This is a brief,
so, the equation,
and we got the characteristic
equation from the last time.
The general topic for today is
going to be oscillations,
which are extremely important
in the applications and in
everyday life.
But, the oscillations,
we know, are associated with a
complex root.
So, they correspond to complex
roots of the characteristic
equation.
r squared plus br plus k equals
zero.
I'd like to begin.
Most of the lecture will be
about discussing the relations
between these numbers,
these constants,
and the various properties that
the solutions,
oscillatory solutions,
have.
But, before that,
I'd like to begin by clearing
up a couple of questions almost
everybody has at some point or
other when they study the case
of complex roots.
Complex roots are the case
which produce oscillations in
the solutions.
That's the relation,
and that's why I'm talking
about this for the first few
minutes.
Now, what is the problem?
The complex roots,
of course, there will be two
roots, and they occur at the
complex conjugates of each
other.
So, they will be of the form a
plus or minus bi.
Last time, I showed you,
I took the root r equals a plus
bi,
which leads to the solution.
The corresponding solution is a
complex solution which is e to
the at, (a plus i b)t.
And, what we did was the
problem was to get real
solutions out of that.
We needed two real solutions,
and the way I got them was by
separating this into its real
part and its imaginary part.
And, I proved a little theorem
for you that said both of those
give solutions.
So, the real part was e to the
a t times cosine b t,
and the imaginary
part was e to the at sine b t.
And, those were the two
solutions.
So, here was y1.
And, the point was those,
out of the complex solutions,
we got real solutions.
We have to have real solutions
because we live in the real
world.
The equation is real.
Its coefficients are real.
They represent real quantities.
That's the way the solutions,
therefore, have to be.
So, these, the point is,
these are now real solutions,
these two guys,
y1 and y2.
Now, the first question almost
everybody has,
and I was pleased to see at the
end of the lecture,
a few people came up and asked
me, yeah, well,
you took a plus bi,
but there was another root,
a minus bi.
You didn't use that one.
That would give two more
solutions, right?
Of course, they didn't say
that.
They were too smart.
They just said,
what about that other root?
Well, what about it?
The reason I don't have to talk
about the other root is because
although it does give to
solutions, it doesn't give two
new ones.
Maybe I can indicate that most
clearly here even though you
won't be able to take notes by
just using colored chalk.
Suppose, instead of plus bi,
I used a minus bi.
What would have changed?
Well, this would now become
minus here.
Would this change?
No, because e to the minus ibt
is the cosine of
minus b, but that's the same as
the cosine of b.
How about here?
This would become the sine of
minus bt.
But that's simply the negative
of the sine of bt.
So, the only change would have
been to put a minus sign there.
Now, I don't care if I get y2
or negative y2 because what am I
going to do with it?
When I get it,
I'm going to write y,
the general solution,
as c1 y1 plus c2 y2.
So, if I get negative y2,
that just changes that
arbitrary constant from c2 to
minus c2, which is just as
arbitrary a constant.
So, in other words,
there's no reason to use the
other root because it doesn't
give anything new.
Now, there the story could
stop.
And, I would like it to stop,
frankly, but I don't dare
because there's a second
question.
And, I'm visiting recitations
not this semester,
but in previous semesters.
In 18.03, so many recitations
do this.
I have to partly inoculate you
against it, and partly tell you
that some of the engineering
courses do do it,
and therefore you probably
should learn it also.
So, there is another way of
proceeding, which is what you
might have thought.
Hey, look, we got two complex
roots.
That gives us two solutions,
which are different.
Neither one is a constant
multiple of the other.
So, the other approach is,
use, as a general solution,
y equals, now,
I'm going to put a capital C
here.
You will see why in just a
second, times e to the (a plus b
i) times t.
And then, I will use the other
solution: C2 times e to the (a
minus b i) t.
These are two independent
solutions.
And therefore,
can't I get the general
solution in that form?
Now, in a sense,
you can.
The whole problem is the
following, of course,
that I'm only interested in
real solutions.
This is a complex function.
This is another complex
function.
It's got an i in it,
in other words,
when I write it out as u plus
iv.
If I expect to be able to get a
real solution out of that,
that means I have to make,
allow these coefficients to be
complex numbers,
and not real numbers.
So, in other words,
what I'm saying is that an
expression like this,
where the a plus bi and a minus
bi are complex roots of that
characteristic equation,
is formally a very general,
complex solution to the
equation.
