U1S7V10 Modeling oscillations.txt

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We told you that sine and cosine are
used in modeling oscillations.
And you may have noticed this pumpkin oscillating
on the end of a slinky.
In this video, we're going to see
how the sine and cosine arise in modeling
the motion of this pumpkin.
Let's get started.
Physics or, rather, classical mechanics
tells us that the motion is governed
by Newton's second law, which says ma is equal to F.
That is the mass of the pumpkin--
we're going to assume the slinky is essentially massless-- times
the acceleration at any time, t, is
equal to the restoring forces that are acting
on the pumpkin at that time.
The next step is that I'm going to choose a frame of reference
for this problem.
So I'm going to let y be the position
along this vertical axis.
And I want to choose where the zero lies on this access.
To do this, I'll use the quick experiment.
There is a position where if I let
the pumpkin go with a zero velocity,
it doesn't really move at all.
This means that the sum of the forces must be zero.
I'm going to go ahead and let this horizontal position be
the zero on my y-axis.
Now, because all motion happens in this vertical direction,
the acceleration is the second derivative of our variable y.
It's the second derivative of position.
So we can write this as ma is equal to m
times y double prime.
Finally, we need to model the restoring forces acting
on this pumpkin.
The simplest possible model that we can assume,
which is actually a pretty good first model,
is that f is proportional to the displacement of the pumpkin
away from the equilibrium.
That's saying that f is equal to negative k times y.
This says that when y is greater than zero--
when we've compressed the spring--
the net force is negative, pointing down.
And when y is less than zero-- when
the spring is expanded-- the net force is positive, pointing up.
Simplifying this, I see that I get
an equation in terms of only y which says that m y double
prime is equal to negative k y.
An equation that involves a variable and its derivatives
is called a differential equation.
So because this involves y and its second derivative,
this is a differential equation.
And differential equations, like any equation, must be balanced,
so the units of every term must be equal.
It's a really good habit to check units as you model.
We're not going to do that.
I'll leave that as an exercise for you.
To simplify this differential equation,
I'm going to divide through by m.
Now, this equation looks very similar
to the second derivatives of sine and cosine.
Remember that if f of t is equal to sine t,
then f double prime is equal to negative f.
And similarly, if g of t is equal to cosine t,
then g prime prime is equal to negative g.
The only difference between these equations
and what we have found is this constant term.
Well, it turns out that a solution to this differential
equation is given by y of t is equal to A,
some constant, times sine of the square root of k over m times t
plus B, another constant, times cosine of the square root of k
over m times t.
To see that this is a solution, all we need to do
is differentiate it twice and see that the second derivative
satisfies the relationship with this equation
by plugging it into our differential equation.
Of course, we do have a tiny little problem.
Even though the square root of k over m is a constant,
at this point you don't have the tools to actually differentiate
this function.
But fortunately, this marks the end
of the first unit about what the derivative is.
And in the next unit, we're going
to develop a toolbox for differentiating
any kind of function that you could ever encounter,
including this one.
So good luck on the homework.
And we look forward to seeing you in the next unit.