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Welcome back.

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And if you were covering your
eyes because you didn't want

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to see calculus, I think you
can open your eyes again.

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There shouldn't be any
significant displays of

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calculus in this video.

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But just to review what we went
over, we just said, OK if

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we have a spring-- and I drew it
vertically this time-- but

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pretend like there's no gravity,
or maybe pretend like

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we're viewing-- we're looking at
the top of a table, because

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we don't want to look
at the effect of

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a spring and gravity.

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We just want to look at
a spring by itself.

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So this could be in deep space,
or something else.

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But we're not thinking
about gravity.

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But I drew it vertically just
so that we can get more

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intuition for this curve.

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Well, we started off saying is
if I have a spring and 0-- x

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equals 0 is kind of the natural
resting point of the

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spring, if I just let this
mass-- if I didn't pull on the

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spring at all.

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But I have a mass attached to
the spring, and if I were to

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stretch the spring to point A,
we said, well what happens?

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Well, it starts with very little
velocity, but there's a

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restorative force, that's going
to be pulling it back

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towards this position.

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So that force will accelerate
the mass, accelerate the mass,

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accelerate the mass, until
it gets right here.

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And then it'll have a lot of
velocity here, but then it'll

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start decelerating.

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And then it'll decelerate,
decelerate, decelerate.

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Its velocity will stop, and
it'll come back up.

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And if we drew this as
a function of time,

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this is what happens.

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It starts moving very
slowly, accelerates.

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At this point, at x equals 0,
it has its maximum speed.

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So the rate of change of
velocity-- or the rate of

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change of position is fastest.
And we can see the slope is

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very fast right here.

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And then, we start slowing
down again, slowing down,

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until we get back to
the spot of A.

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And then we keep going up and
down, up and down, like that.

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And we showed that actually,
the equation for the mass's

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position as a function of time
is x of t-- and we used a

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little bit of differential
equations to prove it.

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But this equation-- not that I
recommend that you memorize

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anything-- but this is a pretty

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useful equation to memorize.

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Because you can use it to
pretty much figure out

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anything-- about the position,
or of the mass at any given

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time, or the frequency of this
oscillatory motion, or

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anything else.

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Even the velocity, if you know
a little bit of calculus, you

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can figure out the velocity
at anytime, of the object.

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And that's pretty neat.

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So what can we do now?

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Well, let's try to figure
out the period of

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this oscillating system.

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And just so you know-- I know
I put the label harmonic

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motion on all of these-- this
is simple harmonic motion.

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Simple harmonic motion is
something that can be

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described by a trigonometric
function like this.

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And it just oscillates back
and forth, back and forth.

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And so, what we're doing
is harmonic motion.

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And now, let's figure out
what this period is.

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Remember we said that after T
seconds, it gets back to its

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original position, and then
after another T seconds, it

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gets back to its original
position.

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Let's figure out
with this T is.

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And that's essentially
its period, right?

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What's the period
of a function?

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It's how long it takes to get
back to your starting point.

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Or how long it takes for the
whole cycle to happen once.

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So what is this T?

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So let me ask you a question.

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What are all the points--
that if this is a

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cosine function, right?

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What are all of the points at
which cosine is equal to 1?

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Or this function would
be equal to A, right?

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Because whenever cosine is
equal to 1, this whole

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function is equal to A.

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And it's these points.

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Well cosine is equal to 1 when--
so, theta-- let's say,

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when is cosine of theta
equal to 1?

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So, at what angles
is this true?

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Well it's true at theta
is equal to 0, right?

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Cosine of 0 is 1.

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Cosine of 2 pi is
also 1, right?

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We could just keep going around
that unit circle.

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You should watch the unit circle
video if this makes no

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sense to you.

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Or the graphing trig
functions.

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It's also true at 4 pi.

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Really, any multiple of
2 pi, this is true.

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Right?

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Cosine of that angle
is equal to 1.

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So the same thing is true.

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This function, x of t, is equal
to A at what points?

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x of t is equal to A whenever
this expression-- within the

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cosines-- whenever this
expression is equal to 0, 2

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pi, 4 pi, et cetera.

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And this first time that it
cycles, right, from 0 to 2

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pi-- from 0 to T, that'll
be at 2 pi, right?

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So this whole expression will
equal A, when k-- and that's

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these points, right?

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That's when this function
is equal to A.

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It'll happen again over
here someplace.

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When this little internal
expression is equal to 2 pi,

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or really any multiple
of 2 pi.

