I will start today with a very interesting phenomenon,
which is known as a beat phenomenon.
Suppose you have two simple harmonic motions with the same
amplitude but with different frequencies.
So, we have X1 is A times cosine omega one T,
and you have X2, which is A times omega 2T.
And, the two are different, same amplitude.
If you sum them, X1 plus X2, you get 2A times
the cosine of half the sum omega one plus omega two divided by
two times T times the cosine of half the difference.
And, when you look at this, you have here a frequency which
is high compared to this frequency.
You can best see that if omega one is very close to omega two.
And, you could simply replace that by omega.
Then, this is simply saying cosine omega T.
and so, if you change from the notation of omega in your head
to frequency in hertz, which is a lot easier to think
of, omega is 2 pi times the frequency, F.
This is in hertz, and this is in radians per
second. Suppose you have an F1,
which is 256 Hz. And, you have an F2,
which is 254 Hz. If you take these numbers now
as a working example, then the period of this
oscillation is 1/256ths of a second, but the period of this
oscillation is only one second because it's the difference
between the two divided by two. And so, what happens is that if
you make a plot of X as a function of time,
I will first sketch this very slow term.
So, this is very slow compared to this one, which is very fast.
I will first make a sketch of the very slow one.
So, this is the slow one. And, I will make also a dotted
line here to guide my hand when I'm going to make the plot.
So, what you're going to see now is that the faster go [MAKES
NOISE], and the slow one goes like this.
And, the net result is this.
This is the fast one. And so, you see right here is
when the slow one is zero. So, the slow one kills the
amplitude. You can think of this as
amplitude modulation that the A is multiplied by that cosine
function. That is just a matter of
definition. We define this as the beat
period when the two are in phase.
And here they are again in phase.
The two are here 180° out of phase.
We define this as the beat period.
For the working example that we have on the blackboard there,
the 256 Hz and the 254 Hz, from here to here would be one
second. That's the time for the slow
one to go 360°. And so the beat period would,
in that case, be half a second.
So, if you want to see in your head what is happening,
then if one oscillation is like this, that when they are in
phase, each one has an amplitude, A.
When they are in phase, you get 2A.
But when they are out of phase, you have to add this 180° out
of phase. Then, the net result is zero.
And, that's the case here. And that's the case there,
and that's the case there. And, this is known in the
literature as a beat phenomenon. And, the frequency of the beat,
so, F beat is then defined as delta F, in our case,
two hertz. Indeed, you see that the beat
period is half a second. A very nice way to demonstrate
this is to make you listen to it.
I have here a tuning fork of 256 Hz.
And, I have another one which is also 256 Hz.
But, I can change this one by putting a little weight on
there. But I can make the frequency of
little lower. I'll put a weight here,
and I will simultaneously make you listen to them.
And if somehow I managed to give them the same amplitude,
your eardrum will reach the signal, which is 256.
And then, there comes a time that you hear nothing.
That means sound plus sound makes silence,
an interesting concept called interference.
You can hear that. So, listen carefully.
That's the beat. If I make the frequency
different larger, the beats come faster.
Can all of you hear this, also in the back?
So now I will make to offset larger.
It's unpleasant, makes you feel like throwing up
or so, very high beat frequency.
If the amplitude of the two are not the same,
then you don't have to be a rocket scientist to anticipate
what will happen. You just never get the zero
here. But you always have some
residual. And so, you will get something
that looks like this. You still have an oscillation
here. And, I can also make you see
that. It's harder to make you hear
that because your ears cannot really tell the difference
between a little bit of sound and no sound.
You hear [SINGS] and that's all you hear.
But I can make you see that. And I'm going to make you see
that right there when we pick up the signals of these two tuning
forks, there's a microphone. And, we will just play it there
with an oscilloscope. So, you will hear it and you'll
see it at the same time. And, it is unlikely when I
strike them that the amplitude has received by the microphone
will be exactly the same. So, it's not likely you will
see it go through zero. But I can, of course,
change the distance through the microphone and try to get it as
close as I can to zero.
Not bad. Now it's close to zero.
I'll do it once more.
Now, I purposely make amplitudes different.
I bring one very close to the microphone.
This is what you saw there on the board.
If I make amplitudes the same, then you get --
You get the zero back.
Beat phenomenon are actually quite common.
When you see an airplane on the runway and the airplane has two
engines, perhaps you have noticed something irritating at
times that is some crazy humming sound that goes up in volume and
then it goes down and then goes up.
That is almost certainly the two engines beating against each
other. They're not rotating at exactly
the same frequency, but slightly off.
That's my explanation why you have that strangely rising
volume and then decaying volume. It's very common.
When I was a student, I remember there were two
[faucets?] in my bedroom. And, no matter how much I
tightened the faucets, they were leaking.
They were dripping. And, they were dripping in such
a way that the frequency with which the drops came out was
almost the same. So, it went like this:
clunk, clunk, clunk.
But, the frequency was never exactly the same.
So, after awhile it would go clunk, clunk,
clunk, clunk,
and then after a while clunk, clunk.
