WEBVTT
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GILBERT STRANG: OK,
it's time to move on
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to second order equations.
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First order equations,
we've done pretty carefully.
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Second order equations
are a step harder.
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But they come up in nature,
they come in every application,
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because they include
an acceleration,
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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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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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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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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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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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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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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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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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Well, that's going to
involve the initial velocity,
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the derivative, at t equals 0.
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Because the
derivative of the sine
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is the cosine, which
equals 1 at t equals 0,
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the derivative of this
is the sine, which is 0.
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So that when I'm looking
at y prime, the derivative,
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I'm looking here at t equals 0.
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And I want y prime of 0.
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But I don't just
want y prime of 0.
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Do you see that that doesn't
have the right derivative,
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at t equals 0?
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Because when I take the
derivative and omega
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n-- that constant, you remember
that constant-- the derivative
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of this will bring
out an omega n.
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So I better have an omega
n down here to cancel it.
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And now I've got it.
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That tells me the
motion, forever and ever.
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Energy is constant.
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Potential energy
plus kinetic energy,
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I could speak about energy.
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But I won't.
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That motion continues
forever, free harmonic motion.
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OK, it goes on and on.
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OK.
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And again, I could write this in
terms of complex exponentials.
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But I'm pretty happy
with that form.
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It's hard to beat that form.
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OK, so what else to do here.
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First of all, we're going
to have cosine omega
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t's for quite awhile.
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I better draw the graph of
that simple, familiar function.
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And so here's a graph
of cosine omega t.
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So here's t.
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Here's cosine of omega.
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Here's 0.
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Here's 2-- well, let
me see what I get.
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So I go cosine of omega, cosine
of omega t is what I'm drawing,
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not cosine t.
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Cosine t would go 0 to 2 pi.
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But I have cosine of omega t.
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So cosine of omega t is
what I want to graph.
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So it starts at 1,
and it comes back.
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It drops, comes back
up, comes back to 1.
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But what is this t final?
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The period, this t is the
period of the oscillation.
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It's the time it takes for
the swing to go up and back.
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And what is that?
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That would be-- so I'm
graphing cosine of omega t,
00:13:59.910 --> 00:14:00.850
is what I'm graphing.
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So omega t starts out
at 0, where t is 0.
00:14:08.070 --> 00:14:14.230
And it goes up to-- so I want
omega t, when I get here,
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this omega t should be 2 pi.
00:14:19.150 --> 00:14:21.420
Right?
00:14:21.420 --> 00:14:31.310
Then I've completed the cosine,
one rotation around the circle,
00:14:31.310 --> 00:14:34.080
one movement of the
pendulum back and forth,
00:14:34.080 --> 00:14:36.140
is in that picture now.
00:14:36.140 --> 00:14:40.270
OK, so omega t is 2 pi, right?
00:14:40.270 --> 00:14:41.920
Right.
00:14:41.920 --> 00:14:43.130
The period is t.
00:14:43.130 --> 00:14:47.120
I think of omega as
the circular frequency.
00:14:47.120 --> 00:14:56.915
Omega is in radians per second.
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That's the units.
00:15:04.500 --> 00:15:08.810
And units truly are
important to keep track.
00:15:08.810 --> 00:15:12.810
Omega, this is omega,
radians per second,
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when I multiply by the
period, the t in seconds,
00:15:16.980 --> 00:15:19.290
I get 2 pi radians.
00:15:19.290 --> 00:15:24.200
OK now, in engineers
and in everyday use,
00:15:24.200 --> 00:15:29.970
there's another frequency called
f, for frequency probably.
00:15:29.970 --> 00:15:34.570
OK, so you should know about f.
00:15:34.570 --> 00:15:39.080
And its frequency is in hertz.
00:15:39.080 --> 00:15:49.550
So f is measured in hertz,
H-E-R-T-Z, named after the guy.
00:15:49.550 --> 00:15:55.770
Not the tomato ketchup guy,
but-- oh, that's Heinz anyway.
