ω zero t. Then, I've got another solution x two of t, which is equal to any other number B, e to the minus i ω zero t. Now, I think you guys can see in your head that if you take
24:29
this and you put it in the equation, it works. And if you take this and put it in the equation, that also works. Because when you take two derivatives, i ω whole squared will be minus ω squared, minus i ω not whole
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squared will also be minus ω not squared, so both will work. Now, we have to ask, how do I decide between that solution and that solution? How do I decide between both? It turns out that we don't have to pick. We can pick both, and I'll tell you what I mean by we can pick both.
25:02
Now, this is a very, very important property. For all of you who are going into economics or engineering or chemistry, this or other disciplines which are mathematical, the property I'm going to mention of this equation is very important. So, please pay attention. This is called a linear equation. A linear equation obeys a
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certain property called superposition. I'll have to tell you what it is. If I give you a differential equation, you know, 96 derivative of x plus 52 times the 37 derivative of x, blah, blah, blah, adds up to 0, what makes it a linear equation is that throughout an
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equation, either you find the function x or its derivatives, but never the squares or derivatives. The cubes at higher powers of x are the derivatives, okay? The function appears to first power, not to second power. For example, if this was the equation you were trying to solve,
26:03
that is not a linear equation. That's called a non-linear equation. If you have a linear equation, that's a very, very profound consequence and I'm going to tell you what it is, and that lies at the heart of so many things we do. So, let me write down two solutions. One of them obeys x1 double dot plus ω
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naught squared x1 equal to 0. The second one obeys x2 double dot plus ω naught squared x2 equal to 0. You don't even have to look at these two solutions. Take a general problem where you have a linear equation, there are two solutions. Add the two equations.
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On the left-hand side, I get the second derivative of x1 plus the second derivative of x2. Remember, that is in fact the second derivative. Now, I go back to the old notation of x1 plus x2. This has to do with the fact that the derivative of the sum is the sum of the derivatives, and the sum of the derivative is the derivative of the sum.
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Yes? I'm coming to that in a minute. I'm first doing a more modest goal of showing that the sum is also a solution. In the second term, I get ω naught squared plus ω naught squared plus
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x1 plus x2 equal to 0. You stare at this equation. Look, this plus this implies what I've written down, simply by adding. But say in words what you found out. What you found out is that if x1 satisfies the equation, x2 satisfies the equation,
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x1 plus x2, let's call it x plus, is x1 plus x2 also satisfies the equation. And the proof is right in front of you. And the key was the derivative of a sum is the sum of the derivatives. Now, I'm going to generalize it more and say, suppose I multiply all of the
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first equation by a, and all of the second equation by b. Well, that's certainly still true, right? You take something equal to zero, multiply both sides by a, it's still going to be zero. Now, imagine adding this to this, add this to this, and what do you get? You will find d squared over dt squared of
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ax1 plus bx2 plus ω0 squared, ax1 plus bx2 is zero. So, here is the punch line. If you give me two solutions to this equation,
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I can manufacture another one by taking the first one times any number I like, plus the second one times any other number I like. So, the story is a lot more detailed, more complicated than it looked. It looked like there are two solutions, actually. There's an infinite number of solutions, because you can pick a and b any way you like. So, linear equations typically
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have an infinite number of solutions, and you build them up by taking a few building blocks. They're like unit vectors i and j. You can take any multiple of this unit vector, any multiple of that unit vector, and this combination will also solve the equation. Just so you don't think this
29:21
is going to happen all the time, let me remind you, you don't have to write this down, but at least follow what I'm saying. Suppose this quantity here was not x1, but the square of x1, and here. Okay, you don't have to write this, because this is not a linear equation. We're not interested in this. But notice, this is nonlinear, and you guys should be able to pick it up. It's nonlinear because I've got the square of the unknown. Now, can you see,
29:43
if I add these two equations, forget a and b, let them both be one, I get x1 squared plus x2 squared. But that is not equal to x1 plus x2, the whole thing, squared. So, even though the derivative, the second derivative of the sum is the sum of the second derivative,
30:02
the sum of the squares is not the square of the sum. So, for a nonlinear equation, you cannot add solutions. For a linear equation, you can combine solutions with arbitrary coefficients. That's the lesson you have learned today. So, our harmonic oscillator equation is a linear equation, so feel free to
30:21
combine them and get the following solution, x of t is equal to A e to the i ω zero t plus B e to the minus i ω zero t. A and B are whatever you like. ω zero is the original root of k over m.
