So, last time, we had driven damped
oscillations, and we did steady-state
solutions. And the steady-state solutions
have no adjustable constants. That is strange because at time
T equals zero, I can put the object in a
certain position. I can give it a certain
velocity, and so I have two free choices.
And yet, those free choices did not show up in our steady-state
solution. In other words,
the system had lost its memory over what happened at time T
equals zero. And therefore,
I discussed with you that you have to wait to go into
steady-state. And that is what I will address
today. There must be something missing
what we did so that the time T equals zero adjustables do show
up. So, let's return to our spring
system. So, we have here spring
constant K mass M. And, this object can be driven.
You could do that with a force directly on it here where you
can shake on the left side, omega zero squared equals K
over M. I will call it that way,
and gamma equals B over M. So, that's the damping.
Now, let us assume that I don't drive it, that I just give it a
kick, time T equals zero. I let it do its own thing.
We discussed that two lectures ago, so we get an un-driven
situation, not driven. And then, the solution,
you should remember, has this exponential decay in
it. [Is it?] X equals some value X
that follows from the initial conditions times E to the minus
gamma over two times T times cosine omega prime T plus alpha?
And these, the X and the alpha can be found if you know at T
equals zero what X is, and what X dot is.
Omega prime is a frequency which is the square root of
omega zero squared minus gamma squared over four,
just a hair under omega zero, but that depends,
of course, on gamma. So, this is the situation when
we are not driving. Let's now take a situation that
we do drive. So, we drive,
for instance, with a force that we apply
directly to this object, which is one way of driving it.
And, that force is F zero times cosine omega T.
This omega is my omega. I force that omega on that
system. I don't give a damn what omega
prime is [or?] the term of what omega is.
And, that's tough luck for this object.
And, we know that in the steady-state solution,
only this omega survives. That one will go.
And, the solution that we have seen that we derived last time
is some amplitude, A, times the cosine of omega T
minus delta. That is my omega that is
nonnegotiable. This is the steady-state.
And, we derived that the tangent of delta was omega gamma
divided by omega zero squared minus omega squared.
And, we found for A this monstrous solution.
Remember, upstairs we had F zero divided by M.
And, downstairs we have the square root omega zero squared
minus omega squared plus omega gamma squared.
And we spent the entire lecture on dealing with this A.
We changed omega, and we evaluated omega equals
zero, omega at residents, omega at very high values.
There is no adjustable constant in the solution.
But, there was one in that solution.
Now, let us look at the differential equations that we
were solving. And so, first we go through the
un-driven system. Un-driven means at T equals
zero, you give it a kick, and you just let it do its own
thing. The differential equation that
we then had was X double lot plus gamma X dot plus omega zero
squared X equal zero. That's the one we had.
And we solved it and we found that solution there.
Then, we were driving it. So, now we drive it.
Now, what's the differential equation?
Well, we had an X double dot, gamma X dot plus omega zero
squared times X. And now, we had here this
driving term. In the case that we directly
put the force on the object, then we had here F zero divided
by M times cosine omega T because the M comes in because
you divide by M, right?
You had Newton's Second Law as MA.
But, in the case that you shake the left side with your hand,
with this [eta?] zero cosine omega T, that is another way of
effectively driving the object. We worked that out.
Then we have here eta zero times omega zero squared times
cosine omega T. But in any case,
you see here a driving term, and we solve our equations.
And, if you take this case, then this is the solution.
If you take this case, then this takes the place of
that. Suppose now I take this
solution and a substitute that solution in this differential
equation. Then I get the result is zero
because, look, forget this.
So, I put it in this part. I get a zero because it fits
this differential equation. So, what is wrong with adding
zero? Therefore, if I add these two
solutions, it must be solution to this differential equation
because you just add zero. And, if you take a 18.03 and
they use very nice terms. They say, yes,
of course, if you have the special solution which is this
one, you have to add the homogenous solution.
And this word homogeneous means that you put the zero in.
And so, the general solution is really the sum of the two.
By adding this one, you effectively add zero.
So, you add nothing, but you get something in
return. What you get in return are your
two adjustable constants. So, now you can deal with the
situation that at T equals zero, you know exactly where that
object is and what its velocity is.
And so, I'll write down out the general solution,
which is the one that governs the major part of the lecture
today. So, X now, a function of time,
is the steady-state solution, A cosine omega T minus delta.
This is my omega. This is my will.
That's Walter Lewin's omega plus X times E to the minus
gamma over two times T times cosine omega prime T plus alpha.
And, this is the will of the oscillator.
And this is my will. They are two different omegas.
And now you can see what happens.
You see that this term will never die out.
It will last forever, and ever, and ever.
But this one is going to die. It's a one over EDK time of two
over gamma. And so, if two over gamma
happens to be ten hours, then you have to wait ten hours
for this one to be down by a factor of E.
But, if T over gamma happens to be 1 ms, then all you have to do
is rate 1 ms for that term to go down by a factor of E.
So, this is the one that will die out.
