- [Instructor] Welcome to a lesson
on free, undamped motion of a spring system.
We can also call this simple harmonic motion.
In this video, we will solve and interpret
an initial value problem modeling free, undamped motion.
Let's begin by setting up our spring system.
We'll start with a spring of length l that's not stretched,
this length here.
Next, we'll attach a mass to this spring,
which will elongate the spring.
And therefore, s will be the spring elongation,
or the amount the spring is stretched.
Next, the mass will be moved above or below
the equilibrium position, which we'll call displacement.
So, in this case, looking at this diagram here,
the displacement would be x.
If the displacement is below the equilibrium position,
x is positive.
And if it's displaced above the equilibrium position,
x will be negative.
And again, this video, we're considering
simple harmonic motion or free, undamped motion.
Free, undamped motion is motion that is not affected
by forces such as gravity,
air resistance, friction, and et cetera.
The motion is only affected by the spring
and the mass attached to the spring.
However, there may be an initial velocity upward or downward
that will affect the starting motion.
So, to get a better idea of free, undamped motion,
let's look at an animation.
This is free, undamped motion.
This is not affected by friction, gravity, air resistance.
And you can see that this motion
would conceptually go on forever.
Now, if we compare this to damped motion,
this system is being affected by
air resistance, gravity, friction, and et cetera.
So, this is damped motion.
And this is free, undamped motion.
So, this is the motion we're considering in this lesson.
We can use a linear second order
homogeneous differential equation with constant coefficients
to describe free, undamped motion.
It can be given in several forms.
These three differential equations are equivalent,
but I think this form here, makes the most sense.
We have m times the second derivative of x,
with respect to x, plus kx equals zero.
Or,
the mass times the acceleration of the displacement,
plus k, the spring constant,
times x, the displacement,
must equal zero.
But you'll often see this equation written
in this form here, where we divide through by m.
And then sometimes, you'll see k divided by m
is replaced with omega squared.
And we'll see why this is the case
when we solve our first differential equation.
We want to solve and interpret the initial value problem,
and then also find the period and frequency.
So, looking at the given information,
we first need to recognize
that the given differential equation
does fit the form for free, undamped motion.
And since x of zero equals eight,
the initial position, or initial displacement,
would be eight units below the equilibrium position.
So, going back down to our diagram,
that means
our mass is stretched eight units downward,
or this distance here would be eight units
below the equilibrium position.
Next, x prime of zero equals zero
means the initial velocity of our mass is zero.
So, once the mass is stretched down eight units,
it's simply released.
It is not given a velocity, either up or downward,
from this position.
So, let's begin by solving this differential equation.
And because we have a linear
second order homogeneous differential equation
with constant coefficients, for review,
we can solve this using a characteristic equation,
as described here.
And then, based upon the types of solutions,
we can determine the form of the general solution.
So, for the characteristic equation,
notice that a would be one,
b would be zero, and c would be 36.
So, this would give us r squared, plus 36, equals zero.
So, if we subtract 36 on both sides,
and then take the square root of both sides,
we can see that r is equal to plus or minus six i.
So, the solutions to the characteristic equation
are complex solutions in the form of
zero, plus or minus six i,
or alpha, plus or minus beta i, from our previous notes.
And therefore, the general solution will be in
this form here, where alpha equals zero,
and beta equals six.
Since alpha is equal to zero,
notice how we have e to the zero here, which is one,
as well as here.
So, we have x of t.
Our position function is equal to c sub one, cosine,
six t,
plus c sub two, sine,
six t.
Now, there is something else I want to point out here.
Notice, in this form here, omega squared would be 36.
So, sometimes you'll see the solution written as
c sub one, cosine
omega t,
plus c sub two, sine omega t.
Again, because omega is squared equals 36,
we can see that omega would be equal to six.
But again, I think it's important
to make the relationship
that we've already solved these types of equations
when we were solving linear second order
homogeneous differential equations
with constant coefficients.
Now, we'll use these two initial conditions here
and find the particular solution,
which will be our position function.
Let's go ahead and do this on the next slide.
If x sub zero equals eight,
that means eight must be equal to
c sub one, cosine zero,
plus c sub two, sine zero,
but sine zero is zero and cosine zero is equal to one.
So now we know that eight must equal c sub one,
and therefore, x of t must equal
eight cosine,
six t,
plus c sub two, sine six t.
And now, since x prime of zero equals zero,
we'll have to find the first derivative to use
this initial condition.
So, x prime of t would be equal to
the derivative of eight cosine six t
would be eight times negative sine six t,
times six, or negative 48,
sine six t,
plus the derivative of c sub one, sine six t,
would be six, c sub two, cosine six t.
And now, since x prime of zero equals zero,
we know that zero must equal negative 48,
sine zero,
plus six c sub two,
cosine zero.
And once again, sine zero is zero.
And cosine zero is one.
So, we have zero equals six, c sub two,
divide both sides by six,
c sub two must equal zero,
which means the position function
of our spring system would be x of t,
equals
eight,
cosine,
six t.
Now, we're not quite done yet.
We were asked to find the period and frequency.
So, let's finish by doing that.
To find the period,
we'll use the formula two pi divided by omega.
This tells the interval on which the cycle repeats.
Then the frequency is a reciprocal of the period,
or omega divided by two pi,
and this tells how many cycles over an interval of time.
So, to make sense of these two calculations,
let's go ahead and think of t as time in seconds.
And let's define x as inches.
And then, to use these two formulas,
we need to recognize that omega squared
would be equal to 36,
and therefore, omega is equal to six.
So, the period t is equal to two pi divided by six,
which would be pi divided by three.
Let's get our decimal approximation for this.
It'd be approximately 1.047.
So again, this tells the interval
in which the cycle repeats.
So, we could say it takes approximately 1.047 seconds
for the mass to complete one cycle,
which means, if you think
of the mass starting here at the bottom,
it would go up and come back down to this same position
in approximately 1.047 seconds.
And then, for the frequency,
we would have six divided by two pi, or three divided by pi.
Once again, notice how these are reciprocals.
This would be approximately 0.955.
Again, this is the number of cycles
over an interval of time.
So, we can say that in this system,
0.955 cycles
are completed every second.
Again, we weren't given these units,
but I think including the units
help us makes sense of what these actually mean.
Let's finish
by looking at the graph of our position function.
Here it is.
We have x of t, equals eight, times cosine six t.
If we analyze this graph closely, notice how
on the interval of length pi over three seconds, here,
notice how we do see one period of our graph.
And three divided by pi,
well that'd probably be somewhere in here.
Where if we go up to the graph,
notice how we can see, over this time interval,
we're completing approximately
0.955 cycles,
just a little bit shy of one complete cycle.
I hope you found this helpful.
In the next few videos, we'll look at some more examples,
where we have to write the differential equation
from given information.