30:46
But if I gave you this solution, will you be happy to take that as a solution for the mass spring system? And if not, what is it you don't like?
31:00
Yes? Anybody have a view? I mean, would you take that as a good solution? Yes. In order to achieve what?
31:20
Well, he's raising a point. Let me repeat his point. A and B, of course, cannot be, in general, they're arbitrary. For this mass and this spring, you can pick any A and B you like. But on a given day, when you pull it by 9 centimeters and release it, the answer has to be chosen so that at t equal to zero, x becomes 9, and the velocity, let's say, was zero when you
31:40
released it, so when you take the derivative of this, the velocity should vanish at t equal to zero. Or maybe it'll have some other velocity. But you can fit initial coordinate, initial velocity, the two numbers, by picking these two numbers. Do you understand that? But that's not enough. I've got another problem. Yes?
32:01
Yes. Yes, that should bother you. The answer is manifestly not real, okay? And we know x is a real function. That is not a mathematical requirement of the equation, but a physical requirement that when you pull a mass by 9
32:20
centimeters and you release it, it's going to oscillate with some real x and not a complex x, so you've got to fix that. So, to say that x is real really means the following. You remember in the complex world of complex numbers, a complex number x plus iy has a complex conjugate x minus iy,
32:41
and the property of real numbers is that when you take that complex conjugate, nothing happens. Because i goes to minus i, if the number was purely real, it satisfies the condition z equal to z star. So, real numbers are their own complex conjugates. In general, a complex number is
33:02
not its own conjugate, but if you want me to draw you a picture, that's where z is, and that's where z star is, but fellows on the x-axis have the property z is the same as z star. z star is a reflection on the x-axis of z, and therefore if the number is real, it's its own reflection. So, I'm going to demand that
33:22
this solution, in addition to satisfying the basic equation, also is real. To do that, I'm going to find the complex conjugate, which is denoted with x star of t. Now, you want to conjugate everything inside. The complex conjugate of A, I'm going to call A star. The complex conjugate of
33:41
e to the i ωt is e to the minus i ωt. Why? Because the i goes to minus i, ω0 and t are real numbers. Nothing happens to them. And B becomes B star, and this becomes e to the plus i ω0t. And I demand that these two fellows are equal.
34:03
If they're both equal, then I take this exponential and demand that those be the same. That means I demand that A be the same as B star, because if A is the same as B star, this will go into this. And then I will demand that B
34:22
is the same as A star. If these conditions are satisfied, x star will become x. So, the trick is to take the function, take its complex conjugate, equate it to itself, and look at the consequence. Now, I want to tell you that if A is equal to B star, then you don't have to worry
34:42
about this as an extra condition, because I am your complex conjugate, you are my complex conjugate. Because the way it works is, if you took A and changed every i to minus i, right, or if you took B, whatever its complex form was, change i to minus i, you get A. But if you do it one more time, you come back to where you are.
35:01
In other words, for any complex number, if you start it and you start it one more time, you come back to where you are. Therefore, if A is equal to B star, take the complex conjugate of the left-hand side, you'll find A star is equal to B star star, which is B. So, you don't need these two conditions, you do need A equal to B star.
35:24
Okay, so we have to put that extra condition now. So, in the real world, for people who want a real answer, you want to write it as A, e to the iω0t plus A star, e to the minus iωt. In other words,
35:41
B is not an independent number. B has to be the complex conjugate of A. Then, all of you can see at a glance that this is now real, because whatever this animal is, this is the complex conjugate of that. It's got all the i's turned into minus i. So, when you add them, all the i's will cancel. The answer will be real. But A is not necessarily real.
