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00:00:00,499 --> 00:00:03,350
GILBERT STRANG: OK,
it's time to move on

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00:00:03,350 --> 00:00:05,560
to second order equations.

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00:00:05,560 --> 00:00:10,220
First order equations,
we've done pretty carefully.

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00:00:10,220 --> 00:00:13,990
Second order equations
are a step harder.

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00:00:13,990 --> 00:00:18,080
But they come up in nature,
they come in every application,

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because they include
an acceleration,

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00:00:21,196 --> 00:00:23,360
a second derivative.

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OK, so this would be a second
order equation, because

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of that second derivative.

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00:00:31,670 --> 00:00:35,430
I'm often going to have
constant a, b, and c.

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We have enough
difficulties to it

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without allowing
those to change.

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So a, b, c constants, and
doing the null solution

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to start with.

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Later, there'll be a forcing
term on the right-hand side.

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But today, for this
video, null solutions.

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And the point is, now,
what's new is that there

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are two null solutions.

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So y null will be a
combination of e--

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and they're both exponentials.

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Having constant
coefficients there

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means exponentials
in the solution.

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So e to the sum
exponent, and another one

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to a, hopefully,
different exponent.

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Sometimes if s1 equals s2,
that'll be a special case

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and we'll have a slight change.

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But this is typical.

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So we have two constants, c1
and c2, in the null solutions.

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And we need two
initial conditions

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to determine those constants.

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So previously, for a first order
equation, we were given y of 0.

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Now, when we have acceleration,
we give the initial velocity,

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y prime of 0.

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May I use prime as a
shorthand for derivative?

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y prime of 0 is dydt, at 0.

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So two condition will,
at the right time,

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determine those two constants.

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And just to say again,
so second derivative

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will be y double prime.

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And that represents,
in physics, it

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represents acceleration-- change
in velocity, change in y prime,

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is y double prime.

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And in a graph of a function,
y double prime shows up

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in bending of the graph.

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Because bending is
a change in slope.

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Bending is a change
in the slope.

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And the slope is y prime,
the first derivative.

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00:03:03,660 --> 00:03:06,670
So to measure changes
in y prime, which

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will bend the graph, my chalk
would be a tangent line.

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00:03:12,300 --> 00:03:15,400
But if that changes,
that gives us

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a y double prime, a
second derivative.

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OK, so I'm ready
for some examples.

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And the first example-- the
most basic equation of motion

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in physics and engineering,
I would say-- it's

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called harmonic motion.

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And b is 0, that's
the key point.

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b is 0.

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It's Newton's law.

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And so a will be the mass m.

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y double prime,
second derivative.

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b is 0.

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Later, b will be a damping term,
a friction term, a resistance

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term.

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But let's have that 0.

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00:04:05,280 --> 00:04:08,500
So we're going to have
perpetual motion, here.

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Plus the force-- so
this is Newton's law.

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ma is f, f equal ma.

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The force is proportional--
with a minus sign,

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so it's going to come on this
side as a plus-- proportional

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to y.

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There's the equation.

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No y prime term.

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my double prime,
plus ky equals 0.

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Starting from an
initial position,

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and an initial velocity,
it's like a spring

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going up and down, or a clock
pendulum going back and forth.

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And we'll see it, so
it's going to be-- well,

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we want to solve that equation.

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So do we see solutions to that?

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So if m and k were 1,
suppose m and k were 1.

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I'm looking for a
second derivative plus

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the function is 0.

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Second derivative is
minus the function.

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I immediately think of
sine t, and cosine t.

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Sines and cosines, because
the second derivative--

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the first derivative
of a sine is cosine.

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The second derivative
is minus the sine.

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It gives us the minus sign
here, the plus sign there, 0.

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So the special solutions here
are y is-- this is the null.

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I'm finding the null solution.

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Is, let me call them
c1 times a cosine.

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And now I have to figure
out the cosine of what?

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I want the cosine to satisfy,
to be a null solution,

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satisfy my equation.

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Let me put it in.

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If there's a square
root of k over mt.

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And you have to
see that if I take

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two derivatives of the
cosine, that will produce

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00:06:20,120 --> 00:06:23,170
minus the cosine, which I want.

