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It’s now time to take a look at an application of second order
differential equations. We’re going to
take a look at mechanical vibrations. In
particular we are going to look at a mass that is hanging from a spring.
Vibrations can occur in pretty much all branches of
engineering and so what we’re going to be doing here can be easily adapted to
other situations, usually with just a change in notation.
Let’s get the situation setup. We are going to start with a spring of length
l, called the natural length, and
we’re going to hook an object with mass m
up to it. When the object is attached to
the spring the spring will stretch a length of L. We will call the
equilibrium position the position of the center of gravity for the object as it
hangs on the spring with no movement.
Below is sketch of the spring with and without the object
attached to it.
As denoted in the sketch we are going to assume that all
forces, velocities, and displacements in the downward direction will be
positive. All forces, velocities, and
displacements in the upward direction will be negative.
Also, as shown in the sketch above, we will measure all
displacement of the mass from its equilibrium position. Therefore, the u = 0 position will correspond to the center of gravity for the
mass as it hangs on the spring and is at rest (i.e. no movement).
Now, we need to develop a differential equation that will
give the displacement of the object at any time t. First, recall Newton’s Second Law of
Motion.
In this case we will use the second derivative of the displacement,
u, for the acceleration and so Newton’s Second Law
becomes,
We now need to determine all the forces that will act upon
the object. There are four forces that
we will assume act upon the object. Two
that will always act on the object and two that may or may not act upon the
object.
Here is a list of the forces that will act upon the object.
- Gravity, Fg
The force due to gravity will
always act upon the object of course.
This force is
- Spring, Fs
We are going to assume that Hooke’s
Law will govern the force that the spring exerts on the object. This force will always be present as well and
is
Hooke’s Law tells us that the force
exerted by a spring will be the spring constant, k > 0, times the displacement of the spring from its natural
length. For our set up the displacement
from the springs natural length is L + u
and the minus sign is in there to make sure that the force always has the
correct direction.
Let’s make sure that this force
does what we expect it to. If the object
is at rest in its equilibrium position the displacement is L and the force is simply Fs
= kL which will act in the upward position
as it should since the spring has been stretched from its natural length.
If the spring has been stretched
further down from the equilibrium position then will be positive and Fs will be negative acting to pull the object back up as
it should be.
Next, if the object has been moved
up past it’s equilibrium point, but not yet to it’s natural length then u will be negative, but still less than L and so L + u will be positive and once again Fs will be negative acting to pull the object up.
Finally, if the object has been
moved upwards so that the spring is now compressed, then u will be negative and greater than L. Therefore, L + u will be negative and now Fs will be positive acting to
push the object down.
So, it looks like this force will
act as we expect that it should.
- Damping, Fd
The next force that we need to
consider is damping. This force may or
may not be present for any given problem
Dampers work to counteract any
movement. There are several ways to
define a damping force. The one that we’ll
use is the following.
where, γ >
0 is the damping coefficient. Let’s
think for a minute about how this force will act. If the object is moving downward, then the
velocity ( ) will be positive and so Fd will be negative and acting
to pull the object back up. Likewise, if
the object is moving upward, the velocity ( ) will be negative and so Fd will be positive and acting
to push the object back down.
In other words, the damping force
as we’ve defined it will always act to counter the current motion of the object
and so will act to damp out any motion in the object.
- External Forces, F(t)
This is the catch all force. If there are any other forces that we decide
we want to act on our object we lump them in here and call it good. We typically call F(t) the forcing function.
Putting all of these together gives us the following for Newton’s Second Law.
Or, upon rewriting, we get,
Now, when the object is at rest in its equilibrium position
there are exactly two forces acting on the object, the force due to gravity and
the force due to the spring. Also, since
the object is at rest (i.e. not
moving) these two forces must be canceling each other out. This means that we must have,
Using this in Newton’s
Second Law gives us the final version of the differential equation that we’ll
work with.
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(2)
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Along with this differential equation we will have the
following initial conditions.