That's why we have to wait. And so, this is called the
[trangent?]. And it will die out faster the
higher gamma is. And this is called,
then, the steady-state solution, which ultimately
survives. Suppose I told you that at T
equals zero, X equals zero, and X dot also equals zero.
And, I was so nasty to say, oh, by the way when you solve
for X and alpha would be a very nice thing to do.
But, it would take you 15 minutes of grinding;
not so fast because, remember, if you have an X dot,
you have to take the time derivative of this entire
function. You have a T here or you have a
T there, and you have to substitute in there time T
equals zero. And then, you have to make that
equal zero. And, it takes you 15 minutes.
And, out pops a [UNINTELLIGIBLE],
a value for X, and a value for alpha.
You haven't learned much physics when you do it,
but you've learned some algebra.
And so, I decided not to spend my time on doing that.
But, in principle, you must agree with me now that
if I specify the initial conditions, I don't have too
call this zero. I could call this X zero.
I can do anything I want to. I can give X dot any value I
want to. And they get unique values for
X and for alpha. And, that is ultimately,
then, what the solution is. I must add the two.
The bottom line, and that's really where the
physics is, and that has to do with problem two-five that you
have this week on your plate is then the following.
If I make a plot of X as a function of time,
then the solution is really the sum of these two.
This is the steady-state solution which has been
amplitude, capital A, which is nonnegotiable.
It has nothing to do with the initial conditions.
It has its own omega, Walter Lewin's omega.
So, let's assume that this is the period, T.
And so, out of that pops a nice cosinusoidal,
sinusoidal curve. Let me put it in here.
And, this never changes. This goes on forever,
and ever, and ever. And this here,
this time, T, is 2 pi divided by omega,
my omega. However, there is also this
one, but this one dies out. And so, I will now put in some
kind of an exponential decay, something like this.
It has its own frequency, omega prime.
I can choose anything. I can make omega larger,
omega prime larger than omega. I can make it smaller so,
I just pick one, and let's suppose that the zero
crossings are here, and here, and here,
and here, and here. And so, for instance,
the curve than that has to be added could be something like
this.
And this time it's T prime is 2 pi divided by omega prime.
And so, you see that the sum of the two, which I will not try to
sketch, is the solution. But it is the pink one that's
going to die out. That's why sometimes you have
to be patient, except if it is 1 ms two over
gamma, you don't have to be very patient.
Now, let us look at the situation.
And I can arrange that. And I'm going to arrange that,
that omega and omega prime are close together.
I can choose that. The system cannot choose omega
prime. The system is stuck.
This is omega prime. The system has no choice.
But, I have a choice. I can make omega any value I
want to. So, I can make it very close to
omega prime. What do you think is going to
happen now? When I turn this system on,
all of a sudden the driver, and my own omega is very close
to that omega prime. It's true that the trangent
will die out, but let's say we take a system
with a pretty high Q. So, it doesn't die out so fast.
What do you expect you're going to see?
Excellent. You're going to see beats
because now you have two harmonic oscillations which have
to be added. But, the frequencies are a
little different. And, if this one survives long
enough, there comes a time that they are in phase.
There comes a time they are out of phase.
You see very low amplitude. And so, you are going to see a
bit phenomenon. And that is to do what I want
you to see. For that, we need a system
preferably with a high IQ. And then, the driving frequency
has to be close to the omega prime.
And for that, I have chosen this system here.
This is an air track. And, we can make the damping
very low, unpleasantly low, believe me, a very low value of
gamma that when I turn on the airflow, so here is a spring
constant, K. Here is one spring constant,
K, and here is the mass. And, there's very little
friction. And now I'm going to drive it
just a little bit off resonance, a little bit below the
resonance frequency. And, what you are going to see
now is the sum of these two. But since gamma is so low,
it will take a long time for this trangent to die out.
And that is exactly what I want you to see.
In addition, you're going to see beats.
And as long as you see beating, you know that you haven't
reached the steady-state solution yet.
But, if you are patient and I'm patient, we probably will see it
go into steady-state, but it may take several
minutes. So, you ready for that?
I'm going to drive it here. And, I start the driving out.
So, relax and look at the amplitude of this object.
And see what happens. Hey, the amplitude is going
down, hey, hey, hey, I call that a beat.
Did you see that? Did you see the amplitude go
down? The two frequencies were being
against each other. Now it's picking up again.
It's nowhere near a steady state: very low gamma,
very high Q. There we go again,
amplitude way down picks up again.
Just be patient. Let's be patient and see
whether we have the privilege of seeing it go into steady-state.
It's a very high Q system. Since I am just below
resonance, the driver in the car will be in phase when I am below
resonance in steady-state solution.
So, this delta will be very close to zero,
below resonance, above resonance,
180° out of phase. But, I am just below resonance.
So, when we go into steady-state,
we also will see that we are very close to a delta of zero.
Now, let's see what the amplitude is now,
and whether the amplitude is changing.
Well, we are getting there. It pays off to be patient.
Later today, I will do an experiment where
two over gamma is two many seconds.
So, all you have to do is wait 4 ms.
So, I'll make up for the fact that now we have to wait to
long. Let's take a look at this now.