And therefore,
the problem becomes,
how, from this expression,
do I get the real solutions?
So, the problem is,
I accept these as the complex
solutions.
My problem is,
to find among all these guys
where C1 and C2 are allowed to
be complex, the problem is,
which of the green solutions
are real?
Now, there are many ways of
getting the answer.
There is a super hack way.
The super hack way is to say,
well, this one is C1 plus i d1.
This is C2 plus i d2.
And, I'll write all this out in
terms of what it is,
you know, cosine plus i sine,
and don't forget the e to the
at.
And, I will write it all out,
and it will take an entire
board.
And then, I will just see what
the condition is.
I'll write its real part,
and its imaginary part.
And then, I will say the
imaginary part has got to be
zero.
And, then I will see what it's
like.
That works fine.
It just takes too much space.
And also, it doesn't teach you
a few things that I think you
should know.
So, I'm going to give another.
So, let's say we can answer
this two ways:
by hack, in other words,
multiply everything out.
Multiply all out,
make the imaginary part equal
zero.
Now, here's a better way,
in my opinion.
What I'm trying to do is,
this is some complex function,
u plus iv.
How do I know when a complex
function is real?
I want this to be real.
Well, the hack method
corresponds to,
say, v must be equal to zero.
It's real if v is zero.
So, expand it out,
and see why v is zero.
There's a slightly more subtle
method, which is to change i to
minus i.
And, what?
And, see if it stays the same
because if I change i to minus i
and it turns out,
the expression doesn't change,
then it must have been real,
if the expression doesn't
change when I change I to minus
I.
Well, sure.
But you will see it works.
Now, that's what I'm going to
apply to this.
If I want this to be real,
I phrase the question,
I rephrase the question for the
green solution as change,
so I'm going to change i to
minus i in the green thing,
and that's going to give me
what conditions,
and that will give conditions
on the C's.
Well, let's do it.
In fact, it's easier done than
talked about.
Let's change,
take the green solution,
and change.
Well, I better recopy it,
C1.
So, these are complex numbers.
That's why I wrote them as
capital letters because little
letters you tend to interpret as
real numbers.
So, C1 e to the (a plus b i)t,
I'll recopy it quickly,
plus C2 e to the (a minus b i).
Okay, we're going to change i
to negative i.
Now, here's a complex number.
What happens to it when you
change i to negative i?
You change it into its-- Class?
What do we change it to?
Its complex conjugate.
And, the notation for complex
conjugate is you put a bar over
it.
So, in other words,
when I do that,
the C1 changes to C1 bar,
complex conjugate,
the complex conjugate of C1.
What happens to this guy?
This guy changes to e to the (a
minus b i) t.
This changes to the complex
conjugate of C2 now,
times e to the (a plus b i) t.
Well, I want these two to be
the same.
I want the two expressions the
same.
Why do I want them the same?
Because, if there's no change,
that will mean that it's real.
Now, when is that going to
happen?
That happens if,
well, here is this,
that.
If C2 should be equal to C1
bar, that's only one condition.
There's another condition.
C2 bar should equal C1.
So, I get two conditions,
but there's really only one
condition there because if this
is true, that's true too.
I simply put bars over both
things, and two bars cancel each
other out.
If you take the complex
conjugate and do it again,
you get back where you started.
Change i to minus i,
and then i to minus i again.
Well, never mind.
Anyway, these are the same.
This equation doesn't say
anything that the first one
didn't say already.
So, this one is redundant.
And, our conclusion is that the
real solutions to the equation
are, in their entirety,
I now don't need both C2 and
C1.
One of them will do,
and since I'm going to write it
out as a complex number,
I will write it out in terms of
its coefficient.
So, it's C1.
Let's just simply write it.
C plus i times d,
that's the coefficient.
That's what I called C1 before.
And, that's times e to the (a
plus b i) t.
There's no reason why I put bi
here and id there,
in case you're wondering,
sheer caprice.
And what's the other term?
Now, the other term is
completely determined.
Its coefficient must be C minus
i d times e to the
(a minus b i) t.
In other words,
this thing is perfectly
general.
Any complex number times that
first root you use,
exponentiated,
and the second term can be
described as the complex
conjugate of the first.
The coefficient is the complex
conjugate, and this part is the
complex conjugate of that.
Now, it's in this form,
many engineers write the
solution this way,
and physicists,
too, so, scientists and
engineers we will include.
Write the solution this way.
Write the real solutions this
way in that complex form.
Well, why do they do something
so perverse?
