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Current time:0:00Total duration:10:03

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