I think it looks terrific. It looks terrific.
In phase, I don't see much beating anymore.
It looks like the amplitude is constant.
I think we've killed this one, and I think this one has
survived. If you increase the damping,
this would happen, of course, earlier that you go
into the steady-state solution.
It looks great. I don't see any change anymore
in the amplitude. So, that's the A that you have
there, capital A, and they are nicely in phase.
It's a very high Q system, so the change to go from delta
zero to delta pi over two at resonance takes place over an
extremely narrow range of frequency.
So, they are still in phase. All right, if we are driving
this system with a force, say, directly on the object,
F zero cosine omega T, then in steady state there is
energy dissipation because there is friction.
And, where there is friction, there is heat.
And that means energy. That means I had to do work to
provide that energy to the steady-state situation.
So, as the thing is never changing, it's A just going on
forever and ever. While that happens,
I must put in energy, which comes out in the form of
heat. So, let us return to the good
old days of 8.01. And, I want to remind you that
work is the dot product between a force and a displacement,
DX. It's a dot product.
It's a scalar work. A little bit of work is done by
this force if it moves over a distance, DX.
If the two are perpendicular to each other, then no work is
done. Satellite into circular orbit
around the Earth: no work is done.
And so, now I can calculate what the power is because the
power is DW DT, how many joules per second I
had to put into the system. And so, if I take a derivative
times the derivative, then I get F dotted with V
because DX DT is simply the velocity.
Now, if I have a one-dimensional system,
and what I mean by that is the force is either in this
direction or in this direction in the velocity is either in
this direction or in this direction.
That's what I mean by a one-dimensional system.
Then I can delete the dot, and then the signs will
automatically take care of the direction.
The minus V is then this. Plus V is that,
and same for force. So, I can kill the dots.
So, now, I have to know what the velocity is.
In the steady-state solution. Well, that's easy,
because I go to the steady-state solution here and I
calculate what X dot is. So I'll put that here,
X dot, which is V. So, the derivative of cosine
omega T is minus omega times the sine.
So, I get minus omega times A times the sine omega T minus
delta. That's the velocity.
But, I know what the force is. That is F zero cosine omega T.
So, there we go, F zero cosine omega T,
but I'm going to put these in also now.
So, I get omega A, and then I put in the cosine
omega T, and now I'm going to put in this one the sine of
omega T minus delta. So, take a deep breath.
F zero cosine omega T is the force.
See that there? And the velocity has been
derived from the first derivative of the steady-state
solution, gives me a minus omega A, sine omega T minus delta.
Now, this, what we have here between brackets can be written
as the sine of omega T cosine delta minus the cosine omega T
times the sine of delta. So, what you see here between
brackets is the same as what you see there.
And I really interested in knowing every moment in time
what exactly the power is? Not really.
Most of the time I'm really interested in knowing what the
average power is, that is required to keep the
thing going, average over one oscillation or over hundreds of
oscillations. That's really what I'm
interested in, not necessarily the
instantaneous power. In other words,
most of the time my interest is really in what this is,
that is, the time average value.
And, think of it as being one oscillation that is fine.
But you can think of it also as days.
Now, I need some experts and audience.
I see here with cosine omega T, and I see here with sine omega
T. What is the time average value
of that product? The time average value [between
the?] sine omega T and the cosine omega T,
if I average it over one cycle or two cycles or three cycles.
Come on, high school, yeah?
What is the time average of sine omega T times cosine omega
T, time average over one period? Zero.
This one times this one times average gives me zero.
You don't believe it, go back to high school.
Ask your high school teacher. He will agree with me.
Here I see a cosine omega T, and I see a cosine omega T
there. What is the time average of
cosine squared omega T? Ah, you guys are waking up.
It's one half. So, therefore,
for this product, sorry, for this product,
I can write one half. And so now we get,
notice there is a minus here. And so this minus picks up this
minus. So, it becomes a plus.
So, I get F zero. I get the half that is here.
Then I get omega. I get my A, and then I get the
sine of delta. Are we happy with that?
So, you see, it collapses into something
that is relatively simple. What is sine delta?
Well, I remember what tangent delta is.
If I know tangent delta, can I then calculate sine
delta? And, the answer is yes,
of course. If this is delta,
and if this is omega gamma, and this is omega zero squared
minus omega squared, then it only takes Pythagoras
to calculate what this is. And so I know that the sine of
delta must be omega gamma divided by the square root of
omega zero squared minus omega squared, squared,
plus omega gamma squared. So, yes, I do know what the
sine of delta is. So, now, I can come to a close
by substituting in here what the sine of delta is.
And, I can substitute in here what A is.
Here is A. You know what is nice?
Look at the downstairs here. It's the same as the downstairs
there. So, if I multiply them,
the square root goes away. So, if I can write down now
what P averages, go slowly, so we get F zero
divided by two. I get an omega.
I will go to the A very shortly.
I will first pick omega gamma here, which is my sine delta.
So, that makes this a square. And, I get a gamma.
And, now I turn to the A, which is an F zero over M.
So, I get a square here. And, I get an M here.