You will have to ask them.
But, in fact,
when we studied Fourier series,
we will probably have to do
something, have to do that at
one point.
If you work a lot with complex
numbers, it turns out to be,
in some ways,
a more convenient
representation than the one I've
given you in terms of sines and
cosines.
Well, from this,
how would I get,
suppose I insisted,
well, if someone gave it to me
in that form,
I don't see how I would convert
it back into sines and cosines.
And, I'd like to show you how
to do that efficiently,
too, because,
again, it's one of the
fundamental techniques that I
think you should know.
And, I didn't get a chance to
say it when we studied complex
numbers that first lecture.
It's in the notes,
but it doesn't prove anything
since I don't think it made you
use it in an example.
So, the problem is,
now, by way of finishing this
up, too, to change this to the
old form, I mean the form
involving sines and cosines.
Now, again, there are two ways
of doing it.
The hack way is you write it
all out.
Well, e to the (a plus b i)t
turns into e to the a t times
cosine this plus i sine that.
And, the other term does,
too.
And then you've got stuff out
front.
And, the thing stretches over
two boards.
But you group all the terms
together.
You finally get it.
By the way, when you do it,
you'll find that the imaginary
part disappears completely.
It has to because that's the
way we chose the coefficients.
So, here's the hack method.
Write it all out:
blah, blah, blah,
blah, blah, blah,
blah, and nicer.
Nicer, and teach you something
you're supposed to know.
Write it this way.
First of all,
you notice that both terms have
an e to the a t
factor.
Let's get rid of that right
away.
I'm pulling it out front
because that's automatically
real, and therefore,
isn't going to affect the rest
of the answer at all.
So, let's pull out that,
and what's left?
Well, what's left,
you see, involves just the two
parameters, C and d,
so I'm going to have a C term.
And, I'm going to have a d
term.
What multiplies the arbitrary
constant, C?
Answer: after I remove the e to
the a t,
what multiplies it is,
e to the b i t plus e to the --
e to the b i t.
Let's write it i b t.
And, the other term is plus e
to the negative i b t.
See how I got that,
pulled it out?
And, how about the d?
What goes with d?
d goes with,
well, first of all,
there's an I in front that i
better not forget.
And then, the rest of it is i.
So, it's i d times,
it's e to the b i t,
e to the i b t minus,
now, e to the minus i b t.
So, that's the way the solution
looks.
It doesn't look a lot better,
but now you must use the magic
formulas, which,
I want you to know as well as
you know Euler's formula,
even better than you know
Euler's formula.
They're a consequence of
Euler's formula.
They're Euler's formula read
backwards.
Euler's formula says you've got
a complex exponential here.
Here's how to write it in terms
of sines and cosines.
The backwards thing says you've
got a sine or a cosine.
Here is the way to write it in
terms of complex exponentials.
And, remember,
the way to do it is,
cosine a is equal to e to the i
a t, i a, plus e to the negative
i a divided by two.
And, sine of a is almost the
same thing, except you use a
minus sign.
And, what everybody forgets,
you have to divide by i.
So, this is a backwards version
of Euler's formula.
The two of them taken together
are equivalent to Euler's
formula.
If I took cosine a,
multiply this through by i,
and added them up,
on the right-hand side I'd get
exactly e to the ia.
I'd get Euler's formula,
in other words.
All right, so,
what does this come out to be,
finally?
This particular sum of
exponentials,
you should always recognize as
real.
You know it's real because when
you change i to minus i,
the two terms switch.
And therefore,
the expression doesn't change.
What is it?
This part is twice the cosine
of bt.
What's this part?
This part is 2 i times the sine
of bt.
And so, what does the whole
thing come to be?
It is e to the a t times 2C
cosine bt plus i times,
did I lose possibly a,
no it's okay,
minus i times i is minus,
so, minus 2d times the sine of
bt.
Shall I write that out?
So, in other words,
it's e to the a t times 2C
cosine b t minus 2d times the
sine of b t,
which is, since 2C and negative
2d are just arbitrary constants,
just as arbitrary as the
constants of C and d themselves
are.
This is our old form of writing
the real solution.
Here's the way using science
and cosines, and there's the way
that uses complex numbers and
complex functions throughout.
Notice they both have two
arbitrary constants in them,
C and d, two arbitrary
constants.
That, you expect.
But that has two arbitrary
constants in it,
too, just the real and
imaginary parts of that complex
coefficient, C plus i d.
