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Harmonic Motion Part 2 (calculus) : We test whether Acos(wt) can describe the motion of the mass on a spring by substituting into the differential equation F=-kx
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- 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 sign of whatever's in
- the inside.
- I'll put the minus outside.
- So it's minus sign 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.