And now, I get this downstairs times this downstairs.
The square root disappears, and so I get omega zero squared
minus omega squared, squared, plus omega gamma
squared. Almost end of story.
I'm going to rewrite it a little.
And Tony French, in his book,
rewrites it in, again, a different way.
He loves to work with Q's. He puts the Q's in there.
I'm going to divide upstairs and downstairs by omega squared.
And, when I do that, I got F zero squared.
I've got a gamma here. I get two M here.
And then, I get here omega zero squared divided by omega minus
omega squared plus gamma squared.
That's what I get. That's not the only way you can
write it, but that's one way you can write it.
Let me check that. I have five zeroes squared.
I have gamma, two M here, and we have this
downstairs. So, the time has come now to
try to see through this equation.
You remember last lecture, we spent the whole lecture not
to look at this dumb equation, but to see through it.
And we were able to see through it, see all its idiosyncrasies.
Let's look at the idiosyncrasies of this average
power over one cycle, over a multiple of cycles.
Well, let us first make gamma infinitely large.
Don't look even at the equation.
That means there is an infinite amount of friction.
The system never gets going. There's no way it will ever
move. So, clearly the average power
must go to zero. And, indeed,
you see that. The gamma is downstairs.
If gamma goes to infinity, the power is zero.
Let's make the mass infinitely large.
If the mass in infinitely large, you have an infinity high
inertia. Nothing will ever get going.
No force will ever get the mass going.
Well, that means you expect that the power goes to zero.
And, indeed, you see here,
if M goes to infinity, the power goes to zero.
Let's say the force goes to zero.
We are not even driving it. Well, if we are not even
driving it, I hope you will agree with me you don't have to
put in any work, right?
Nothing gets going. So, clearly you expect,
then, that the power will go to zero.
And, indeed, if F zero goes to zero,
you see that the power is zero. Suppose you make omega zero.
So that means there will never be any velocity.
You never pick up any velocity. It takes infinitely long.
Omega is zero. So, clearly,
if nothing ever moves the power will go to zero.
Now, look. If omega makes zero,
this is zero. But this one goes to infinity.
And, therefore, the power goes to zero.
So, you need no equations for that, just common sense to
immediately conclude that this has to be the case.
Suppose you go to infinity with omega.
Very, very fast. Well, if you go infinitely
fast, because of the inertia of the system, it can never react,
can never get going. So I will predict that the
power must go to zero. And, if you put omega infinity
in here, this goes to zero. This goes to infinity,
and so the power goes to zero. So, all this is complete common
sense. All of this you could have
predicted without that equation. But, isn't it nice that the
equation supports my intuition? So, now comes the question,
if omega goes to omega zero, and that is the reason why I
wrote it in this way. Notice when omega goes to zero,
when omega goes to omega zero, this one goes to zero.
And therefore, that is the frequency at which
the average power is the maximum [ever?], can never be any higher
because it's independent of gamma, right,
my omega? So, I have an omega equals
omega zero. We reach the maximum value for
power. We can never go any higher.
It's exactly at omega zero. And, this is zero.
You lose one gamma. And so you find that this
value, then, becomes F zero squared divided by 2M gamma.
And, you can write that, rewrite have a little bit.
I do that because Tony French likes Q's.
And so, I write it with a Q in there.
Remember that Q is omega zero divided by gamma.
So, I can rewrite this as Q times F zero squared upstairs.
So, I get the Q in there, which is always nice,
divided by 2M omega zero. So, this is the same thing.
So, if I now plot, make a curve for you of P
average, not P maximum, but P average as a function of
frequency, so here is omega and here is P average.
Here is omega zero. It goes through a maximum
exactly at omega zero. It starts at zero,
you see, and it ends at zero. And, it sweeps up to a maximum,
and then it goes down again. And, for reasonable values of
Q, these curves look extremely symmetric.
And so, this value here is then the average max,
which is that value, this value.
If we look at the width of this curve, at half maximum of the
power, so this is one half times P
maximum, then you can show, and it is not so difficult
algebraically, but I will not attempt it,
you can show that the width at half maximum is very close to
gamma. Remember, gamma and omega have
the same unit. It's one divided by second.
In other words, if you go to half maximum,
this point here is omega zero minus gamma over two.
And, this point here is omega zero plus gamma over two.
And so, you see immediately, which of course makes sense,
that if gamma is very small, that the peak gets very narrow.
And, if gamma is very high, the peak gets very broad.
That's intuitively quite pleasing.
High Q systems, very narrow peaks.
And, that's the way that Tony French likes to plot his data.
I will show you that on the overhead here,
this is just a picture from your book.
What Tony does here, he plots not omega here,
but he plots omega divided by omega zero.
So, that means the resonance is at one.
And, he doesn't plot the average value for P here,
but he plots it into strange units into the unit F zero
squared divided by 2M omega zero.