36:02
The complex number A, which is some complex number, has a length and it can have a phase, because every complex number can be written in this form. So, if you remember that, then you find x looks like absolute value of A, e to the iω0t
36:22
plus φ plus absolute value of A, e to the minus ω0t plus φ. Let me write it ω0t. Maybe that's too fast for you, so let me repeat what I did. In the place of A,
36:43
write the absolute value times e to the iφ. e to the iφ will combine with e to the iωt and form this exponential. Second term will be just the conjugate of the whole thing. You don't even have to think. Now, what is this function I have? I want you to think about it. Do you recognize this creature
37:04
here as something familiar now? Yes? Do you have any idea? Yeah. Pull out the A.
37:23
You remember this great identity. Where is it here? This one. e to the iθ plus e to the minus iθ is 2 times cosine θ. So, this becomes 2 times absolute value of A cosine ωt
37:41
plus φ. And if you want, you can call 2A as another number, c cosine ω0t plus φ.
38:04
Oh, but for any complex number, you can pull certainly this number out of both. You pull it out and you've got e to the i something plus e to the minus i something. So, this is a long and difficult way to get back the answer. You've got this by fiddling
38:21
around and guessing and arguing physically. But I want to show you that with exponential function, you would have come to this answer anyway. So, your point of view is, you know, we don't need this. We've got enough problems in life. We're doing very well with cosines. Thank you. Why do you bring this exponential on us? Well, now I'm going to give you a problem where you cannot
38:42
talk your way out of this by just turning it into a word problem. The word problem I meant. We asked a word question, find me a function whose second derivative looks like itself. And you can either start with exponentials and differentiate twice, or sines and cosines. So, you don't need exponentials. But look at the following problem.
39:00
This is a problem of a mass, m, force constant k, moving on a surface with friction. The minute you've got friction, you have an extra force. So, you find mx double dot equal to minus kx, which is the old force due to the spring.
39:20
Okay? Friction also exerts a force which has got some coefficient b, but it's multiplied by the velocity. We know that. If you're moving to the right, the force is the left. If you're moving to the left, the force is the right. Frictional force is velocity dependent. So, the equation you want to solve when you've got friction is really mx double dot plus bx dot plus kx equal
39:49
to 0. I'm going to divide everything by m. Take that m, put it here, and put it here. Then, I'm going to rewrite for
40:01
us the equation we want to solve with friction. x double dot plus γx dot plus ω zero squared x equal to zero, where γ is just b over m. This is the equation you want to solve. Can you solve this by your
40:27
usual word problem? It's going to be difficult because you want a function which, when I take two derivatives, looks like the function plus some amount of its own derivative. If you take cos ωt, it won't do it.
40:41
If you take sin ωt, it won't do it. So, we can still solve it because even this equation you want to solve by the same mindless approach, which is to say, let x look like a e to the αt and it'll work. Let's see how it'll work.
41:05
It has to work. You can see why because when I take two derivatives of x, I will get α squared. Let me pull the a out of everything, plus α γ plus ω zero squared e to the αt equal to
41:20
zero. So, what we learn is, yes, there are solutions of this form to this equation. Once again, a can be whatever number you like because if a vanishes, you kill the whole solution.
41:42
e to the αt is not going to vanish at every instant in time, so the only way is for this to vanish. That means α that you put into this guess is not any old number, but the solution to this quadratic equation, α squared plus α γ plus ω zero squared is equal to zero.
42:03
So, we want α squared plus α γ plus ω zero squared equal to zero. So, α must be the root of this equation. And we all know a quadratic equation will have two roots and we will get two solutions and we can add them with any coefficient you like. There will also be a solution.