It drove me crazy. But, of course,
I take pride in knowing that it's a big phenomenon,
except that there, sound plus sound never made
silence because when they were 180° out of phase you hear both
drops. You're kidding,
but that's the way it is. Another example of [bit?]
phenomenon whereby two oscillators very closely the
same period, give it enough time, they go out of phase and
then they go back into phase. We have made an effort here
with these two pendulums, really an effort,
about 60 cm length to really make these lengths the same.
And so, when I give them both a certain amplitude,
they will go in unison. They are in phase,
but no matter how hard you try, they will never be exactly the
same. So, as I continue the lecture,
you'll see that they gradually go out of phase.
And, it may take five minutes for them to be out of phase.
That would be, then, the time from here to
here. That would be half a [bit?]
period. So, I'll show you that,
and I'm not going to wait for them to be 180° out of phase
because I can take a long time since the periods are so closely
the same. If you looked at them now,
you'd say, oh boy, are they in unison!
Keep an eye on it. And if you see at minutes from
now, they will no longer be exactly in phase.
When you think of a conductor for an orchestra,
with the conductor is really doing: he is making sure that
all the instruments in phase. He's really a phase locker;
he's a phase keeper. If he didn't do that,
then each person would play at his own pace.
Then, of course, you would end up with chaos.
So, now perhaps you know what the expression comes from:
keep the phase. Now, I go to the heart of this
lecture. And, the heart of this lecture
is damping. You see it's already out of
phase a little? Not completely out of phase,
but they no longer, just keep an eye on them.
You'll see that they grow. At the heart of this lecture is
damping. When we have a string or we
have a pendulum and we let it oscillate, we all note that the
oscillation will come to a halt. That means there is friction or
there is air drag, and it will ultimately die out
or it will stop just like these will stop.
The frictional force always opposes velocity.
And the frictional force in its most general form as a vector is
minus C1 times V, whereby C is a positive number.
So you see that it opposes velocity, minus C2 times V
squared. And, I put here the unit
vector, V, to remind you that it's opposing the velocity.
This term has a name. This is called the viscous
term. And the other term has a name
which is called the pressure term.
One is proportional with the velocity, and the other one is
proportional with the square of that.
In my lecture number 12, which is on OCW which is of my
1999 Newtonian mechanics lectures, I spent an entire
lecture on this equation. And I discussed when the
viscous term is dominant and when the square term is
dominant. I do demonstrations in both
domains. If the velocity is low,
this term dominates. If the velocity is high,
if the speed is high, that term dominates.
When you drive your car 10 to 20 miles an hour,
there is the V squared term that dominates the air drag.
It slows you down. A raindrop that falls,
it's the V squared term that dominates, not this one.
But, if you take ball bearings and you drop them in syrup,
then the velocity is always very low.
The speed is very low, and that this term dominates.
And that's one of the experiments I do during this
lecture to show you that, indeed, the predictions that
follow from this part can really be verified quite beautifully.
So, if you have the patience, I would strongly advise you to
watch that lecture. To manage the mass,
it becomes almost impossible, except where you want to do it
numerically, to take both terms into account.
To be nice to you and 8.03, we always ignore this term,
and we only take that one into account.
And, of course, that is only valid if the
velocities are low, which is often the case,
but of course not always. I will start with an extremely
pedestrian example. I have here a spring on a
horizontal surface. And as it moves there is
friction. And we assume that the friction
has this form. The spring constant is K.
The mass is M. And if I move it the distance X
away from equilibrium, then there is a spring force
that drives it back to equilibrium.
In addition, there is this frictional force,
and I have no idea in which direction it is because if the
object moves in this direction, but personal force is like so.
If the object at this moment moves in this direction,
then the frictional force is so.
So, I cannot put it in there. I have to know the velocity in
order to put in the frictional force, so I won't do that.
But what I can do, I can write down Newton's
second Law, which says that MX double dot is minus KX.
That is this force when X is positive.
The force is in the negative direction, so the minus sign
takes into account that we are dealing with vectors.
And then we have minus this term.
Now, when we deal with it in 8.03, we will call this C1 just
B. That's a tradition.
We won't call it C1. We call it B.
[NOISE OBSCURES] times X dot. X dot is a factor,
is a velocity. And if the velocity is
positive, then the force is in this destruction.
If the velocity is negative, then the frictional force is in
that direction. So, that now is the
differential equation that we have to solve.
I will introduce some shorthand notations that K over M is omega
zero squared. Remember, that is the angular
frequency of the spring oscillating in the absence of
any friction. We've seen that last time.
And, I will also introduce the symbol B over M equals gamma.
That's also a tradition. Gamma is the same units as
omega, seconds to the power minus one.
And, so if I rewrite this differential equation,
then I get X double dot plus gamma times X dot plus omega
zero squared times X equals zero.
And you see, now our task ahead of us,
to solve this differential equation, what is X as a
function of time? Without starting the math,
you can sort of in your head see what's going to happen.
An oscillation that goes on forever, and ever,
and ever is going to come to a halt.
So, instead of a nice cosine that continues,
the cosine will gradually decay away.
And then, this thing will stop. I take the spring,
I offset it, starts to oscillate,
and it comes to a halt. So, the amplitude must die out,
must decay. That's a prediction.