So, now, he effectively can compare the vertical axis with
the Q value because he likes the fact that it is Q times higher
than something. And, he has plotted this in
terms of that something. And so, if you take a curve for
Q equals ten, which has the peak here in
power, you see, indeed, that he finds that very
close on his scale to ten. Notice, also,
the nice symmetry. And, you see for lower values
of Q, which are curves here that, indeed,
the peak gets broader. The width of this peak is one
over Q because, remember, this axis is omega
divided by omega zero. So, if the width is gamma on
that peak, it is now gamma divided by omega zero in this
plot. And, gamma divided by omega
zero is one over Q. So, here, the width in this
presentation is directly inversely proportional to Q.
So, if Q is ten, then the width there is one
tenth. He then shows you another plot
whereby he does what I did there.
He plots it as a function of omega, not as a function of
omega divided by omega zero. And then, he emphasizes the
fact that the width here is that gamma that I mentioned at half
the maximum power you get here [the width/with?] of gamma.
And, this is a picture that I chose verbatim from your book.
This is the best moment for the break.
That means the mini quiz. I realize it's a bit early;
we are only 40 minutes into the lecture.
But it's a natural point. You will see what comes
afterwards. It's better that we make the
break now. So, therefore,
I need some help from people who are willing to hand out the
mini quiz. It would be nice if I can find
the mini quizzes. I have them here,
but someone took them. Oh, no, they are still there.
I have a nice conspiracy. Afterwards, after the break,
we will collect them this time in some boxes so that it's a
little bit more organized. And so, I'm returning to an RLC
circuit, which we discussed earlier, the good old days of
8.02. I'm going to drive it now,
not with the battery but with an alternating power supply.
V zero cosine omega T.
Yeah, put it in here, thank you.
So, here's the circuit, resistor R, self inductance L,
capacitor C. And, I have to write down,
now, the differential equation. I will adopt a positive current
in this direction. That will be my positive
current. The charge here on this right
plate I will call Q. And, therefore,
by that definition, I is then DQ DT.
Sign sensitive. I call the potential difference
over this capacitor, in going from the right side to
the left side, I call that V of C.
That, then, is Q divided by C. All of that is sign sensitive.
I go a round this circuit and I want to calculate the closed
loop integral of E dot DL. And that closed loop integral
of E dot DL is not zero, which many books tell you,
even many professors tell you. It is not zero,
but it is minus D phi DT. This is [UNINTELLIGIBLE] Law,
and this runs our economy. Because of the magnetic flux
change in closed loops, we can generate induced [EMS?],
which run our economy. Look at the lights.
Luckily, this is not zero. This [fires?] in magnetic flux
that goes through a surface, any surface that you can attach
to this closed loop. So, I've done this before.
So, I can do it a little faster.
I go from here to here. So, that is IR.
There is no electric field inside this ideal self inductor
because a superconducting wire cannot be an [E?] field.
So, that is zero going from here to there when I go over the
capacitor. I get my VC,
and here, depending upon the phase, if I assume this plus and
this minus, but you can reverse that.
Then, I would get, when I walk into this
direction, I would get minus V zero cosine omega T.
But, if you feel like reversing it, I have no problem with that.
That's just a matter of 180° phase.
[It's known here from?] physics.
And, this now equals minus D phi DT.
The only thing where you apply [UNINTELLIGIBLE] law is you
should always integrate in the direction that you have your
current assumed, that it is minus L DI DT.
If you do it in the opposite direction, then it is plus L DI
DT. I have learned a certain
discipline in my life. It took me many years.
So, you have a long way to go. And I always go in the
direction of I, so I never have to think that
this is minus L DI DT. This now covers the minus D phi
DT. So, now, what do I do?
I bring the L in. And, I take one more time the
derivative. And so, I get L times I double
dot plus R times I dot plus VC. But I take the time derivative.
So, the Q dot becomes I. So, I get I divided by C.
And that now becomes the timed derivative of this function.
But, it goes to the right side, which makes the minus sign a
plus. But, when I take the derivative
of the cosine omega T, I get a minus omega out.
So, I get here minus P zero times omega times the sign of
omega T. This is the differential
equation that has to be solved. And, I will divide this out by
L. I will divide everything by L.
I will put the C a little higher, and so with R over L
gamma, and with omega zero squared equals one over LC,
this becomes, then, I double dot plus gamma
times I dot plus omega zero squared times I.
That now equals minus V zero divided by L times omega times
the sign of omega T. But, here, you see a
differential equation. And that differential equation
looks amazingly similar to this one.
And so, you should be able to solve that.
In fact, you wouldn't even want to solve it.
You can write down immediately the answer.
You're going to get an I, which is an I zero,
which takes the place of that A there.
This is a steady state solution.
I only go for steady state solution times the sign of omega
T minus delta: no adjustable constants.
It's a steady-state solution that I have, steady state.
And, I will leave you to find me I zero, and you can work out
what delta is. That is part of your problem
set anyhow. But, with the knowledge that
you have here, you could write it down in a
manner seconds. So, without my telling you what
I zero is, at least working it out algebraically,
we can talk 8.02. And then, we can make all kinds
of predictions without even looking at the equation.
So, that's interesting. So, everything that I'm going
to tell you now I do without knowing what I zero is.
And, it better work out that way.
