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# Introduction to Harmonic Motion: Intuition behind the motion of a mass on a spring (some calculus near the end).

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- Let's see if we can use what we know about springs now to
- get a little intuition about how the
- spring moves over time.
- And hopefully we'll learn a little bit
- about harmonic motion.
- We'll actually even step into the world of differential
- equations a little bit.
- And don't get daunted when we get there.
- Or just close your eyes when it happens.
- Anyway, so I've drawn a spring, like I've done in the
- last couple of videos.
- And 0, this point in the x-axis, that's where the
- spring's natural resting state is.
- And in this example I have a mass, mass m,
- attached to the spring.
- And I've stretched the string.
- I've essentially pulled it.
- So the mass is now sitting at point A.
- So what's going to happen to this?
- Well, as we know, the force, the restorative force of the
- spring, is equal to minus some
- constant, times the x position.
- The x position starting at A.
- So initially the spring is going to pull
- back this way, right?
- The spring is going to pull back this way.
- It's going to get faster and faster and faster and faster.
- And we learned that at this point, it has a lot of
- potential energy.
- At this point, when it kind of gets back to its resting
- state, it'll have a lot of velocity and a lot of kinetic
- energy, but very little potential energy.
- But then its momentum is going to keep it going, and it's
- going to compress the spring all the way, until all of that
- kinetic energy is turned back into potential energy.
- Then the process will start over again.
- So let's see if we can just get an intuition for what x
- will look like as a function of time.
- So our goal is to figure out x of t, x as a function of time.
- That's going to be our goal on this video and
- probably the next few.
- So let's just get an intuition for what's happening here.
- So let me try to graph x as a function of time.
- So time is the independent variable.
- And I'll start at time is equal to 0.
- So this is the time axis.
- Let me draw the x-axis.
- This might be a little unusual for you, for me to draw the
- x-axis in the vertical, but that's because x is the
- dependent variable in this situation.
- So that's the x-axis, very unusually.
- Or we could say x of t, just so you know x is a function of
- time, x of t.
- And this state, that I've drawn here, this is at time
- equals 0, right?
- So this is at 0.
- Let me switch colors.
- So at time equals 0, what is the x position of the mass?
- Well the x position is A, right?
- So if I draw this, this is A.
- Actually, let me draw a line there.
- That might come in useful.
- This is A.
- And then this is going to be-- let me try to make it
- relatively-- that is negative A.
- That's minus A.
- So at time t equals 0, where is it?
- Well it's at A.
- So this is where the graph is, right?
- Actually, let's do something interesting.
- Let's define the period.
- So the period I'll do with a capital T.
- Let's say the period is how long it takes for this mass to
- go from this position.
- It's going to accelerate, accelerate, accelerate,
- accelerate.
- Be going really fast at this point, all kinetic energy.
- And then start slowing down, slowing down, slowing down,
- slowing down.
- And then do that whole process all the way back.
- Let's say T is the amount of time it takes to do that whole
- process, right?
- So at time 0 today, and then we also know that at time T--
- this is time T-- it'll also be at A, right?
- I'm just trying to graph some points that I know of this
- function and just see if I can get some intuition of what
- this function might be analytically.
- So if it takes T seconds to go there and back, it takes T
- over 2 seconds to get here, right?
- The same amount of time it takes to get here was also the
- same amount of time it takes to get back.
- So at T over 2 what's going to be the x position?
- Well at T over 2, the block is going to be here.
- It will have compressed all the way over here.
- So at T over 2, it'll have been here.
- And then at the points in between, it will be at x
- equals 0, right?
- It'll be there and there.
- Hopefully that makes sense.
- So now we know these points.
- But let's think about what the actual function looks like.
- Will it just be a straight line down, then a straight
- line up, and then the straight line down, and then a
- straight line up.
- That would imply-- think about it-- if you have a straight
- line down that whole time, that means that you would have
- a constant rate of change of your x value.
- Or another way of thinking about that is that you would
- have a constant velocity, right?
- Well do we have a constant velocity this entire time?
- Well, no.
- We know that at this point right here you have a very
- high velocity, right?