Now, let me try your intuition. Let us assume that the system
will oscillate with a unique frequency omega.
Do you think that this omega will be larger than omega zero,
smaller than omega zero, or equal to omega zero?
Or to put it in the other way, will a whole oscillation,
under the influence of friction, take longer or shorter
with the same amount of time than in the absence of friction?
So, who thinks that if I had damping that the whole period of
a whole oscillation will take longer?
Who thinks it will take shorter?
Who thinks it will be the same? OK, we'll see what comes out of
this. No one thinks that it takes
shorter? Your intuition is very good
there. Now I'm going to give you some
advice, some fatherly advice. During the next 18 minutes,
take no notes because during the next 18 minutes I'm going to
solve this differential equation for you, and I have decided
which I do not always do, to do it exactly the way that
[French?] does it: very elegant.
So, if you take notes, you're wasting your time
because you have in your book. And I would much rather prefer
that you follow each step that I make so you have more time to
concentrate on the methods. So, don't write anything down.
That is my recommendation. 18 minutes: you can time me if
you don't believe me. We're going to jump through the
complex plane. This is a beautiful example
where the complex solutions really come in handy.
And so, I'm going to change this equation to Z,
which is now a complex notation, Z double dot plus
gamma, Z dot plus omega zero squared times Z equals zero.
And then Z, my trial function, is some amplitude A times E to
the power JPT, plus alpha.
Now, clearly this has the smell of an amplitude.
And, this here has the smell of a frequency.
Anything E to the power JPT has the smell of the frequency.
But you'll see there you are in for a surprise.
What happens is this P. Once we have found the proper
solution to Z, we take the real part,
and we're in business. Then we come back to X.
I'm going to take the second derivative of this equation.
So, JP comes out twice so that this J squared times P squared,
and so I'm going to see now minus P squared,
that is the second derivative of this function.
Then the Z comes later. Then I have plus gamma.
And then, what pops out is JP because it's only the first
derivative: JP. You have to be very careful
that you can distinguish your J's from your gammas.
And then you get plus omega zero squared and the whole thing
times Z. And that now must be equal to
zero. Well, if you have an equation
like this, you may think it's one equation,
but it's really two equations because if something like this
has to be zero, and of course Z is zero is not
an acceptable solution because that's the oscillation,
if you have an equation like this which has a real part and
it has an imaginary part, and both must independently be
zero for obvious reasons. These are apples and these are
oranges, and five apples minus five oranges is never zero
oranges, is never zero apples. You cannot subtract apples from
coconuts. So, this has to be zero and
this has to be to be zero. Now, if you made gamma zero,
you'd have no problem. There's no damping.
So, that's ridiculous. If you make P zero,
and you no longer have any oscillation going on because you
smell that PT, that's going to be the
oscillatory thing. So, that's also unacceptable.
And so, this won't work. Therefore, P itself must be
complex. And so, we're going one step
further now and we're going to say the only way that we could
make this work is if we say P equals N plus JS whereby N and S
then, I hope, are real.
So, let's first calculate what P squared is so we go slowly.
So, we don't get into trouble with the algebra.
So, P squared is N squared plus this one squared.
The J squared is minus one. So, that is minus S squared
plus 2N times J times S. So, now I can go back to this
forum and have minus P squared. So, I get minus N squared.
That is this minus P squared plus S squared minus 2N J times
S. So, I'm done here.
Now, I get gamma JP plus gamma times J.
And, now I have to multiply that by P which is N plus gamma
times J times JS. And then, I have plus omega
squared, and this now has to be zero.
I have here J squared. J squared is minus one.
So, I can erase this and replace it by gamma S.
You cannot do that. That's the price you pay if you
didn't take my advice not to take any notes.
So, this is minus gamma S. And now, this has to be zero.
That means the apples must add up to zero.
These are the apples, and the oranges have to be
zero. This is an orange;
this is an orange; this is an orange;
and this is an orange. For this to be zero,
there is a J here. There's a J here.
There's an N here. There's an N here.
2S must be gamma. So, gamma over two is S.
How? We found S.
For this to be zero, we find that N squared minus N
squared plus gamma squared over four because we know what S is
now, so, plus gamma squared over four minus gamma over two times
minus gamma S, which is minus gamma squared
over two minus gamma squared over two plus omega zero squared
equals zero. And, that means that N squared
is omega zero squared minus gamma squared over four.
So, we also have N. So, we now have N and we have
S. And so, now I can go back to my
trial function and I can substitute the values that we
have found in that complex function.
So, we get Z equals A times E to the power J.
And, now we have to put in BT plus alpha.
But, B is N plus JS. So, NT plus JST plus alpha,
yeah, I go slowly. I don't skip too many steps.
But, I know that S is gamma over two.
But, I know that J squared is minus one.
So, therefore, this part here and this part
here can be brought out. So, we get A times E to the
minus gamma over two times T. Here, we have the T.
The J squared makes it a minus, and S is gamma over two.
And then, I get E to the power J times NT plus alpha.
The first thing that you see here, which is amazing,
this term here is an exponential decay in time.
There is nothing oscillatory about it.
It means that this A, which is the original
amplitude, is gradually dying out with a decay time,
[ta?], which is two divided by gamma.