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(3)
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Note that we’ll also be using (1) to
determine the spring constant, k.
Okay. Let’s start
looking at some specific cases.
Free, Undamped
Vibrations
This is the simplest case that we can consider. Free or unforced vibrations means that F(t) = 0 and undamped vibrations means
that γ = 0.
In this case the differential equation becomes,
This is easy enough to solve in general. The characteristic equation has the roots,
This is usually reduced to,
where,
and ω0 is called the natural
frequency. Recall as well that m > 0 and k > 0 and so we can guarantee that this quantity will be
complex. The solution in this case is then
We can write (4) in
the following form,
where R is the
amplitude of the displacement and δ is the phase shift or phase angle of the
displacement.
When the displacement is in the form of (5)
it is usually easier to work with.
However, it’s easier to find the constants in (4)
from the initial conditions than it is to find the amplitude and phase shift in
(5)
from the initial conditions. So, in
order to get the equation into the form in (5) we
will first put the equation in the form in (4),
find the constants, c1 and
c2 and then convert this
into the form in (5).
So, assuming that we have c1 and c2
how do we determine R and δ ? Let’s start with (5) and
use a trig identity to write it as
Now, R and δ are constants and so if we compare (6)
to (4)
we can see that
We can find R in
the following way.
Taking the square root of both sides and assuming that R is positive will give
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(7)
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Finding δ is just as easy. We’ll start with
Taking the inverse tangent of both sides gives,
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(8)
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Before we work any examples let’s talk a little bit about
units of mass and the British vs. metric system differences.
Recall that the weight of the object is given by
where m is the
mass of the object and g is the
gravitational acceleration. For the
examples in this problem we’ll be using the following values for g.
This is not the standard 32.2 ft/s2 or 9.81 m/s2,
but using these will make some of the numbers come out a little nicer.
In the metric system the mass of objects is given in
kilograms (kg) and there is nothing
for us to do. However, in the British
system we tend to be given the weight of an object in pounds (yes, pounds are
the units of weight not mass…) and so we’ll need to compute the mass for these
problems.
At this point we should probably work an example of all this
to see how this stuff works.
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Example 1 A
16 lb object stretches a spring ft by itself. There is no damping and no external forces
acting on the system. The spring is
initially displaced 6 inches upwards from its equilibrium position and given
an initial velocity of 1 ft/sec downward.
Find the displacement at any time t,
u(t).
Solution
We first need to set up the IVP for the problem. This requires us to get our hands on m and k.
This is the British system so we’ll need to compute the
mass.
Now, let’s get k. We can use the fact that mg = kL to find k. Don’t forget that we’ll
need all of our length units the same.
We’ll use feet for the unit of measurement for this problem.
We can now set up the IVP.
For the initial conditions recall that upward
displacement/motion is negative while downward displacement/motion is
positive. Also, since we decided to do
everything in feet we had to convert the initial displacement to feet.
Now, to solve this we can either go through the
characteristic equation or we can just jump straight to the formula that we
derived above. We’ll do it that
way. First, we need the natural
frequency,
The general solution, along with its derivative, is then,
Applying the initial conditions gives
The displacement at any time t is then
Now, let’s convert this to a single cosine. First let’s get the amplitude, R.
You can use either the exact value here or a decimal
approximation. Often the decimal
approximation will be easier.
Now let’s get the phase shift.
We need to be careful with this part. The phase angle found above is in Quadrant
IV, but there is also an angle in Quadrant II that would work as well. We get this second angle by adding onto the first angle. So, we actually have two angles. They are
We need to decide which of these phase shifts is correct,
because only one will be correct. To
do this recall that
Now, since we are assuming that R is positive this means that the sign of cosδ will be the same as the sign of c1 and the sign of sinδ will be the same as the sign of c2. So, for this particular case we must have
cosδ < 0 and sinδ > 0. This means that the phase shift must be in
Quadrant II and so the second angle is the one that we need.
So, after all of this the displacement at any time t is.
Here is a sketch of the displacement for the first 5
seconds.
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