Well, that took half the
period, and it was a long,
I don't consider it a
digression because learning
those ways of dealing with
complex numbers of complex
functions is a fairly important
goal in this course,
actually.
But let's get back now to
studying what the oscillations
actually look like.
Okay, well, I'd like to save a
little time, but very quickly,
you don't have to reproduce
this sketch.
I remember very well from
Friday to Monday,
but I can't expect you to for a
variety of reasons.
I mean, I have to think about
this stuff all weekend.
And you, God forbid.
So, here's the picture,
and I won't explain anymore
what's in it,
except there's the mass.
Here is the spring constant,
the spring with its constant
here.
Here's the dashpot with its
constant.
The equation is from Newton's
law: m x double,
so this will be x,
and here's, let's say,
the equilibrium point is over
here.
It looks like m x double prime;
we derived this last time,
plus c x prime plus k x equals
zero.
And now, if I put that in
standard form,
it's going to look like x
double prime plus c over m x
prime plus k over m times x
equals zero.
And, finally,
the standard form in which your
book writes it,
which is good,
it's a standard form in general
that is used in the science and
engineering courses.
One writes this as,
just to be perverse,
I'm going to change x back to
y, okay, mostly just to be
eclectic, to get you used to
every conceivable notation.
So, I'm going to write this to
change x to y.
So, that's going to become y
double prime.
And now, this is given a new
name, p, except to get rid of
lots of twos,
which would really screw up the
formulas, make it 2p.
You will see why in a minute.
So, there's 2p times y prime,
and this thing we
are going to call omega nought
squared.
Now, that's okay.
It's a positive number.
Any positive number is the
square of some other positive
number.
Take a square root.
You will see why,
it makes the formulas much
pretty to call it that.
And, it makes also a lot of
things much easier to remember.
So, all I'm doing is changing
the names of the constants that
way in order to get better
formulas, easier to remember
formulas at the end.
Now, we are interested in the
case where there is
oscillations.
In other words,
I only care about the case in
which this has complex roots,
because if it has just real
roots, that's the over-damped
case.
I don't get any oscillations.
By far, oscillations are by far
the more important of the cases,
I mean, just because,
I don't know,
I could go on for five minutes
listing things that oscillate,
oscillations,
you know, like this.
So they can oscillate by going
to sleep, and waking up,
and going to sleep,
and waking up.
They could oscillate.
So, that means we're going to
get complex roots.
The characteristic equation is
going to be r squared plus 2p.
So, p is a constant,
now, right?
Often, p I use in this position
to indicate a function of t.
But here, p is a constant.
So, r squared plus 2p times r
plus omega nought squared is
equal to zero.
Now, what are its roots?
Well, you see right away the
first advantage in putting in
the two there.
When I use the quadratic
formula, it's negative 2p over
two.
Remember that two in the
denominator.
So, that's simply negative p.
And, how about the rest?
Plus or minus the square root
of, now do it in your head.
4p squared minus 4 omega nought
squared.
So, there's a four in both of
those terms.
When I pull it outside becomes
two.
And, the two in the denominator
is lurking, waiting to
annihilate it.
So, that two disappears
entirely, and it will we are
left with is,
simply, p squared minus omega
nought squared.
Now, whenever people write
quadratic equations,
and arbitrarily put a two in
there, it's because they were
going to want to solve the
quadratic equation using the
quadratic formula,
and they don't want all those
twos and fours to be cluttering
up the formula.
That's what we are doing here.
Okay, now, the first case is
where p is equal to zero.
This is going to explain
immediately why I wrote that
omega nought squared,
as you probably already know
from physics.
If p is equal to zero,
the mass isn't zero.
Otherwise, nothing good would
be happening here.
It must be that the damping is
zero.
So, p is equal to zero
corresponds to undamped.
There is no dashpot.
The oscillations are undamped.
And, the equation,
then, becomes the solutions,
then, are, well,
the equation becomes the
equation of simple harmonic
motion, which,
I think you already are used to
writing in this form.
And, the reason you're writing
in this form because you know
when you do that,
this becomes the circular
frequency of the oscillations.
The solutions are pure
oscillations,
and omega nought is
the circular frequency.
So, right away from the
equation itself,
if you write it in this form,
you can read off what the
frequency of the solutions is
going to be, the circular
frequency of the solutions.
Now, the solutions themselves,
of course, look like,
the general solutions look like
y equal, in this particular
case, the p part is zero.
This is zero.