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And because of that, the chain
will bring out this square root

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twice.

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So it will bring out the factor
k over m for y double prime.

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And that factor, k over
m, the m's will cancel

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and I'll have the k
that matches that k.

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It's a solution.

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And the other solution
is just like it.

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It's the sine.

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Sine of this square
root of k over m, t.

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That's worth putting
a box around.

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That's what I mean by
free harmonic motion.

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Something is just oscillating.

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In rotation problems,
something is just

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going around a circle
at a constant speed.

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And notice, these are
not the same as those.

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Cosines are related
to exponentials,

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but not identical.

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So I could write the answer
this way, using cosine and sine.

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Or as you'll see, I can
write the same formula

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using exponentials,
complex exponentials.

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Everybody remembers
the big formula

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that allows complex
numbers in here

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is Euler's formula, that
the exponential of i omega t

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is the cosine plus i
times the sine of omega t.

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I'll write it again.

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That's the solution.

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It's got two null solutions.

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They're independent.

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They're different.

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And we've got two constants.

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Because our equation
is linear, we

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can safely multiply by any
constant, and add solutions,

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and they stay solutions because
we have a linear equation,

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and 0 on the right-hand side.

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Good.

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Of course, we can't write
this square root of k over m

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forever.

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Let me do what everybody
does-- introduce omega.

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It's omega natural.

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The n here stands for
the natural frequency,

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the frequency that
that clock is going at.

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And that is the square
root of k over m.

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So our equation, we could
rewrite that equation.

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Let me rewrite that
equation to make it simple.

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I'll divide by m.

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No problem, to divide by m.

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So then I have y
double prime, plus k

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over m, which is
omega n squared,

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the natural frequency squared.

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y equals 0.

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Let's put a box around that one,
because you couldn't be better

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than that.

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The constant a is 1.

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The constant b is 0.

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The constant c is a known
omega n squared, depending

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on the pendulum itself.

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OK, and the solutions then,
I'll just copy this solution.

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y null is c1 cosine of omega nt.

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Of course, omega n
is that square rood.

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And c2 sine of omega nt.

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00:10:00,740 --> 00:10:02,560
Oh, well, wait a minute.

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I can figure out
what c1 and c2 are ,

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coming from the initial
conditions, right?

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The initial conditions,
if I plug in t equals 0,

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then I want to get
the answer y of 0.

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The known initial
condition, the place

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the pendulum started swinging
from, the place the spring,

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you pull the spring
a distance y of 0.

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You let go.

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At t equals 0, so I'm
plugging in t equals 0, at t

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equals 0, that's 0.

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Forget the sine.

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This is 1.

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So I discover that
c1 should be y of 0.

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Simple.

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00:10:48,840 --> 00:10:50,400
c1 is y of 0.

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Because that gives me the
right answer, at t equals 0.

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And what about c2?

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Can I figure out c2?

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00:11:04,090 --> 00:11:08,520
Well, that's going to
involve the initial velocity,

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00:11:08,520 --> 00:11:12,220
the derivative, at t equals 0.

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Because the
derivative of the sine

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00:11:14,170 --> 00:11:19,190
is the cosine, which
equals 1 at t equals 0,

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00:11:19,190 --> 00:11:21,970
the derivative of this
is the sine, which is 0.

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00:11:21,970 --> 00:11:26,630
So that when I'm looking
at y prime, the derivative,

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00:11:26,630 --> 00:11:29,470
I'm looking here at t equals 0.

186
00:11:29,470 --> 00:11:33,000
And I want y prime of 0.

187
00:11:33,000 --> 00:11:35,370
But I don't just
want y prime of 0.

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00:11:35,370 --> 00:11:39,050
Do you see that that doesn't
have the right derivative,

189
00:11:39,050 --> 00:11:40,400
at t equals 0?

190
00:11:40,400 --> 00:11:42,530
Because when I take the
derivative and omega

191
00:11:42,530 --> 00:11:49,040
n-- that constant, you remember
that constant-- the derivative

192
00:11:49,040 --> 00:11:51,850
of this will bring
out an omega n.