Suppose I make omega go to zero.
Remember 8.02? Remember the word reactants
that a capacitor has a certain reactant, which is one over
omega C, which has units of ohms?
If omega goes to zero, these reactants go to infinity.
No current can ever flow. So, I predict that I zero will
go to zero. Suppose my omega goes to omega
zero. Now, your memory may fail you
here on 8.02. But, we have a wonderful
demonstration in 8.02 that at resonance, one over omega C,
which is the reactant of the capacitor minus omega L,
which is the reactance of the inductor, is zero.
That determines, actually, the resonance.
And, when this is the case, perhaps you remember that the
system doesn't even know there is a capacitor,
and it doesn't even know there is a self inductor.
The two at all moments in time exactly cancel each other.
And therefore, Ohm's law holds.
There is no L. There is no C.
There is only the power supply and the resistor.
And so, since Ohm's law says V equals IR, you must get I zero
equals V zero divided by R. That is what you must get at
resonance. It's nonnegotiable.
Now, if omega goes to infinity, omega L says,
ha, ha, yeah, over my dead body.
No current ever, imagine, a very fast changing
signal. That's what the whole self
inductance is about. It doesn't want any changes.
It's conservative like you and me.
And so, the self inductance says, sorry, the reactant is
infinitely high, no current.
And so, I zero goes to zero. And so, I make these
predictions. And that's always nice that you
can use the knowledge to make predictions without even ever
having looked at this differential equation.
Any of these predictions I made to not come out of my knowledge
of that differential equation. So, if we make now a plot of
the current I zero, that is not the current,
but that is the maximum possible current.
It is that I zero that you see here without having solved it.
I can look now what it's going to do exactly at I omega zero.
It will go to a maximum. It will start at zero.
I will go to a maximum, and then it will fall off at
zero. And, this value here is V zero
divided by R. This is not power.
I have now plotted current. The power, of course,
would go as I squared R. That is the heat that you
dissipate in the resistor. So, that will go as I squared.
I have plotted here I as a function of not T,
but this is I as a function of omega.
You know when you catch an error that I make,
you get partial credit for this course.
So please, when you see me make a mistake, scream.
So, this is omega. Here is zero.
And, this is what I want to demonstrate, now.
I'm not going to show you I zero only.
But, what I'm going to do is the following.
I'm going to show you what I is as a function of omega,
here being the resonance. Let us suppose I pick this
omega. Well, then, this is my
solution. So, yes, the amplitude is I
zero. Here is the sine omega T minus
delta. So, all you will see,
then, is this.
Plus zero, minus plus zero, if I do it here,
then it goes [MAKES NOISE]. And, the demonstration that we
have prepared for you is one whereby we will sweep omega from
zero to a value which I remember in terms of Hertz is about 2,000
hertz, and we'll do that in one sixth of a second.
And so, what you see is you see this as an envelope,
which is the I zero envelope. But, you will see this.
And then, [it speeds?] back. And so, you see two things.
You'll see the sine omega T term.
But, as omega changes, you will see it go through
resonance. And then you'll see it go over
resonance. So, I will give you the values
that we have chosen. R equals 50 ohms.
L equals 50 millihenry. And, we choose C 0.5 µF,
which is substantially higher than what we did before.
And, we choose C so high because we want a low Q system.
Omega zero is now 6.3 times ten to the third.
It is in radians per second. So, F zero is about 1,000 Hz.
And, we're going to sweep it from zero over 1,000,
which is resonant to about 2,000.
And then we sweep it back. And so, the Q of this system,
which is omega zero divided by gamma, is about 6.3.
And, what the V zero is of that circuit is not so important.
But it is [four fold?], but that's not so important.
And so, what we're going to show you, we measure the
potential difference over a very small resistor,
which is somewhere in that circuit, 1.7 ohms I believe.
And so, that potential difference over that resistor is
IR. And, R is a constant,
and that's what we're going to show you.
So, we're going to show you something that is directly
linear proportional with I. It's not power.
It's I. We want to know power.
You have to square it. I square R is the power.
And, we're going to sweep it 1/6 of a second this way,
and 1/6 of a second back. Why did I only take into
account the steady state solution?
Why don't I have to also include the trangent solution,
which with this experiment took us five minutes to finally
arrive at the steady state solution?
Why am I leaving it out? I can't hear you.
Where is the sound coming from? Yes.
What is two over gamma, which is the decay time?
One over EDK time? Well, there is two over gamma.
I didn't write down what gamma is.
Gamma is 1,000. Gamma is R over L,
right? Gamma is R over L is 1,000.
So, two over gamma, which is the one over EDK time,
is 2 ms. Or, another way of putting it
is that in about two oscillations,
I'm down by a factor, E.
Remember, Q over pi oscillations will reduce the
amplitude by a factor of E. So, in two oscillations,
already the trangent phenomenon is effectively killed.
And so, I don't have to take it into account.
You won't even notice it. So, I told you earlier,
when the decay time here was very long.
I'll show you another experiment where the decay time
is extremely short. While I'm at it,
the experiment is set up there. You will see it here.