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So we could say, so x of t is
equal to A when the square

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root of k over m times
t, is equal to 2 pi.

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Or another way of thinking about
it, is let's multiply

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both sides of this equation
times the inverse of the

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square root of k over m.

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And you get, t is equal to 2 pi
times the square root-- and

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it's going to be the inverse
of this, right?

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Of m over k.

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And there we have the period
of this function.

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This is going to be equal
to 2 pi times the square

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root of m over k.

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So if someone tells you, well I
have a spring that I'm going

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to pull from some-- I'm going to
stretch it, or compress it

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a little bit, then I let go--
what is the period?

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How long does it take for the
spring to go back to its

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original position?

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It'll keep doing that, as we
have no friction, or no

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gravity, or any air
resistance, or

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anything like that.

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Air resistance really is just
a form of friction.

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You could immediately-- if you
memorize this formula,

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although you should know where
it comes from-- you could

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immediately say, well I know
how long the period is.

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It's 2 pi times m over k.

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That's how long it's going to
take the spring to get back--

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to complete one cycle.

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And then what about
the frequency?

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If you wanted to know cycles per
second, well that's just

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the inverse of the
period, right?

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So if I wanted to know the
frequency, that equals 1 over

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the period, right?

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Period is given in seconds
per cycle.

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So frequency is cycles
per second, and this

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is seconds per cycle.

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So frequency is just going
to be 1 over this.

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Which is 1 over 2 pi times the
square root of k over m.

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That's the frequency.

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But I have always had trouble
memorizing this, and this.

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You always [UNINTELLIGIBLE]

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k over m, and m over
k, and all of that.

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All you have to really
memorize is this.

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And even that, you might
even have an intuition

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as to why it's true.

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You can even go to the
differential equations if you

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want to reprove it
to yourself.

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Because if you have this, you
really can answer any question

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about the position of the
mass, at any time.

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The velocity of the mass, at any
time, just by taking the

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derivative.

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Or the period, or the frequency
of the function.

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As long as you know how to take
the period and frequency

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of trig functions.

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You can watch my videos, and
watch my trig videos, to get a

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refresher on that.

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One thing that's pretty
interesting about this, is

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notice that the frequency
and the period, right?

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This is the period of the
function, that's how long it

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takes do one cycle.

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This is how many cycles it does
in one second-- both of

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them are independent of A.

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So it doesn't matter, I could
stretch it only a little bit,

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like there, and it'll take the
same amount of time to go

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back, and come back like
that, as it would if I

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stretch it a lot.

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It would just do that.

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If I stretched it just a little
bit, the function would

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look like this.

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Make sure I do this right.

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I'm not doing that right.

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Edit, undo.

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If I just do it a little bit,
the amplitude is going to be

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less, but the function is going
to essentially do the

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same thing.

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It's just going to do that.

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So it's going to take the same
amount of time to complete the

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cycle, it'll just have
a lower amplitude.

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So that's interesting to me,
that how much I stretch it,

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it's not going to make it take
longer or less time to

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complete one cycle.

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That's interesting.

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And so if I just told you, that
I actually start having

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objects compressed, right?

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So in that case, let's
say my A is minus 3.

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I have a spring constant
of-- let's say k is,

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I don't know, 10.

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And I have a mass of 2
kilograms. Then I could

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immediately tell you what the
equation of the position as a

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00:08:47,100 --> 00:08:48,630
function of time at
any point is.

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It's going to be x of t will
equal-- I'm running out of

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space-- so x of t would equal--
this is just basic

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subsitution-- minus 3 cosine of
10 divided by 2, right? k

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over m, is 5.

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So square root of 5t.

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I know that's hard to read,
but you get the point.

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I just substituted that.

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But the important thing to
know is this-- this is, I

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think, the most important
thing-- and then if given a

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trig function, you have trouble
remembering how to

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00:09:19,690 --> 00:09:21,690
figure out the period or
frequency-- although I always

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just think about, when does
this expression equal 1?

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And then you can figure out--
when does it equal 1, or when

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00:09:29,952 --> 00:09:32,416
does it equal 0-- and you can
figure out its period.

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00:09:32,416 --> 00:09:33,360
If you don't have it,

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you can memorize this formula
for period, and this formula

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for frequency, but I think that
might be a waste of your

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00:09:38,830 --> 00:09:39,830
brain space.

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Anyway, I'll see you
in the next video.