193
00:11:51,850 --> 00:11:56,240
So I better have an omega
n down here to cancel it.

194
00:11:56,240 --> 00:11:58,680
And now I've got it.

195
00:11:58,680 --> 00:12:05,250
That tells me the
motion, forever and ever.

196
00:12:05,250 --> 00:12:07,070
Energy is constant.

197
00:12:07,070 --> 00:12:09,370
Potential energy
plus kinetic energy,

198
00:12:09,370 --> 00:12:10,940
I could speak about energy.

199
00:12:10,940 --> 00:12:13,770
But I won't.

200
00:12:13,770 --> 00:12:19,980
That motion continues
forever, free harmonic motion.

201
00:12:19,980 --> 00:12:22,740
OK, it goes on and on.

202
00:12:22,740 --> 00:12:23,750
OK.

203
00:12:23,750 --> 00:12:29,890
And again, I could write this in
terms of complex exponentials.

204
00:12:29,890 --> 00:12:33,180
But I'm pretty happy
with that form.

205
00:12:33,180 --> 00:12:35,120
It's hard to beat that form.

206
00:12:35,120 --> 00:12:40,030
OK, so what else to do here.

207
00:12:40,030 --> 00:12:44,950
First of all, we're going
to have cosine omega

208
00:12:44,950 --> 00:12:48,930
t's for quite awhile.

209
00:12:48,930 --> 00:12:54,160
I better draw the graph of
that simple, familiar function.

210
00:12:54,160 --> 00:12:59,500
And so here's a graph
of cosine omega t.

211
00:12:59,500 --> 00:13:01,930
So here's t.

212
00:13:01,930 --> 00:13:03,350
Here's cosine of omega.

213
00:13:03,350 --> 00:13:04,540
Here's 0.

214
00:13:04,540 --> 00:13:07,690
Here's 2-- well, let
me see what I get.

215
00:13:07,690 --> 00:13:12,130
So I go cosine of omega, cosine
of omega t is what I'm drawing,

216
00:13:12,130 --> 00:13:13,920
not cosine t.

217
00:13:13,920 --> 00:13:16,920
Cosine t would go 0 to 2 pi.

218
00:13:16,920 --> 00:13:20,110
But I have cosine of omega t.

219
00:13:20,110 --> 00:13:24,120
So cosine of omega t is
what I want to graph.

220
00:13:24,120 --> 00:13:27,725
So it starts at 1,
and it comes back.

221
00:13:27,725 --> 00:13:33,510
It drops, comes back
up, comes back to 1.

222
00:13:33,510 --> 00:13:37,290
But what is this t final?

223
00:13:37,290 --> 00:13:45,400
The period, this t is the
period of the oscillation.

224
00:13:45,400 --> 00:13:51,590
It's the time it takes for
the swing to go up and back.

225
00:13:51,590 --> 00:13:53,340
And what is that?

226
00:13:53,340 --> 00:13:59,910
That would be-- so I'm
graphing cosine of omega t,

227
00:13:59,910 --> 00:14:00,850
is what I'm graphing.

228
00:14:03,900 --> 00:14:08,070
So omega t starts out
at 0, where t is 0.

229
00:14:08,070 --> 00:14:14,230
And it goes up to-- so I want
omega t, when I get here,

230
00:14:14,230 --> 00:14:16,400
this omega t should be 2 pi.

231
00:14:19,150 --> 00:14:21,420
Right?

232
00:14:21,420 --> 00:14:31,310
Then I've completed the cosine,
one rotation around the circle,

233
00:14:31,310 --> 00:14:34,080
one movement of the
pendulum back and forth,

234
00:14:34,080 --> 00:14:36,140
is in that picture now.

235
00:14:36,140 --> 00:14:40,270
OK, so omega t is 2 pi, right?

236
00:14:40,270 --> 00:14:41,920
Right.

237
00:14:41,920 --> 00:14:43,130
The period is t.

238
00:14:43,130 --> 00:14:47,120
I think of omega as
the circular frequency.

239
00:14:47,120 --> 00:14:56,915
Omega is in radians per second.