42:23
So, let me write it down. In other words, I'm going to explicitly solve this quadratic equation. So, go back to the old days and remember that the solution of that equation will be α equals minus γ plus or minus square root of
42:40
γ squared minus 4 ω r squared over 2. I would like to rewrite this as minus γ over 2 plus or minus square root of γ over 2 squared minus ω r squared. And let's call the two roots,
43:04
one with a plus sign and one with a minus sign, as α plus and minus. This is a shorthand for this whole combination. You understand that if I give you the mass and the spring constant, and I give you the coefficient of friction,
43:20
the number B, α plus and α minus are uniquely known. So, you'll get two precise numbers coming out of this whole game. And your answer then to this problem will be x of t is any number A times e to the minus α plus t, with any number B times e to the minus α minus t,
43:41
but α plus and minus are these two numbers. I want you to notice both numbers are falling exponentials. It's very important because now I'm talking about a mass where I pull the spring, there's a lot of friction, and I let it go.
44:02
I don't want an exponentially growing answer. So, I'm going to assume that that makes no sense. The x should eventually vanish. And that will in fact happen. Let's make sure you understand that. Look at these two roots, minus γ over 2, plus and minus. Well, that is this solution, minus γ over 2 squared,
44:23
minus ωo squared. I'm going to assume in this calculation that γ over 2 is bigger than ωo. Let's do that first. Then, you can see that that's a negative number, that's a negative number, the whole thing is negative.
44:41
So, α minus is definitely, let me see how I wrote it. Oh, I'm sorry, I made a mistake here. I think these are called just plus and minus α. α has two roots and I have to write them as they are.
45:02
Okay, I'm sorry. So, please remove this one. I don't want to sign there, because they are the two answers for α. So, I put e to the α plus t. So again, I should change my answer here. I'm really sorry about this one. I should write it as, now I'm going to write it below so you have a second chance. So, here is what it looks like.
45:21
A, e to the minus, γ over 2, plus square root of γ over 2 squared, minus ωo squared t, plus b, e to the minus γ over 2, minus γ over 2 squared, minus ωo
45:44
squared t. Pardon me. I think I did mean this, or did I get it wrong? These are the two values of α.
46:03
So, they both have the first number to be minus γ over 2. And the square root comes to the plus sign for one of them and minus sign for the other one. I probably have a and b mixed up. Let me check that. No, I think even that's okay.
46:25
Yes? I'm willing to be corrected if I got a and b mixed up. I know these are correct. I want to make sure this is what I call b. b has a coefficient α minus, which would be this one. Oh, yes, yes.
46:40
So, what should I do? Call this a, call this b. How is that now? Let's see if that works out. This should be b, e to the α minus t, b to the minus γ over 2, minus that, yes.
47:07
Pardon me. Okay, the point of writing it this way is to show you that both these powers, and I'm going to show you how to do this. This is clearly a positive number, so it's e to the
47:23
minus a negative number. I just wanted to show you that this object, that's all I wanted to emphasize, inside the square brackets is also positive. Because this is γ over 2, you can see this is a number smaller than γ over 2 squared. So, if you take the square root of that number, it will be smaller than γ over 2. So, this cannot overturn the sign of this.
47:42
It will still be a positive. The overall sign will be negative. So, if you draw the picture here, it's a sum of two exponentials, it will just die down after a while. And how do I find a and b? To find a and b, what I have to do is to take extra data.
48:01
One of them may be initial position x of 0 is given to me. If x of 0 is given to me as some number, I take the solution, I put t equal to 0, and have it match this. If I put t equal to 0, I get this exponential. Well, all these exponentials vanish, and I just get a plus b, because e to the 0 is
48:22
1. So, I already have one condition on a and b. Their sum must be the initial position. How about the initial velocity? Go back to this x and take dx dt. What do you get? dx dt is α plus a,
48:41
e to the α plus t, plus α minus b, e to the α minus t. The derivative of my answer is this. Evaluate it at t equal to 0. At t equal to 0, you put t equal to 0 here, you find x dot of 0 equal to α plus a plus α
49:07
minus b. So, here are the two equations you need to find a and b. Solve these two simultaneous equations. In other words, I will tell you the initial position, and I will tell you the initial velocity. You will take then this
49:24
number known, and that number known. This is a linear simultaneous equation for a and b. You will fiddle around and solve for a and b. Yes? Student asks a question. Professor Ramamurti Shankar asks the following question. In this problem, if everything is real,
49:45
see, previously what happened is when I took the complex conjugate of this function, the conjugate turned it into the other function, and the conjugate of that function turned back into this one. That's why a and b were related by complex conjugation. Here, if I take the complex conjugate of x,
50:01
e to the α plus t, if it's real, remains itself. So, a just goes into a star, b goes into b star, and you want the answer to be unchanged. That requires a equal to a star and b equal to b star. Thanks for pointing out that here we do want a equal to a star and b equal to b star.