Now remember, gamma has unit second minus
one. So this really has units time.
So, this one is going to be responsible for that decay that
we discussed earlier. What is this now?
E to the power JNT. N is obviously a frequency.
This is a striking example in a complex plane of a rotating
vector, and angular frequency is N.
But, we know what N is. And so, this N really is
nothing but omega squared. N squared is really omega
squared. And so, I can replace this N
now by whatever you have here. And so, if I write this now AE
minus gamma over two, I can write down now J omega T
plus alpha. And I know exactly now what
that omega is. That is the square root of this
number. And so, therefore,
omega squared is omega zero minus gamma squared over four.
So, those of you in the audience who intuitively sense
that the frequency would go down because of the friction,
they were right. You see, omega is lower than
omega zero. Now, of course,
if gamma is very low, the two could be close
together. It depends on gamma.
That's obvious. If you have a lot of friction,
then omega will be a lot less than omega zero.
If you have very little friction, well,
then they will be very closely the same.
And so, we can now write down in all its glory a real part of
this function which then becomes X equals some amplitude,
A, times E to the minus gamma over two times T times E to the
minus J times omega T plus alpha.
Uh-oh, look what I did. I have an equation in the
complex plane. That's perfectly OK.
And then I went back to the real world going from Z to
complex plane to the real part of Z.
And, what did I write down? This is fine.
But then I wrote down E to the power minus J omega T plus
alpha. First of all,
there should not be a minus sign in front of the J.
That's not a major problem. But, I do not take it out of
the complex plane. I leave it in the complex
plane. So, clearly what I should have
written, which is what I will do now, X equals EA times E to the
minus gamma over two times T times the cosine of omega T plus
alpha. Now, we are in the real world.
You're going to see this equation at least for the next
five or 10 minutes of the tape. It is a thorn in my side,
but there's nothing I can do about it.
You just have to live with it. And, think of it as being this.
And, there are two adjustable constants which depend entirely
on the initial condition. At time T equals zero,
there are two things you can do to the object.
You can bring it to a certain point away from equilibrium and
you can give it a kick. We call that a philosophy.
And you're free to choose any way you want to do that.
And so, it's clear that with those two choices that you have
that in your final solution you end up with two adjustable
constants which depend on the initial condition.
And so, you see that this amplitude is going to die out
with the one over EDK time of two over gamma seconds,
and that the frequency is lower than the frequency of the
un-damped system. I try to make a plot.
So here is T and here is X. And to guide my hand I'm going
to first put in this exponential.
And now, I'm going to put in that oscillatory term here
whereby the frequency is uniquely determined.
Omega is uniquely determined, and the period,
T, is that omega divided by 2 pi.
Oops. How could I?
Look what I wrote. I wrote T equals omega divided
by 2 pi. It was not my day.
Clearly it should have been T equals 2 pi divided by omega.
Now, unfortunately for me, very shortly afterwards,
i.e. erase this from the blackboard
so you won't see it for very long.
In any case, this is what it should be.
And therefore, I'm going to put in here zero
crossings to guide myself so that I don't make the mistake
that you often see that people think that this period is slowly
getting shorter in time. That is not true.
It is amplitude that decays away.
And so, if we now try to put in the oscillation,
then this is what's going to happen.
And so, you see that the oscillation dies away in time,
but the period, T, is uniquely determined.
And, that depends on how much friction there is.
It is not uncommon to introduce a quality factor,
Q, that is high if there is little damping,
and that is low if there is a lot of damping.
And, that Q, which is dimensionless,
is omega zero divided by gamma. You see, immediately if gamma
is high, then Q is low. It's a low-quality oscillator.
If you introduce that, you can go back to your omega
squared equation. And that omega squared becomes
omega zero squared times one minus one divided by 4Q squared.
Is this any different from that?
No, so it's a different way writing it because you introduce
this. What it tells you is that if Q
is about ten, and I bet you the Q's are much
higher for these pendulums, then you have here one divided
by 400 that is one quarter of a percent.
But, since omega is the square root of that,
it's only 1/8 of a percent. So, for Q of ten,
omega is only 1/8 of a percent lower than omega zero.
Even if you make the Q as low as two, the frequencies are only
off by about 3.2%. So I want you to appreciate
that most of the time, but not always is omega very
close to omega zero. We can look at the decay in a
different way but we cannot do that here.
I will do that here on the centerboard because we don't
need that anymore. Instead of saying I have to
wait two over gamma seconds for the amplitude to go down by a
factor of E, I can ask myself, how many oscillations do I have
to wait for the amplitude to go down by a factor,
E? How many oscillations?
Well, N oscillations would take this long, T being the period of
one oscillation. But the reasonable values of Q,
T and T zero are the same like omega and omega zero are closely
the same. So, I can write for this that
this is approximately N times T zero.
So, that's approximately N times two pi divided by omega
zero. I can now substitute this time
in this T, and only concentrate on that decay portion,
that early part. So, what I find,
then, that A after N oscillations is A times E to the
minus gamma over 2T, E to the minus gamma over two
times this time after N oscillations,
N divided by 2 pi divided by omega zero.