It's simply,
so, in this case,
r is equal to omega nought i
times omega naught plus or
minus, but as before we don't
bother with the minus sign since
one of those roots is good
enough.
And then, the solutions are
simply c1 cosine omega nought t
plus c2 sine omega nought t.
That's if you write it out in
the sign, and if you write it
using the trigonometric
identity, then the other way of
writing it is a times the cosine
of omega nought t.
But now you will have to put it
a phase lag.
So, you have those two forms of
writing it.
And, I assume you remember
writing the little triangle,
which converts one into the
other.
Okay, so this justifies calling
this omega nought squared
rather than k over m.
And now, the question is what
does the damp case look like?
It requires a somewhat closer
analysis, and it requires a
certain amount of thinking.
So, let's begin with an epsilon
bit of thinking.
So, here's my question.
So, in the damped case,
I want to be sure that I'm
getting oscillations.
When do I get oscillations if,
well, we get oscillations if
those roots are really complex,
and not masquerading.
Now, when are the roots going
to be really complex?
This has to be,
the inside has to be negative.
p squared minus omega squared
must be negative.
p squared minus omega nought
squared must be less than zero
so that we are taking a square
root of negative number,
and we are getting a real
complex roots,
really complex roots.
In other words,
now, this says,
remember these numbers are all
positive, p and omega nought are
positive.
So, the condition is that p
should be
less than omega nought.
In other words,
the damping should be less than
the circular frequency,
except p is not the damping.
It's half the damping,
and it's not really the damping
either because it involved the
m, too.
You'd better just call it p.
Naturally, I could write the
condition out in terms of c,
m, and k.
So, your book does that,
but I'm not going to.
It gives it in terms of c,
m, and k, which somebody might
want to know.
But, you know,
we don't have to do everything
here.
Okay, so let's assume that this
is true.
What is the solution look like?
Well, we already experimented
with that last time.
Remember, there was some
guiding thing which was an
exponential.
And then, down here,
we wrote the negative.
So, this was an exponential.
In fact, it was the
exponential, e to the negative
pt.
And, in between that,
the curve tried to do its
thing.
So, the solution looks sort of
like this.
It oscillated,
but it had to use that
exponential function as its
guidelines, as its amplitude,
in other words.
Now, this is a truly terrible
picture.
It's so terrible,
it's unusable.
Okay, this picture never
happened.
Unfortunately,
this is not my forte along with
a lot of other things.
All right, let's try it better.
Here's our better picture.
Okay, there's the exponential.
At this point,
I'm supposed to have a lecture
demonstration.
It's supposed to go up on the
thing, so you can all see it.
But then, you wouldn't be able
to copy it.
So, at least we are on even
terms now.
Okay, how does the actual curve
look?
Well, I'm just trying to be
fair.
That's all.
Okay, after a while,
the point is,
just so we have something to
aim at, let's say,
okay, here we are going to go,
we're going to get down through
there.
Okay then, this is our better
curve.
Okay, so I am a solution,
a particular solution
satisfying this initial
condition.
I started here,
and that was my initial
velocity.
The slope of that thing gave me
the initial velocity.
Now, the interesting question
is, the first,
in some ways,
the most interesting question,
though there will be others,
too, is what is this spacing?
Well, that's a period.
And now, it's half a period.
I clearly ought to think of
this as the whole period.
So, let's call that,
I'm going to call this pi over,
so this spacing here,
from there to there,
I will call that pi divided by
omega one because this,
from here to here,
should be, I hope,
twice that, two pi over omega
one.
Now, my question is,
so this, for a solution,
it's, in fact,
is going to cross the axis
regularly in that way.
My question is,
how does this period,
so this is going to be its half
period.
I will put period in quotation
marks because this isn't really
a periodic function because it's
decreasing all the time in
amplitude.
But, it's trying to be
periodic.
At lease it's doing something
periodically.
It's crossing the axis
periodically.
So, this is the half period.
Two pi over omega one
would be its full
period.
What I want to know is,
how does that half period,
or how does-- omega one is
called its pseudo-frequency.
This should really be called
its pseudo-period.
Everything is pseudo.
Everything is fake here.
Like, the amoeba has its fake
foot and stuff like that.
Okay, so this is its
pseudo-period,
pseudo-frequency,
pseudo-circular frequency,
but that's hopeless.
I guess it should be circular
pseudo-frequency,
or I don't know how you say
that.
I don't think pseudo is a word
all by itself,
not even in 18.03,
circular.