I'm going to make this 100 ohms.
I'm going to double it to show you that this point will exactly
go down by a factor of two because, remember,
the peak is V zero divided by R.
And, I'm not changing P zero. And so, when I double R,
you will see it come down to here, and you will see it get
broader. And then, I will go to 150
ohms. That's easy to do for us.
So, we go to 150 ohms. So, at 100 ohms,
the Q is 3.15. And, at 150 ohms,
the Q is three times smaller. It's about 2.1.
So, at 150 ohms, it would be one third of this
height, somewhere here. And the peak will be broader.
And, all of that you get today very easily by changing the
resistance. Again, this is not power.
Power is I squared times R. This is simply the current.
All right, I will give you the correct light setting because we
have to make it a little darker. And, then we,
and there it is. So, noticed the vertical
oscillations go so fast that your eyes cannot even follow
them. But, that's the sine omega T.
And then, when it sweeps over resonance, you see very
dramatically this value which is V zero divided by R.
Now, notice on the scale here that Marcos has set the I zero
at resonance at three scale units.
So, it's nice when we go from our 50 ohms to 100 ohms to go
down by a factor of two. And, you can check that.
And when we go to 150 ohms, it should go down from three
units to one unit. So, it could quantitatively
check this. And, you'll see that the curve
gets broader because it has a low Q.
So, now I will make it 100 ohms.
You see, it's down by a factor of two from here to here.
And, you may have noticed that it also gets broader.
And now, I will go to 150 ohms. And you see,
it's down by a factor of three. And again, it is broader.
So, this is an amazing way how with RLC circuits you can do
wonderful things because you can manipulate omega zero very
easily by changing L and C. And, you can manipulate the
driving frequency also very easily.
So, it is clear that systems respond strongly when they're
exposed to their resonance frequency.
We've seen that for pendulums. We've seen that for springs.
We've seen at 401 glass. And we've seen this now for an
RLC circuit. So, these systems at resonance
absorb a large amount of energy per unit time out of the driver.
I have here two tuning forks, which have an extremely high Q.
My attempts to measure it, I conclude it's way larger than
even 1,000. And they have exactly the same
frequency: both 256 Hz, this one and this one,
both 256 Hz. That's' the way they are
designed to a high degree of accuracy, to better than a
fraction of 1 Hz. But, the Q's are so high that
if you were to plot, if you drive these tuning forks
and you are to plot here this average power as a function of
omega, then you get something like this.
It means you have to drive it exactly at the right frequency.
Otherwise, it will not go into resonance.
Well, we know how to get this going.
You just bang it. That means you dump the whole
spectrum on it. It picks out the frequency that
likes, and now I'm going to show you something remarkable.
When this one generates 256 pressure waves,
this one feels those pressure waves.
And it loves it because it's just at the right frequency.
And so, it starts oscillate. And so, when I stop this one,
you'll hear this one. And that's called resonance
absorption. Let's do that first.
Now, you must understand that the sound waves go from here to
there. Not much power reaches that
point. So, when I stop this one,
you hear sound but it's not overwhelming.
So, you have to be very quiet.
Hear it?
And I can do the same by hitting this one,
and then this one will start to resonate.
Now, if the driving frequency is off by a fraction of the
hertz, 1 Hz is enough, 1 Hz difference because the Q
is so high that this system will not be able to get this one
going. And, I can make this frequency
a little lower than 256 Hz by putting this weight on here.
And, we have measured the frequency at this loaded way.
It's roughly 255 Hz. And, you are somewhere here
because it's so narrow. The resonance absorption peak
is very sharp. By the way, this is power
because what reaches here is joules per second.
That's what gets it going, energy per second.
So it's really a power transfer.
So now, I change the frequency, and there we go.
Nothing. But just change this by 1 Hz
and you hear nothing. Dead.
So, now you see, you get some respect for high
Q's. If you want to get a resonance
absorption in the high Q system. You've got to be dead on that
frequency. So, if you, for instance,
banged all the keys on the piano, and this one would be
nearby, it would only start to resonate if one of those strings
would produce exactly the 256 Hz.
Otherwise, it would not. It ignores everything.
It's only sensitive to that resonance frequency.
You probably in high school have learned a little bit about
atomic physics. And, you probably know that
electrons have discrete energy levels and discrete orbits and
atoms. And, you can excite the atom.
You can bring an electron in a higher orbit,
discrete orbits which cost you a discrete amount of energy.
And, when the atom recombines, when the electron falls back,
you get that energy back, exactly the same amount that
you had to put in. And, that energy that you get
back comes out most of the time in the form of what we call
electromagnetic radiation. I know that in 8.03 we are
going to deal with electromagnetic radiation in the
future. But, it's enough for now that
you know that light, infrared, UV,
gamma rays, x-rays, all of that radio emission,
all of that is electromagnetic radiation.
And so, here I have the energy level, energy increasing,
here with an electron in orbit. That's the energy of that
electron. A higher energy state is when I
bring this electron here. I cannot do anything in
between. Quantum mechanics says it's one
or the other. And, if this difference is
delta E in energy, this E stands now for energy.