And, you lose a two. But, omega zero divided by
gamma is Q. And so, now you have it in a
forum which is minus N times pi over Q.
Now, is this any different from what we had before?
No. Here we say I have to wait two
over gamma seconds for the amplitude to go down by a factor
of E. Here we say if N is Q divided
by pi, then the amplitude goes down by one over E.
So, in one case I ask myself how many seconds do I have to
wait? In the other case I ask myself,
how many oscillations to have to wait?
And so, if Q is ten, it tells you that you have to
wait about three oscillations, roughly, for the amplitude to
go down by a factor of E. And, if Q is 100,
you have to wait more like 32 oscillations for the amplitude
to go down by a factor of E. 18 minutes are up.
So, now you can start taking notes again.
I have here two pendulums. And the pendulums have about
the same length. And the objects have about the
same radius. That means the B,
which is this coefficient in front of the velocity is about
the same. But gamma is not the same
because if they have the same B, but if the mass of the two
objects is very different, it's Styrofoam and this is a
billiard ball, there is a huge difference in
gamma. And, since the period of the
two pendulums is very closely the same, they have the same
length. You see that the Q of the two
systems must be very different because if B is the same,
N is very much higher than the billiard ball.
Then the gamma is much lower. And therefore,
the Q of this system is way higher than the Q of this
system. In fact, if I wanted to wait
how many oscillations it would take for that amplitude to go
down by a factor of E, I may have to wait five or 10
minutes. So, I will not attempt that.
But, I will attempt that to measure Q with this pendulum,
if I bring this here, the separation from equilibrium
is about 27 cm. And, by the time that it has
decayed to this, it's 10 cm.
So, that's about a factor of E. And, I want the students in the
audience who are sitting here who can really see it head-on,
I wanted to say stop. Scream the word stop when the
27 has decayed to 10. And, in the meantime,
we count the number of oscillations.
So, we have then counted the number of oscillations.
And therefore, we know what Q is because we
multiplied the number of oscillations by pi.
So, we have measured Q. We could also have done it
through a time measurement, but I would prefer to do it
this way. Now, you guys over there keep
your mouth shut because there's no way you can see it.
Right? You just can't.
You have a projection effect which is awful.
So, I only want to hear from you.
Are you ready for that? Count and say stop.
Not all of you may say stop at the same moment.
Some of you may say stop after ten oscillations.
Other may say stop after 11 oscillations.
That's fine. That is part of the uncertainty
in the measurement. Ready?
27. There we go.
One, two, you see the decay already.
Three, four, five, six, seven,
eight, nine, ten.
My goodness, are you guys crazy?
You're paying a lot of tuition and you can even pay attention
to demonstration. At nine I think it was already
at ten. When it is here,
then it is ten. I'll give you a second chance.
You have to scream stop.
One, two, three, four, five, six,
seven, eight, nine, ten, 11,
12, OK. Yeah, it is somewhere around
ten or 11. And I don't know why you didn't
scream stop, but that's your problem.
So, N is about 11 maybe plus or minus one, 10% uncertainty.
And so, Q then is about five times higher.
It's a crude measurement, but very roughly you would get,
then, times 11. You get about 35.
That means the frequency damped is almost identical to the
frequency un-damped because, remember, this equation,
the one over four, I erase that perhaps but it is,
you have it up there. If Q is 35, you can see how
close omega is to omega zero. On problem set two,
your very first task is to do a take-home experiment with
something similar to what I did today.
Make sure you pick up a kit today.
You can do that between 11:00 and 1:00 and 3:00 to 5:00 in
room 4335, or you can do it tomorrow between 2:00 and 5:00.
You share one kit, so you choose a partner.
And you can do all these experiments together with your
partner. I want to see in your solutions
uncertainties. Any timing measurements that
you have to do has to come with an estimate of your uncertainty.
And all the conclusions they draw must carry on these
uncertainties just as we did last time when we were exploring
the possibility whether the equation of the spring was
indeed accurate. We were only able to come to
the conclusion that it was not accurate because we had our
proper uncertainties in there. So, I want to see uncertainties
in there. Now comes the mini quiz.
Five minute break: it's a bit early.
But that's the best point today to do.
So, if some of you can help me hand it out, and so after five
minutes bring it back to me. It shouldn't take you more than
one minute. That leaves you still with four
minutes to stretch your legs. [SOUND OFF/THEN ON] I will now
take you back to the good old days of 8.02.
And, I will take you back to an RLC circuit [where?] I have a
battery. This is the plus side and this
is the minus side. Here I have a resistor,
R. I have here a pure self
inductor, L. And here I have a capacitor,
C. And then, I have a switch here
and I can throw the switch. And then the battery is going
to charge up the capacitor. And then you'll get
oscillations. This is a wonderful example of
damped oscillations. The way that I solve these
problems is a strict discipline. I assume that when I started
that there is a current going in this direction,
I. That current will make this
side of the capacitor more positive than that side.
So, I equals DQ DT, Q being the charge on this side
of the capacitor. And now I have to do the closed
loop integral of E dot DL. [UNINTELLIGIBLE] does not hold
the closed loop integral is not zero because we have a magnetic
flux going through a surface attached to this closed loop.
