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Current time:0:00Total duration:9:46

So where I left off in the last
video, I'd just rewritten the spring equation. And I just wrote force is
mass times acceleration. And I was in the process of
saying, well if x is a function of t, what's
acceleration? Well, velocity is this
derivative of x with respect to time, right? Your change in position
over change of time. And acceleration is the
derivative of velocity, or the second derivative of position. So you take the derivative
twice of x of t, right? So let's rewrite this equation
in those terms. Let me erase all this--I actually want to
keep all of this, just so we remember what we're talking
about this whole time. Let me see if I can
erase it cleanly. That's pretty good. Let me erase all of this. All of this. I'll even erase this. That's pretty good, all right. Now back to work. So, we know that-- or hopefully
we know-- that acceleration is the second
derivative of x as a function of t. So we can rewrite this as
mass times the second derivative of x. So I'll write that as-- well,
I think the easiest notation would just be x prime prime. That's just the second
derivative of x as a function of t. I'll write the function
notation, just so you remember this is a function of time. Is equal to minus
k times x of t. And what you see here, what
I've just written, this is actually a differential
equation. And so what is a differential
equation? Well, it's an equation where,
in one expression, or in one equation, on both sides of
this, you not only have a function, but you have
derivatives of that function. And the solution to a
differential equation isn't just a number, right? A solution to equations that
we've done in the past are numbers, essentially, or a set
of numbers, or maybe a line. But the solution to differential
equations is actually going to be a function,
or a class of functions, or a set
of functions. So it'll take a little time to
get your head around it, but this is as good an example as
ever to be exposed to it. And we're not going to solve
this differential equation analytically. We're going to use our intuition
behind what we did earlier in the previous video. We're going to use that to guess
at what a solution to this differential equation is. And then, if it works out, then
we'll have a little bit more intuition. And then we'll actually know
what the position is, at any given time, of this mass
attached to the spring. So this is exciting. This is a differential
equation. When we drew the position-- our
intuition for the position over time-- our intuition tells
us that it's a cosine function, with amplitude A. So we said it's A cosine omega
t, where this is the angular velocity of-- well, I don't want
to go into that just yet, we'll get a little bit more
intuition in a second. And now, what we can do is,
let's test this expression-- this function-- to see if it
satisfies this equation. Right? If we say that x of t is equal
to A cosine of wt, what is the derivative of this?
x prime of t. And you could review
the derivative videos to remember this. Well, it's the derivative of the
inside, so it'll be that omega, times the
outside scalar. A omega. And then the derivative-- I'm
just doing the chain rule-- the derivative of cosine of t is
minus sine of whatever's in the inside. I'll put the minus outside. So it's minus sine of wt. And then, if we want the second
derivative-- so that's x prime prime of t. Let me do this in a different
color, just so it doesn't get monotonous. That's the derivative
of this, right? So what's the derivative
of-- these are just scalar values, right? These are just constants. So the derivative of the
inside is an omega. I multiply the omega times
the scalar constant. I get minus A omega squared. And then the derivative of
sine is just cosine. But the minus is still there,
because I had the minus to begin with. Minus cosine of omega t. Now let's see if this is true. So if this is true, I should
be able to say that m times the second derivative of x of
t, which is in this case is this, times minus Aw
squared cosine wt. That should be equal to minus k
times my original function-- times x of t. And x of t is a cosine wt. I'm running out of space. Hopefully you understand
what I'm saying. I just substituted x prime
prime, the second derivative, into this, and I just
substituted x of t, which I guess that's that, in here. And now I got this. And let me see if
I can rewrite. Maybe I can get rid of
the spring up here. I'm trying to look for space. I don't want to get rid of this,
because I think this gives us some intuition
of what we're doing. One of those days that I wish
I had a larger blackboard. Erase the spring. Hopefully you can remember
that image in your mind. And actually, I can
erase that. I can erase that. I can erase all of this, just so
I have some space, without getting rid of that nice curve I
took the time to draw in the last video. Almost there. OK. Back to work. Make sure my pen feels
right, OK. So all I did is I took-- we
said that by the spring constant, if you rewrite force
as mass times acceleration, you get this. Which is essentially a
differential equation, I just rewrote acceleration as
the second derivative. Then I took a guess, that this
is x of t, just based on our intuition of the drawing. I took a guess. And then I took the second
derivative of it. Right? This is the first derivative,
this is the second derivative. And then I substituted the
second derivative here, and I substituted the function here. And this is what I got. And so let me see if I can
simplify that a little bit. So if I rewrite there, I get
minus mAw squared cosine of wt is equal to minus
kA cosine of wt. Well it looks good so far. Let's see, we can get rid of the
minus signs on both sides. Get rid of the A's
on both sides. Right? We can divide both sides by A. Let me do this in black, just
so it really erases it. So if we get rid of A on both
sides, we're left with that. And then-- so let's see, we
have mw squared cosine of omega t is equal to k
cosine of omega t. So this equation holds
true if what is true? This equation holds true if mw
squared-- or omega squared, I think that's omega. I always forget my--
is equal to k. Or another way of saying
it, if omega squared is equal to k over m. Or, omega is equal to the
square root of k over m. So there we have it. We have figured out what
x of t has to be. We said that this differential
equation is true, if this is x of t, and omega is
equal to this. So now we've figured out the
actual function that describes the position of that spring
as a function of time. x of t is going to be equal to--
we were right about the A, and that's just intuition,
right, because the amplitude of this cosine function is A--
A cosine-- and instead of writing w, we can now write the
square root of k over m. The square root of k over m t. That to me is amazing. We have now, using not too
sophisticated calculus, solved a differential equation. And now can-- if you tell me at
5.8 seconds, where is x, I can tell you. And I just realized that I am
now running out of time, so I will see you in the
next video.