- You have a very high velocity.
- We know at this point you have a very low velocity.
- So you're accelerating this entire time.
- And you actually, the more you think about it, you're
- actually accelerating at a decreasing rate.
- But you're accelerating the entire time.
- And then you're accelerating and then you're decelerating
- this entire time.
- So your actual rate of change of x is not constant, so you
- wouldn't have a zigzag pattern, right?
- And it'll keep going here and then you'll have a point here.
- So what's happening?
- When you start off, you're going very slow.
- Your change of x is very slow.
- And then you start accelerating.
- And then, once you get to this point, right here, you start
- decelerating.
- Until at this point, your velocity is exactly 0.
- So your rate of change, or your slope, is going to be 0.
- And then you're going to start accelerating back.
- Your velocity is going to get faster, faster, faster.
- It's going to be really fast at this point.
- And then you'll start decelerating at that point.
- So at this point, what does this point correspond to?
- You're back at A.
- So at this point your velocity is now 0 again.
- So the rate of change of x is 0.
- And now you're going to start accelerating.
- Your slope increases, increases, increases.
- This is the point of highest kinetic energy right here.
- Then your velocity starts slowing down.
- And notice here, your slope at these points is 0.
- So that means you have no kinetic
- energy at those points.
- And it just keeps on going.
- On and on and on and on and on.
- So what does this look like?
- Well, I haven't proven it to you, but out of all the
- functions that I have in my repertoire, this looks an
- awful lot like a trigonometric function.
- And if I had to pick one, I would pick cosine.
- Well why?
- Because when cosine is 0-- I'll write it down here--
- cosine of 0 is equal to 1, right?
- So when t equals 0, this function is equal to A.
- So this function probably looks something like A cosine
- of-- and I'll just use the variable omega t-- it probably
- looks something like that, this function.
- And we'll learn in a second that it looks
- exactly like that.
- But I want to prove it to you, so don't just
- take my word for it.
- So let's just figure out how we can figure out what w is.
- And it's probably a function of the mass of this object and
- also probably a function of the spring
- constant, but I'm not sure.
- So let's see what we can figure out.
- Well now I'm about to embark into a little bit of calculus.
- Actually, a decent bit of calculus.
- And we'll actually even touch on differential equations.
- This might be the first differential equation you see
- in your life, so it's a momentous occasion.
- But let's just move forward.
- Close your eyes if you don't want to be confused, or go
- watch the calculus videos at least so you know what a
- derivative is.
- So let's write this seemingly simple equation, or let's
- rewrite it in ways that we know.
- So what's the definition of force?
- Force is mass times acceleration, right?
- So we can rewrite Hooke's law as-- let me switch colors--
- mass times acceleration is equal to minus the spring
- constant, times the position, right?
- And I'll actually write the position as a function of t,
- just so you remember.
- We're so used to x being the independent variable, that if
- I didn't write that function of t, it might get confusing.
- You're like, oh I thought x is the independent variable.
- No.
- Because in this function that we want to figure out, we want
- to know what happens as a function of time?
- So actually this is also maybe a good review
- of parametric equations.
- This is where we get into calculus.
- What is acceleration?
- If I call my position x, my position is equal to x as a
- function of t, right?
- I put in some time, and it tells me what my x value is.
- That's my position.
- What's my velocity?
- Well my velocity is the derivative of this, right?
- My velocity, at any given point, is going to be the
- derivative of this function.
- The rate of change of this function with respect to t.
- So I would take the rate of change with
- respect to t, x of t.
- And I could write that as dx, dt.
- And then what's acceleration?
- Well acceleration is just the rate of change
- of velocity, right?
- So it would be taking the derivative of this.
- Or another way of doing it, it's like taking the second
- derivative of the position function, right?
- So in this situation, acceleration is equal to, we
- could write it as-- I'm just showing you all different
- notations-- x prime prime of t, second derivative of x with
- respect to t.
- Or-- these are just notational-- d squared x over
- dt squared.
- So that's the second derivative.
- Oh it looks like I'm running out of time.
- So I'll see you in the next video.
- Remember what I just wrote. just wrote

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