So, the closed loop integral of E dot DL is minus D phi DT,
phi being that magnetic flux going through that surface.
And so, I want to know what the E vectors are in each one of
these components is the curl is in this direction.
Then the electric field in the resistor is in this direction.
The self inductor is made of superconducting wire.
So, there is never any E field inside a self inductor.
Unlike what many EE people think, there is never any
electric field inside a self inductor.
So, the EE is zero. This plus and this is minus.
So, the potential, electric field is in this
direction. This is plus and this is minus.
And so, the electric field is opposing me if I go around
clockwise. So, if now I go around
clockwise and I call the integral E dot DL from this side
to this side of the capacitor, if I call that V of C,
which is Q divided by C. That's the definition of
capacitor. If you're ready for this,
and I start here, and I go clockwise around,
then from here to here, I get plus IR.
Going through the self inductor, the integral E dot DL
from here to here is zero because there is no electric
field inside the self inductor. Going from here to here I get
plus VC. That is what we call the
potential difference over the capacitor.
Here, the electric field is opposing me.
I walk into the electric field, so E dot DL is negative.
And, this is minus V zero. And now, I go to Mr.
Maxwell [UNINTELLIGIBLE] and he says that this now is minus L DI
DT. [The hell with Kishov?].
Minus L DI DT, and now I have my equation
correct. So, now I'm going to replace I
by DQ DT. And now I'm bringing L to the
left. So, I get L times Q double dot
plus R times Q dot because I is Q dot plus VC,
which is Q divided by C. And that now equals V zero.
And, I'm going to divide by L. And so I get Q double dot plus
R over L times Q dot plus Q over LC equals V zero divided by L.
I'm going to replace R over L by gamma.
You will see very shortly why we do that.
And, one over LC is omega zero squared.
And, when we do that, we get an equation which is
almost identical to the one we had on the blackboard for the
spring. We get Q double dot plus gamma
times Q dot plus omega zero squared times Q equals V zero
divided by L. And, now we have to solve this
differential equation. You recognize that gamma is the
damping. R over L is the same function
as the damping had in the case of the spring.
In the case of the spring, it was B over M.
Here, it is R over L. The larger R is,
the more heat dissipation there is.
Heat dissipation goes in terms of I squared R,
and clearly that means you take energy out of the system.
So, that means there is [tempi?].
And this is, then, the natural frequency of
the RLC circuit if there is no R.
If there is only an L and a C, this is the square of the
frequency. If only this were zero,
then I know the solution because we had it on the board
there. We still have the complex
notation there. Then, Q would be some Q1,
for which we have an A there, times E to the minus gamma over
two times T, gamma is this value, times the cosine of omega
T plus alpha. [And/N?] omega squared is omega
zero squared minus gamma squared over four.
That would be the solution if this were zero.
No, it's not a zero. So, you can take 18.03 and try
to solve if it's not zero. Or, you could think like a
physicist and say, I don't need 18.03.
[LAUGHTER] What is the difference between
this differential equation in the one before is the spring
that we had a zero here. In the case of the zero,
it means that the oscillation would end up at X equals zero
whereas here the oscillation in the charge will end up if we
wait long enough with a fully charged capacitor because when
the oscillation has died out, the capacitor has been fully
charged. And so, clearly,
if you add here, this charge on that capacitor,
then you must be OK. That must take into account
that it is not zero. And, this Q max is simply zero
times C, which means the capacitor is fully charged.
So, do it your 18.03 way, or do it with brains and then
you will immediately agree to this solution to the
differential equation. So, when you see here the
decay, you see the oscillation. And then where you add up,
you must add up with a fully charged capacitor.
You do have to solve depending upon the initial conditions for
alpha. And you have to solve for Q1.
So, that means you have to know your initial conditions.
And, the initial conditions of this problem,
for instance, would be that at T equals zero
when I throw the switch that there's no charge on the
capacitor. There is one way I could do it.
And there is no current flowing.
That could be my initial condition, not the only one
possibly, but that is one possible initial condition.
And now, I will leave you with a little bit of work.
First of all, you substitute in this
equation. Q equal zero when T is zero.
So, you still have Q1, and you still have the cosine
of all. Then you have to take the
derivative of this equation which gives you the current.
And then you have to make that zero.
Now, you've got two terms where you do the derivative.
So, be careful. You get the derivative of this
one times this, and the derivative of this one
times this. So, you have to do it slowly.
When you do that, you'll find that Q1 equals
minus Q max divided by the cosine of alpha.
And you will find that the tangent of alpha equals minus
gamma over two omega. I don't have a very good
feeling for these numbers, but what I do have a good
feeling for it is that I know when Q, say, is larger or equal
to five. Then I know that omega is very
close to omega zero, extremely close.
So, that means I can write for this minus gamma divided by two
omega zero, which is minus one over 2Q.
And, if Q is five or larger, my goodness,
this is one tenth. If we forget the minus sign,
now, so that means that phi is only something like 5.7° with a
minus sign. And, the cosine of 5.7° is one
for a physicist, at least.
So, therefore, for all systems whereby Q is
not absurdly low, you can say that Q1 equals
minus Q max. And I'll put here wiggle,
but it's an extremely good approximation.