50:23
So, what I'm telling you is, go take the solution, conjugate it, and equate it to itself. And remember that now this exponential remains real to begin with, so it doesn't go into anything. It remains itself and I compare this exponential before and after its coefficient that went from a to a star,
50:42
they have to be equal, so a is a star and b is b star. Yes? Well, look at these functions. There is no complex numbers here. That's why everything is real. So, what problems have I solved now?
51:03
First, I put γ equal to 0, no friction, and I just re-derive the harmonic oscillator, cos ωt minus φ or plus φ. Then, I took a problem with friction, then I got a solution that was just exponentially falling, because even though it looks
51:21
like α plus, that's a negative number, that's a negative number, in the end both are falling. So, they will, it says you pull the mass and let it go. It'll relax to its initial position and it'll stop. But all of you must know that there's got to be something in between. In the one case, I have a mass that oscillates forever. In the other case,
51:40
I have a mass, when I pull it and let it go, it goes back to equilibrium and reaches zero at infinite time. But we all know that the real situation you run into all the time, when you pull something and let it go, it vibrates and it vibrates less and less and less and less, and then eventually comes to rest. Where is that solution?
52:02
It's got to come out of this thing. Yes? That's correct. So, you've got to go back to the roots I found for α. It's minus γ over 2, plus or minus square root of γ over 2 squared, minus ωo squared. I had taken γ over 2 to be bigger than ωo.
52:22
So, what I did was I turned on friction, but I didn't turn on a small amount of friction. I turned on a rather hefty amount of friction so that the friction term was bigger than the ωo term. But if you imagine the other limit where you have no friction and you turn on a tiny amount of friction, then the tiny γ over 2
52:40
will be smaller than this ωo. Of course, then you've got the square root of a negative number, right? So, you should really write it as, so let's take γ over 2 less than ωo, then you will write this as minus γ over 2, plus or minus i, times ωo squared, minus γ over 2 squared. In other words,
53:01
you should write the stuff inside as minus of this number now, and take the square root of minus 1 and write it as i. So now, what do the solutions look like? They look like x equal to
53:24
a e to the minus γ over 2. Let me write the exponentials if you like. a e to the minus γ over 2, e to the minus γt, plus i ω
53:40
prime t, but I'm going to call this combination as ω prime. That's why we have ωo and ω prime, plus b e to the minus γ over 2, minus i ω prime t. Now, e to the minus
54:02
γ over 2 is common to both the factors. You can pull it out. Imagine pulling this factor out common to the whole expression. Then you just got a to the i ω prime, plus b to the minus i ω prime. That's a very familiar problem. If you want that to be real, you want a to be equal to the complex conjugate of b.
54:22
Then you repeat everything I did before, so I don't want to do that one more time. You will get some number c times e to the minus γ over 2t, times cosine ω prime t, plus you can add a φ there. So, what does this look like?
54:48
What does this graph look like? It looks like cos ωt, but it's got this thing in the front. If γ vanishes, then forget the exponential completely, this is our oscillating mass. If γ is not zero,
55:02
imagine it's one part in 10,000. Then for the first one second, this number will hardly change from one. Meanwhile, this would have oscillated some number of times. But if you wait long enough, this exponential will start coming into play. And the way to think about it is, it's an oscillation whose amplitude itself falls
55:23
with time. And if you draw a picture of that, I think you will not be surprised that if you draw the picture, it will look like this. It's called a damped oscillation. So, that's the method by which you describe the three cases.