Okay, here's my question.
If the damping goes up,
this is the damping term.
If the damping goes up,
what happens to the
pseudo-frequency?
The frequency is how often the
curve, this is high-frequency,
and this is low-frequency,
okay?
So, my question is,
which way does the frequency
go?
If the damping goes up,
does the frequency go up or
down?
Down.
I mean, I'm just asking you to
answer intuitively on the basis
of your intuition about how this
thing explains,
how this thing goes,
and now let's get the formula.
What, in fact,
is omega one?
What is omega one?
The answer is when I solve the
equation, so,
r is now, so in other words,
if omega one is,
sorry, if I have p,
if p is no longer zero as it
was in the undamped case,
what is the root,
now?
Okay, well, the root is minus p
plus or minus the square root of
p squared, --
-- now I'm going to write it
this way, minus,
to indicate that it's really a
negative number,
omega squared minus p squared.
Now, I'm going to call this,
because you see when I change
this to sines and cosines,
the square root of this number
is what's going to become that
new frequency.
I'm going to call that minus p
plus or minus the square root of
minus omega one squared.
That's going to be the new
frequency.
And therefore,
the root is going to change so
that the corresponding solution
is going to look,
how?
Well, it's going to be e to the
negative pt times,
let's write it out first in
terms of sines and cosines,
times the cosine of,
well, the square root of omega
one squared is omega one.
But, there's an i out front
because of the negative sign in
front of that.
So, it's going to be the cosine
of omega one t
plus c2 times the sine of omega
one t.
Or, if you prefer to write it
out in the other form,
it's e to the minus p t times
some amplitude,
which depends on c1 and c2,
times the cosine of omega one t
minus the phase lag.
Now, when I do that,
you see omega one is
this pseudo-frequency.
In other words,
this number omega one is the
same one that I identified here.
And, why is that?
Well, because,
what are two successive times?
Suppose it crosses,
suppose the solution crosses
the x-axis, sorry,
y-- the t-axis.
For the first time,
at the point t1,
what's the next time it crosses
t2?
Let's jump to the two times
across it.
So, I want this to be a whole
period, not a half period.
What's t2?
Well, I say that t2 is nothing
but 2 pi divided by omega one.
And, you can see that because
when I plug in,
if it's zero,
if I have a point where it's
zero, so, omega one t minus phi,
when will it be zero for the
first time?
Well, that will be when the
cosine has to be zero.
So, it will be some multiple
of, it will be,
say, pi over two.
Then, the next time this
happens will be,
if that happens at t1,
then the next time it happens
will be at t1 plus 2 pi divided
by omega one.
That will also be pi over two
plus how much?
Plus 2 pi, which is the next
time the cosine gets around and
is doing its thing,
becoming zero as it goes down,
not as it's coming up again.
In other words,
this is what you should add to
the first time to get this
second time that the cosine
becomes zero coming in the
direction from top to the
bottom.
So, this is,
in fact, the frequency with
which it's crossing the axis.
Now, notice,
I'm running out of boards.
What a disaster!
In that expression,
take a look at it.
I want to know what depends on
what.
So, p, in that,
we got constants.
We got p.
We got phi.
We got A.
What else we got?
Omega one.
What do these things depend
upon?
You've got to keep it firmly in
mind.
This depends only on the ODE.
It's basically the damping.
It depends on c and m.
Essentially, it's c over 2m
actually.
How about phi?
Well, phi, what else depends
only on the ODE?
Omega one depends
only on the ODE.
What's the formula for omega
one?
Omega one squared.
Where do we have it?
Omega one squared,
I never wrote the formula for
you.
So, we have omega nought
squared minus p squared equals
omega one squared.
What's the relation between
them?
That's the Pythagorean theorem.
If this is omega nought,
then this omega one,
this is p.
They make a little,
right triangle in other words.
The omega one depends on the
spring.
So, it's equal to,
well, it's equal to that thing.
So, it depends on the damping.
It depends upon the damping,
and it depends on the spring
constant.
How about the phi and the A?
What do they depend on?
They depend upon the initial
conditions.
So, the mass of constants,
they have different functions.
What's making this complicated
is that our answer needs four
parameters to describe it.
This tells you how fast it's
coming down.
This tells you the phase lag.
This amplitude modifies,
it tells you whether the
exponential curve starts going,
is like that or goes like this.
And, finally,
the omega one is this
pseudo-frequency,
which tells you how it's
bobbing up and down.