And so, therefore, my solution here can now be
changed by simply replacing the Q1 minus Q max.
And truly it should be approximately,
but the approximation is extremely close.
It depends on Q. If now you want to plot this,
you get, of course, something completely similar to
what you have here except that you do not end up at zero.
The Q is not zero, but it's offset by this Q max.
So, if this is T, if this is zero,
and if this is Q max, which is where ultimately
charge will be if you wait long enough, that to plot now this
equation will then give me a curve like this.
Be very careful. The period is uniquely
determined. The period does not change.
That's determined by that omega.
2 pi divided by this omega is the periods, and so you start
with Q zero.
And you see, if you waited long enough,
you end up with Q max. It is this one,
whereas in the case of the spring, if you wait long enough,
you end up with nothing, X equals zero.
And, this is something that I can demonstrate to you in a
very, very nice way I must say. I find this one of the most
intriguing demonstrations. Instead of having a battery,
and throw the battery, in which case you would only,
for an extremely short amount of time, for an amount of time
which is roughly [UNINTELLIGIBLE] divided by
gamma seconds, which may only be milliseconds,
you'd see this thing. But, we want to give you more
than a few milliseconds. And therefore,
what we do: we replace that battery by 7 ms constant
voltage, and we repeat that. And, every time,
we show you this portion. And so, you will see it on the
Tektronix. We trigger it every time that
it comes up here and that it comes up here.
And you will see them. This will be staring you in the
face. I'm going to give you the
values that I will be using so that you can digest at home
exactly what you're going to see, and what all the constants
are in this problem. The R is going to be 50 ohms.
The L is going to be 50 MH, ten to the minus three.
So, that tells me that gamma is 1,000, is R over L.
That tells me that two divided by gamma, which is the one over
EDK time, is then two milliseconds.
C is going to be 0.06 µF. That would give you omega zero
if you're interested in omega zero.
I never had a good feeling for omegas.
I like frequencies, hertz, much better.
But if you're interested in omega zero, which is one over
the square root of LC, that would be 18.3 times 10 to
the third. And, that is radians per
second. The frequency in hertz which is
2 pi lower is that in 29.06 Hz. And so, now you have also Q,
which is omega zero divided by gamma.
And, that Q is now 18.3. And, that means Q divided by
pi, which is, is there a problem Marcos?
Q divided by pie which is also the number of oscillations that
you have to wait for this function to go down by a factor
of E. Remember, we do that with the
pendulum. That is now about 5.8.
So, we're going to take a look at a curve like that,
and then we're going to indeed check that after roughly 6
oscillations, the amplitude is indeed down by
a factor of E. However, while we are at it,
I'm going to kill two birds with one stone.
I'm that going to make us 100 millihenry a little later,
which changes the gamma. The gamma becomes 500.
The decay time than becomes four milliseconds.
I'm not going to change C. So, the frequency goes down by
the square root of two. That becomes 2,055.
And now, the Q also changes. The Q goes up by a factor
square root of two, becomes 25.8,
and now you have to wait about a point to oscillations for the
amplitude to go down by a factor of E.
And that's what I want to show you now.
And, I had to choose my channel unless Marco has already done
that for me. Ah, the surprise is already
there. So, you see now,
here, the oscillations with about 2,900 Hz.
This is the location where the capacitor is fully charged,
that is, that Q max and Q zero is right here at the left side
at the bottom. That is where Q is zero.
And, let's now count the 5.8 oscillations,
roughly six, one, two, three,
four, five, six. And, if you look,
now, at this amplitude, after six oscillations,
and you compared it with this, you can very easily see that it
is roughly a factor of 2.7. So, that works exactly as
predicted. And, you see this beautiful
exponential decay. If now I change the
self-inductance to hundred millihenries,
then the frequency goes down from here 2,906 to 2,055.
The Q goes up, and now you have to wait eight
oscillations. So, Q is zero is on the left
side of the bottom. This is, again,
fully charged Q max. And now we go one,
two, three, four, five, six, seven,
eight. And, the amplitude now is
roughly the same as it was before when we had six
oscillations. So, you see that this works
like a charm if you do it with an L or a C circuit:
really quite remarkable. So dying oscillation and the
damping is the result of the resistor, which is dissipating
energy. If you want sleepless nights,
which is healthy, you may want to try to find for
yourself a convincing reason why the L also comes in the damping.
The L can never dissipate any energy.
Only the R can. Why is it that the gamma is R
over L? I'll be very honest with you.
When I searched for one, it took me 10 minutes to come
up with the right answer. At first you say,
this is weird. But, if you give it a little
bit more thought, and if you got a B+ or higher
for 8.02, you can do it.
All right.
All this is very nice and dandy, but what now happens if
gamma squared over four is larger than omega zero squared?
What now happens if gamma squared over four is larger than
omega zero squared? So, our solution,
omega squared, equals omega zero squared minus
gamma squared over four is utter nonsense because omega would
then be imaginary. So now, we have to go back to
our original differential equation and find a new
solution. We are closer to a solution
than you may think. Remember, we did have at one
point in time N squared equals omega zero squared minus gamma
squared over four. And then, we recognized that N
really is omega. Well, N is therefore also J
times gamma squared over four minus omega squared because if
you square this, you get that.