55:43
One is no friction, and one is small friction or friction obeying this condition, when you have oscillatory motion whose amplitude is damped with time. Then you have the other case where γ has crossed some threshold, when ωo is smaller than
56:01
γ over 2, then you get two falling exponentials. So, what do you imagine the mass is doing? With no friction, if you pull it, let it go, vibrate forever. That's not very realistic. If you turn a tiny amount of friction, it will do this, which is very ubiquitous. We see it all the time. You pull a swing and you let it go, after a while it comes to rest.
56:22
This tells you it never quite comes to rest, because as long as T is finite, this number is going to be non-zero, but it'll be very small. At some point, you just cannot see it. The third option is over-damped. Over-damped is when γ is bigger than ωo over 2. Then the answer is nothing that oscillates.
56:41
Nothing oscillates. Everything falls exponentially with time. There you should imagine, you pull something, let it go, it comes slowly and stops. When you buy your shocks in your car, the shock absorbers, they're supposed to damp the vibration of the car. It's got a spring, but there's suspending the tires, but it's immersed in some
57:01
viscous medium so that its vibrations are damped. So, when you hit a bump, if your shocks are dead, you vibrate a lot of time and slowly settle down. That's when your shocks are in this regime. When you bought them, they were in this regime. When you bought them, they were doing this. Once you hit a bump and you overshoot from your normal
57:20
position, you come down to zero and you stay there. Okay, so that's the ideal situation for damping. All right, so I'm going to the last topic in oscillation, but is there anything else? Is there anything here that you need clarification?
57:43
Yes? Ah, yes. So, when γ over 2 equals ωo, this thing vanishes and you seem to have only one solution, e to the minus γt over 2. You don't have the plus or minus, right?
58:02
And we sort of know in every problem there must be two solutions, because we should be able to pick the initial position and velocity at will. If a solution is only one free parameter, you cannot pick two numbers at will. Now, that's a piece of mathematics I don't want to do now, but what you can show is in that case, the second solution looks like this.
58:23
It's a new function, not an exponential, but t times an exponential will solve the equation. Those who want to see that this is true can put that into the equation and check. In other words, you can just pick your γ carefully so that the square root vanishes. Only a γ is left,
58:40
but that γ is not independent of ω. That γ is equal to 2ωo. For that problem, you can check that that's a solution, so is that. There are nice ways to motivate that, but I think I don't have time to do that. Yes? Why was A equal to A-star here?
59:02
You mean here? Okay, take this x. So, A and B demand that when you conjugate it, nothing should happen, okay? In this particular example, when α plus and α minus are real numbers, e to the α plus t remains e to the α plus t when you take conjugate. Therefore, A has to remain A-star, and B must also
59:23
remain invariant when you take the conjugation. In other words, if this was an imaginary exponential, like in the other problem, when you take the conjugate, this exponential becomes this guy, and this guy and that exponential becomes that guy, then by matching the solution, you will be able to show this must be the conjugate of that, that must be the
59:41
conjugate of this. But if no imaginary exponentials appear, each term must separately be real. Now, this is something you will have to think about. I'm not saying my repeating it makes it any clearer. So, what I want you to do is take the x, take the complex conjugate of x, equate the two sides.
01:00:00
So, it's like saying that if you've got e to the plus αt, its coefficient should match on both sides, and e to the minus αt should have matching coefficients. It's like saying when two vectors are equal, the coefficient of i should match and the coefficient of j should match. There's a similar theorem that if you take a sum of two independent functions equated to the sum of two other same
01:00:21
independent functions, the coefficient must separately match. If you impose that, you will find what I told you. So, the really interesting problem is this. I got the mass, I got the spring, I got the friction, but I'm going to apply an extra
01:00:41
force, Fℏ cos ωt. This is called a driven oscillator. So far, the oscillator was not driven. In other words, no one is pushing and pulling it. Of course, you pulled it in the beginning and you released it, but once you released it, you took it. No one's touching the oscillator. The only forces on it are due
01:01:01
to internal frictional forces or spring force. But now I want to imagine a case where I am actively driving the oscillator by my hand, exerting a cos ωt force. So, ω here is the frequency of the driving force, okay?