240
00:15:01,180 --> 00:15:04,500
That's the units.

241
00:15:04,500 --> 00:15:08,810
And units truly are
important to keep track.

242
00:15:08,810 --> 00:15:12,810
Omega, this is omega,
radians per second,

243
00:15:12,810 --> 00:15:16,980
when I multiply by the
period, the t in seconds,

244
00:15:16,980 --> 00:15:19,290
I get 2 pi radians.

245
00:15:19,290 --> 00:15:24,200
OK now, in engineers
and in everyday use,

246
00:15:24,200 --> 00:15:29,970
there's another frequency called
f, for frequency probably.

247
00:15:29,970 --> 00:15:34,570
OK, so you should know about f.

248
00:15:34,570 --> 00:15:39,080
And its frequency is in hertz.

249
00:15:39,080 --> 00:15:49,550
So f is measured in hertz,
H-E-R-T-Z, named after the guy.

250
00:15:49,550 --> 00:15:55,770
Not the tomato ketchup guy,
but-- oh, that's Heinz anyway.

251
00:15:55,770 --> 00:15:57,480
Sorry about that.

252
00:15:57,480 --> 00:16:02,950
Hertz -- not the car, that's
what I was trying to say,

253
00:16:02,950 --> 00:16:10,100
but the German guy who
was involved, early,

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with this stuff.

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So what is f?

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f times t is 1.

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00:16:20,630 --> 00:16:24,135
Instead of dealing with 2
pi, which count to radians,

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the 1 just counts complete
loops, complete oscillations.

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So f compared to know omega,
f is smaller by a factor 2 pi.

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00:16:39,350 --> 00:16:43,750
So f times t is
1. f is 1 over t.

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Omega is 2 pi over t.

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00:16:46,290 --> 00:16:48,770
So putting these
together, all I'm saying

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is, omega is 2 pi times f.

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So when we say we're
getting 60 cycles-- so

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that's what I would
measure tf in.

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f would be in cycles,
is cycles per second.

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One cycle, 2 pi radians.

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This isn't big, heavy math.

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00:17:21,440 --> 00:17:24,470
But it's more important
than a lot of math,

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00:17:24,470 --> 00:17:28,170
just to get these
letters straight.

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00:17:28,170 --> 00:17:31,030
So there's capital
T, the period,

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and two measures of frequency.

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00:17:32,990 --> 00:17:36,170
One is omega in
radians per second,

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and the other is f in
full cycles per second.

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So one is 2 pi times the other.

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Good.

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00:17:43,300 --> 00:17:44,070
OK.

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And we have this.

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00:17:45,730 --> 00:17:46,450
OK.

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00:17:46,450 --> 00:17:54,440
I think we've got the
key ideas here, then,

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00:17:54,440 --> 00:18:01,360
for unforced motion, pure
oscillation going on forever.

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00:18:01,360 --> 00:18:07,960
And let me just write
what I already mentioned.

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00:18:07,960 --> 00:18:15,700
A different way to express
yn would be with small c's, e

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to the i omega nt.

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00:18:20,210 --> 00:18:26,400
And c2 e to the
minus i omega nt.

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00:18:26,400 --> 00:18:33,280
All I'm saying is that this
form, with exponentials,

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00:18:33,280 --> 00:18:40,100
is entirely equivalent to this
form, with cosine and sine.

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00:18:40,100 --> 00:18:44,190
That's form allows me two
constants, capital C1 and C2.

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00:18:44,190 --> 00:18:48,080
This form allows me two
constants, little one and c2.

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00:18:48,080 --> 00:18:49,850
And from that, I have this.

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00:18:49,850 --> 00:18:50,950
From this, I have this.

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00:18:50,950 --> 00:18:55,150
So we really do have
exponentials here.

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00:18:55,150 --> 00:18:59,680
And the key message is
that for pure oscillation,

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those exponentials are pure
imaginary exponent, i omega nt.

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00:19:07,070 --> 00:19:11,790
OK, that's the best example,
the simplest example,

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00:19:11,790 --> 00:19:13,500
the first example.

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00:19:13,500 --> 00:19:15,050
Thanks.