I did something weird again. I told you, it's not my day.
We had this earlier in the lecture, which is fine.
Then I said, well --
[SOUND OFF/THEN ON] -- conditions.
[NOISE OBSCURES] Now we get gamma over two,
[NOISE OBSCURES] over four minus omega zero squared to the
power one half times [NOISE OBSCURES] plus A2 times E to the
minus gamma, not minus gamma, minus gamma over two minus
gamma squared over four minus omega zero squared to the power
one half times T. You have to take my word for it
that this is the solution. And when you look at the
solution, nothing is oscillating anymore.
This is a decaying amplitude, K1, and this is a decaying
amplitude, A2. They have different decay
times. This determines one decay time,
and this determines the other decay time.
And, depending upon the initial conditions, will you be able to
find A1 and A2? At T equals zero,
you can specify what X is, and you can specify what X dot
is.
No oscillations: that means I can make some
guess of what the response will be.
Let this be a spring system, and this is X equals zero.
And, I offset the spring to position, X zero,
and I let it go at zero speed. That's my initial condition,
zero speed. So, that means when I let it
go, it must go in this direction, right,
in that diagram? Because DX DT must be zero.
But, there is no oscillation. So will happen is this,
and will asymptotically go to zero with two different decay
times. Now, if you released it here
with a high-speed in this direction, then you might see
this, but not in oscillation. An overshoot:
yes, but not an oscillation. And, you will get a chance in
one of your problems, I think it is 1-7 for this week
where you can play with this a little bit.
There is one more case, which is a very special case.
And that is when omega zero is exactly gamma over two.
And, that's called critical damping.
So, we have three [new means?]: under-damping.
We did that in detail. Then you have over damping
which gives you this solution, and then you have critical
damping. And with critical damping,
the solution, and I give you this without any
derivation, is A plus B times T times E to the minus gamma over
two times T. Only one decay time,
one exponential decay time, but there is something funny
here. There is also a T here.
So, this BT is linearly growing at the beginning,
but of course this one kills it.
And, again, your two adjustable parameters, A and B follow from
the initial conditions. So, now I will show you the
effect of damping. In a case of a mechanical
system, we have a torsional pendulum there.
And, the torsional pendulum has extremely high Q,
very little damping. I'll show it to you there.
But before we show it there, just concentrate on this.
Without any damping, if it oscillates and amplitude,
which is a torsional pendulum is one that has a spring like a
clock, this will die out very, very slowly.
Very high Q, very little damping.
What we have here in electromagnets,
and we can change the current through the electromagnet,
and the copper plate that move through the electromagnet will
now develop any current because there is a change in the
magnetic flux through that metal causes anti-currents,
and the anti-currents will oppose the motion of the metal.
You've seen that with 8.02 where you throw a coin through a
very strong magnetic field. The coin slows down enormously.
Instead of going clunk, a coin can go down this slowly.
It's a very nice demonstration in 8.02.
So, we can also, using the anti-current,
we can introduce artificial damping.
And that's what you're going to see.
One, two, three, this should be it.
Yeah, oh, we have to change the position, but that's not a big
deal. We may have moved it during,
oh, we also need some light on it.
I turned the light off, thank you.
Let's first make sure that we, yeah, there we see it.
So, I will give it a large amplitude.
The amplitude now is in terms of angle.
No damping at all, and if you look closely at the
turning point, you can say,
yeah, yeah, it is decaying. Give it a little bit of time.
But very slowly, you can see it decaying.
You have to wait a long time to see the amplitude go down by a
factor of E because you have such a very high Q system.
But now, I'm going to give it damping.
It's still under-damped. First give it a high amplitude.
There we go. Damping is in now.
Look how fast it's going now.
With this system, I cannot show you an example of
over damping. We did that,
and then the current has to be so high that we begin to smell
smoke. So, we decided not to do that.
But if I go to the point where you just don't smell smoke,
then you are still under-damped,
but you get a huge amount of damping.
And the reason why I can say with confidence that it still
under-damped, I release it at zero speed.
And, if I release it at zero speed, and if it is over-damped,
it can never overshoot. Yet, you will see that there is
a minute overshoot. So, it is still under-damped.
That's getting very close to being critically damped.
And so, I will jack these up to the maximum possible.
I can go way beyond 15. Is that the smoking domain?
It's OK. We don't go to the smoking
domain, right? So, I will first put in the
damping, and then I'll offset it and let it go.
Damping is in now. I'll offset it,
and let it go. Three, two, one,
zero. You see the overshoot?
It's just a teeny-weeny little. That tells you that it is still
under-damped because it's not supposed to overshoot since I
released it at zero speed. Of course, if I give it a push
in that direction, of course it will overshoot.
That's obvious. But if I do it with zero speed,
that solution that we have on the blackboard dictates that it
cannot overshoot. I'll show it once more.
And, you will see the overshoot.
That's it. So, that is the whole story
about damping we have under-damping,
we have critical damping, and over damping.
And, let's not put a damper on it.
I will see you again on Thursday, and make sure you pick 1042 1:21:01,347 up your 8.03 kits.