01:01:21
That's why there are so many ωs in this problem. The ω prime that you saw here is not going to appear too many times. It's a matter of convenience to call this ω prime. But this ω will appear all the time. When I write an ω with no subscript, it's the frequency of the driving force. And the equation we want to solve is xℏ plus
01:01:45
γxℏ plus ωℏ squared x, equal to F over m, because I divided everything by m cos ωt. So, we've got to solve this problem. Now, this is really difficult
01:02:05
because you cannot guess the answer to this by word problem. Now, you can do the following. If the right-hand side had been e to the iωt instead of cos ωt, you would be fine because then you can pick an x that
01:02:20
looks like e to the iωt. Then, when you take two derivatives, it'll look like e to the iωt, one derivative will look like e to the iωt, The x itself will look like e to the iωt. You can cancel it on both sides. But what you have is cos ωt. So, here's what people do. It's a very clever trick. People say, let me manufacture a second
01:02:41
problem. Nobody gave me this problem, okay? This is the problem he gave me. I make up a new problem, the answer to which is called y. But y is the answer to the following problem. The driving force is sin ωt. So, this is what you gave me
01:03:05
to solve. This is my artifact. I introduced a new problem you did not give me, but I want to look at this problem. Now, here is a trick. You multiply this equation by
01:03:23
any number, it's still going to satisfy the equation. Let me multiply by i. Put an i here, put an i here, put an i here, put an i here. Multiply both sides by i and add them.
01:03:42
Then, I have got x plus iy double dot plus γ times x dot plus iy dot plus ω naught squared and x plus iy. Let's introduce a number z, which is x plus iy. It varies with time. Then, this equation, by adding the two equations, looks like z double dot plus γ z dot plus
01:04:04
ω naught squared z. this is f over m, e to the i ωt. So, I have manufactured a problem in which the thing
01:04:23
that's vibrating is not a real number. The force driving it is also not a real number. It's the cos ωt plus i sin ωt. But, if I can solve this problem by some trick, At the end, what do I have to do?
01:04:41
I have to take the real part of the answer, because the answer will look like a real part and an imaginary part. I'll have to dump the imaginary part. That will be the answer to the question I posed. The imaginary part will be the answer to the fictitious question I posed. Now, why am I doing this? The reason is the following.
01:05:01
If the driving force is e to the i ωt, I can make the following ansatz at guess. z is some constant times e to the i ωt. I will show you now solutions of this form do exist, because let me take that assumed form and put it into this equation. Then, what do I get?
01:05:22
I get minus ω squared z naught e to the i ωt, because two derivatives of e to the i ωt give me e to the i ω times i ω. Then, one derivative gives me i ω times γ times
01:05:40
z naught e to the i ωt, and no derivative just leaves it alone, e to the i ωt equals F naught over m e to the i ωt. So, let me rewrite this as follows. Let me rewrite this as minus ω squared plus
01:06:04
i ω γ plus ω naught squared times z naught e to the i ωt is F over m e to the i ωt. This is what I want to be true. Well, e to the i ωt,
01:06:20
whatever it is, can be canceled on two sides because it is not zero. Anything that's not zero you can always cancel. And here is the interesting result you learned. For this equation to be valid, the z naught that you pick here in your guess satisfies the condition z naught equals F over m divided by
01:06:47
ω naught squared minus ω squared plus i ω γ. Now, parts of it may be
01:07:04
easy, parts of it may be difficult. The easy part is to take the guess I made and stick it in the equation, cancel exponentials and get the answer. What you should understand is x was what I was looking for, I brought in a partner y and I solved for z, which is x plus iy,
01:07:22
and z was assumed to take this form with the amplitude which itself could be complex times e to the i ωt. If z naught looks like this, then z, which is z naught e to the i ωt, looks like F over m e to the i ωt divided by this
01:07:44
complex number in the denominator I'm going to call i for impedance. i is called impedance and it's the following complex number, ω naught squared minus ω squared plus i ω γ.