00:15:55.770 --> 00:15:57.480
Sorry about that.
00:15:57.480 --> 00:16:02.950
Hertz -- not the car, that's
what I was trying to say,
00:16:02.950 --> 00:16:10.100
but the German guy who
was involved, early,
00:16:10.100 --> 00:16:11.640
with this stuff.
00:16:11.640 --> 00:16:12.810
So what is f?
00:16:17.870 --> 00:16:20.630
f times t is 1.
00:16:20.630 --> 00:16:24.135
Instead of dealing with 2
pi, which count to radians,
00:16:24.135 --> 00:16:31.770
the 1 just counts complete
loops, complete oscillations.
00:16:31.770 --> 00:16:39.350
So f compared to know omega,
f is smaller by a factor 2 pi.
00:16:39.350 --> 00:16:43.750
So f times t is
1. f is 1 over t.
00:16:43.750 --> 00:16:46.290
Omega is 2 pi over t.
00:16:46.290 --> 00:16:48.770
So putting these
together, all I'm saying
00:16:48.770 --> 00:16:54.190
is, omega is 2 pi times f.
00:16:54.190 --> 00:16:58.960
So when we say we're
getting 60 cycles-- so
00:16:58.960 --> 00:17:02.520
that's what I would
measure tf in.
00:17:02.520 --> 00:17:10.495
f would be in cycles,
is cycles per second.
00:17:13.430 --> 00:17:16.209
One cycle, 2 pi radians.
00:17:19.290 --> 00:17:21.440
This isn't big, heavy math.
00:17:21.440 --> 00:17:24.470
But it's more important
than a lot of math,
00:17:24.470 --> 00:17:28.170
just to get these
letters straight.
00:17:28.170 --> 00:17:31.030
So there's capital
T, the period,
00:17:31.030 --> 00:17:32.990
and two measures of frequency.
00:17:32.990 --> 00:17:36.170
One is omega in
radians per second,
00:17:36.170 --> 00:17:40.830
and the other is f in
full cycles per second.
00:17:40.830 --> 00:17:42.740
So one is 2 pi times the other.
00:17:42.740 --> 00:17:43.300
Good.
00:17:43.300 --> 00:17:44.070
OK.
00:17:44.070 --> 00:17:45.730
And we have this.
00:17:45.730 --> 00:17:46.450
OK.
00:17:46.450 --> 00:17:54.440
I think we've got the
key ideas here, then,
00:17:54.440 --> 00:18:01.360
for unforced motion, pure
oscillation going on forever.
00:18:01.360 --> 00:18:07.960
And let me just write
what I already mentioned.
00:18:07.960 --> 00:18:15.700
A different way to express
yn would be with small c's, e
00:18:15.700 --> 00:18:20.210
to the i omega nt.
00:18:20.210 --> 00:18:26.400
And c2 e to the
minus i omega nt.
00:18:26.400 --> 00:18:33.280
All I'm saying is that this
form, with exponentials,
00:18:33.280 --> 00:18:40.100
is entirely equivalent to this
form, with cosine and sine.
00:18:40.100 --> 00:18:44.190
That's form allows me two
constants, capital C1 and C2.
00:18:44.190 --> 00:18:48.080
This form allows me two
constants, little one and c2.
00:18:48.080 --> 00:18:49.850
And from that, I have this.
00:18:49.850 --> 00:18:50.950
From this, I have this.
00:18:50.950 --> 00:18:55.150
So we really do have
exponentials here.
00:18:55.150 --> 00:18:59.680
And the key message is
that for pure oscillation,
00:18:59.680 --> 00:19:07.070
those exponentials are pure
imaginary exponent, i omega nt.
00:19:07.070 --> 00:19:11.790
OK, that's the best example,
the simplest example,
00:19:11.790 --> 00:19:13.500
the first example.
00:19:13.500 --> 00:19:15.050
Thanks.