Then, when the electron falls back from here to here,
it emits electromagnetic radiation with this energy.
But, if I radiate onto this atom, electromagnetic radiation
with exactly that energy, then this electron can go from
here to there. And that is called resonance
absorption. Now, let us stick,
for now, to visible light. The higher the energy the bluer
the light is, or as modern physicists would
say, the higher the frequency. And the lower the energy,
the redder the light, the lower the frequency.
So, our visible light that we can see with our eyes goes all
the way from the red, low energy, to the violet,
high-energy. Let's go to the sun.
The sun radiates in the visible spectrum all the way from the
red to the violet. But, in the solar atmosphere,
R elements. And, when these elements see
just the right energy from that spectrum to which they are
exposed, they love to take out of that spectrum just the right
energy that gets them into an excited state,
which is called resonance absorption.
And so, that energy is removed from the spectrum.
So, when you look at the solar spectrum, there are bands in the
spectrum, but the colors are missing.
Absorption in the spectrum, dark bends in the spectrum.
They were discovered in 1802 by William Wollaston.
And in 1814, Fraunhofer had cataloged 475 of
these lines. And, they are now referred to
as Fraunhofer absorption lines. Even though they did not
understand the physics, this is a quantum mechanics
picture that came from Niels Bohr, 20th Century,
even though they did not understand what happened they
had noticed that these black lines in the solar spectrum
coincided with emission lines in the spectrum that they can
generate in the laboratory by heating up the various elements.
And so, without understanding why, they were able to say,
ah, I see magnesium in the sun. I see aluminum in the sun,
in the solar atmosphere. And so, that opened a whole new
industry of spectroscopy which allowed astronomers to determine
the chemical composition of the atmospheres of stars.
And, it was in 1868 that Joseph Lockyer found at least one dark
line which did not coincide with any emission line in the
laboratory. There is no element on Earth
that he could say, that must be the cause of that
dark line. And so, he calls it helium
because the Greek word for the sun is helios.
So, it helium is an element that was first discovered on the
sun before it was later found on Earth.
I want to show you resonance absorption on the scale of an
atom. And, the way I'm going to do
that, the setup is here, is as follows.
We have a carbon [arc?]. Think of that as being the sun,
which produces a spectrum, a beautiful continuous
spectrum. I will show you that spectrum
all the way from the red to the violet.
And then we have here a burner. And we're going to put table
salt here on the grid, which dissociates the table
salts, gives me sodium gas. That's what I want because
sodium, when you heat it, can produce an emission line in
yellow. But if you can produce that
emission line when the electron goes from here to there,
it's the 11th electron by the way.
It's the most outer electron of sodium, 11 protons,
11 electrons. So, if it can produce an
emission line when it goes from here to here,
it can also, resonance absorption,
namely when it sees that yellow line, the energy that
corresponds with the yellow line, it sucks it up and it
produces, then, a dark line because when it
absorbs out of here, this yellow line,
it re-emits it almost immediately.
But the re-emission will be in all directions.
And so, what's left over here is very little of that yellow.
And so, a dark line appears in the spectrum.
That then is the absorption line.
Its power, what I'm going to show you, because light
intensity, which I will show you there on that screen is how many
joules per second? So, it is resonance absorption
of power. Now, there is a catch.
And the catch is that probably only you up here will be able to
see it. And, others could come down.
You're going to see the spectrum here first of the sun,
which is my carbon arc. Then I will put in sodium.
And you will see an unbelievable,
unimaginable, beautiful, sharp light like a
razor in the yellow. But you've got to be close.
So, let's first, Marcos, if you manage to open
the gas, he knows exactly where that gas valve is,
then I will ignite it. We won't put it yet in beam.
OK, so we're going to make it completely dark.
So, here are the [sold?] crystals.
And, we're going to show you the spectrum there.
We try to actually make you see it here.
[UNINTELLIGIBLE] that didn't work out well.
OK, so I'm going to turn on the sun.
There's the sun. OK, now we make it completely
dark, and I will give you a minute or so for your eyes to
adjust. So, you see a spectrum here;
you see a spectrum there. How we do that is our problem.
And, you will know how we do that in a month or so when you
will learn about gradings. You will get a grading,
actually, from us. There is a grading here,
a wonderful piece of physics, which decomposes the light in
colors, works way better than a prism.
And, you get one on the right side.
And, you get a mirror image on the left side.
Look here and let your eyes adjust.
And then comes the moment of truth.
I'm going to put in here now the sodium.
Unbelievable. I see a line here,
sharp as a razor blade. Can you see it,
Nicole? Isn't it incredible?
Come closer. All of you, come closer;
look at that line. Come on; come out of your
seats. Look at that line,
and I will move the sodium out. Now I'm at the sodium out.
And now I move it in again. And there it is.
There it is. You see that?
Isn't that amazing? And now I move it out.
And now I move it in.
Isn't that a fantastic line? Look at that line.
Resonance absorption of an
extremely high Q system on an atomic scale.
I hope you can sleep tonight. [LAUGHTER] See you Thursday.