01:08:11
Okay, we're almost done now. So, the z looks like
01:08:20
F naught over m e to the i ωt divided by this complex number i. Imagine this complex number i, what does it look like? It's got a real part and an imaginary part. The imaginary part is i ω γ, the real part is ω naught squared minus ω squared.
01:08:45
This is the complex number i. So, you are told the answer to our problem is to find this number z, then take the real part. Does everybody agree that every complex number i can be written as its absolute value times e
01:09:03
to the i φ, which is going to be here. And I'm going to write it that way because now things become a lot simpler if you write it that way. F naught over m divided by absolute value of i, e to the i φ,
01:09:22
e to the i ωt. Now, this e to the i φ, I can delete it here and put it upstairs as e to the i φ. So, let me write it in that form. Then, it's very easy. F naught over m divided by the magnitude of this i times
01:09:44
e to the i ωt minus φ, where φ is this angle. Well, now that I got z, I know how to find x. x is the real part of z. Now, when I look at real
01:10:02
part, all these are real numbers, so I will keep them as e to the i φ, because they are F naught over m divided by i. The real part of this function, I hope you know by now, is cos ωt minus φ. And that's the answer.
01:10:22
The answer to the problem that was originally given to us is the following. You know the magnitude of the applied force, the amplitude of the applied force, you know the mass of the particle. You need to find this absolute value of i and φ. For that, you construct this complex number whose real part is i φ, and the real part of this whose
01:10:40
imaginary part is i ωγ. Then, the absolute value of i is just by Pythagoras' Theorem, ω naught squared minus ω squared squared plus ω squared γ squared. And the phase φ obeys
01:11:01
the condition tan φ is equal to ω γ divided by ω naught squared minus ω squared squared plus ω squared γ squared. I'm sorry, it's just the imaginary part over real part.
01:11:23
So, this is the answer to the problem that was given to us. But there's one subtle point you should notice, which is the following. Where are the free parameters in this problem?
01:11:40
Everything is determined in this problem. φ, absolute i, all these are known, but we know every equation should have two free numbers. You know what they are? Anybody know? Okay, let me tell you what the answer to that question is. I'm saying I can add to this the solution I got earlier
01:12:02
on without the driving force, because in this problem, when I had a right-hand side with a driving force, let me add a zero to that. That's harmless. And by superposition principle, if that force will produce that displacement, zero will produce what?
01:12:22
Well, we have seen all morning that even when the right-hand side is zero, there are those solutions I got for you, e to the γ over 2 cos ω prime t, etc. So, you can always add to this what's called a complementary function, which is the solution of the equation with no driving force, which is what we were studying earlier in the class.
01:12:41
So, this is the answer. But usually, people don't bother with this, because they all have in them e to the minus γt over 2. So, if you wait long enough, this will die out. This is the only thing that will remain. So, at early times, this is not the full answer. You should add to this the
01:13:02
answer when there is no driving force. And together, they form the full answer. And the numbers a and b that you have there will be chosen to match some initial conditions like initial position, initial velocity. Okay, now, time is up.
01:13:20
I'm going to stop. So, I have not been able to finish some parts of this, so I'm trying to see what I can do. I had asked for you to give this problem set to me on what is this today? On Wednesday, right? So, you don't have enough time. I mean, I haven't taught you a problem set. I'm going to give you five minutes more of work. So, I don't want to keep you
01:13:41
back. So, here's what I will do. I will post on the website some notes on just the missing part here. So, you can read it, and you can do one or two problems you may not be able to do without that. Then, I will come on Wednesday, and I will teach that to you again, but you will be able to hand in your problem set.