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, F_{g}
The force due to gravity will
always act upon the object of course.
This force is
 Spring, F_{s}
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 spring's 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 F_{s}
= 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 F_{s} will be negative acting to pull the object back up as
it should be.
Next, if the object has been moved
up past its equilibrium point, but not yet to its natural length then u will be negative, but still less than L and so L + u will be positive and once again F_{s} 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 F_{s} will be positive acting to
push the object down.
So, it looks like this force will
act as we expect that it should.
 Damping, F_{d}
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 F_{d} 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 F_{d} 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.


(2)

Along with this differential equation we will have the
following initial conditions.


(3)

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 _{ω}_{0} is called the natural
frequency. Recall as well that m > 0 and k > 0 and so we can guarantee that this quantity will not 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, c_{1} and
c_{2} and then convert this
into the form in (5).
So, assuming that we have c_{1} and c_{2}
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


(7)

Finding δ is just as easy. We’ll start with
Taking the inverse tangent of both sides gives,


(8)

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/s^{2} or 9.81 m/s^{2},
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.
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 c_{1} and the sign of sinδ will be the same as the sign of c_{2}. 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.

Now, let’s take a look at a slightly more realistic
situation. No vibration will go on
forever. So let’s add in a damper and
see what happens now.
Free, Damped
Vibrations
We are still going to assume that there will be no external
forces acting on the system, with the exception of damping of course. In this case the differential equation will
be.
where m, δ, and k are all positive constants. Upon solving for the roots of the
characteristic equation we get the following.
We will have three cases here.
1.
In this case we will get a double
root out of the characteristic equation and the displacement at any time t will be.
Notice that as the displacement will approach zero and so the
damping in this case will do what it’s supposed to do.
This case is called critical damping and will happen when
the damping coefficient is,
The value of the damping
coefficient that gives critical damping is called the critical damping
coefficient and denoted by _{γ}_{CR}.
2.
In this case let’s rewrite the
roots a little.
Also notice that from our initial
assumption that we have,
Using this we can see that the
fraction under the square root above is less than one. Then if the quantity under the square root is
less than one, this means that the square root of this quantity is also going
to be less than one. In other words,
Why is this important? Well, the quantity in the parenthesis is now
one plus/minus a number that is less than one.
This means that the quantity in the parenthesis is guaranteed to be
positive and so the two roots in this case are guaranteed to be negative. Therefore the displacement at any time t is,
and will approach zero as . So, once again the damper does what it is
supposed to do.
This case will occur when
and is called over damping.
3.
In this case we will get complex
roots out of the characteristic equation.
where the real part is guaranteed
to be negative and so the displacement is
Notice that we reduced the sine
and cosine down to a single cosine in this case as we did in the undamped
case. Also, since λ < 0 the displacement will approach zero
as and the damper will also work as it’s supposed
to in this case.
We will get this case will occur
when
and is called under damping.
Let’s take a look at a couple of examples here with damping.
Example 2 Take
the spring and mass system from the first example and attach a damper to it
that will exert a force of 12 lbs when the velocity is 2 ft/s. Find the displacement at any time t, u(t).
Solution
The mass and spring constant were already found in the
first example so we won’t do the work here.
We do need to find the damping coefficient however. To do this we will use the formula for the
damping force given above with one modification. The original damping force formula is,
However, remember that the force and the velocity are
always acting in opposite directions.
So, if the velocity is upward (i.e.
negative) the force will be downward (i.e.
positive) and so the minus in the formula will cancel against the minus in
the velocity. Likewise, if the
velocity is downward (i.e.
positive) the force will be upwards (i.e.
negative) and in this case the minus sign in the formula will cancel against
the minus in the force. In other
words, we can drop the minus sign in the formula and use
and then just ignore any signs for the force and velocity.
Doing this gives us the following for the damping
coefficient
The IVP for this example is then,
Before solving let’s check to see what kind of damping
we’ve got. To do this all we need is the
critical damping coefficient.
So, it looks like we’ve got critical damping. Note that this means that when we go to
solve the differential equation we should get a double root.
Speaking of solving, let’s do that. I’ll leave the details to you to check that
the displacement at any time t is.
Here is a sketch of the displacement during the first 3
seconds.

Notice that the “vibration” in the system is not really a
true vibration as we tend to think of them.
In the critical damping case there isn’t going to be a real oscillation
about the equilibrium point that we tend to associate with vibrations. The damping in this system is strong enough
to force the “vibration” to die out before it ever really gets a chance to do
much in the way of oscillation.
Example 3 Take
the spring and mass system from the first example and this time let’s attach
a damper to it that will exert a force of 17 lbs when the velocity is 2
ft/s. Find the displacement at any
time t, u(t).
Solution
So, the only difference between this example and the
previous example is damping force. So
let’s find the damping coefficient
So it looks like we’ve got over damping this time around
so we should expect to get two real distinct roots from the characteristic
equation and they should both be negative.
The IVP for this example is,
This one’s a little messier than the previous example so
we’ll do a couple of the steps, leaving it to you to fill in the blanks. The roots of the characteristic equation
are
In this case it will be easier to just convert to decimals
and go that route. Note that, as
predicted we got two real, distinct and negative roots. The general and actual solution for this
example are then,
Here’s a sketch of the displacement for this example.

Notice an interesting thing here about the displacement
here. Even though we are “over” damped
in this case, it actually takes longer for the vibration to die out than in the
critical damping case. Sometimes this
happens, although it will not always be the case that over damping will allow
the vibration to continue longer than the critical damping case.
Also notice that, as with the critical damping case, we
don’t get a vibration in the sense that we usually think of them. Again, the damping is strong enough to force
the vibration do die out quick enough so that we don’t see much, if any, of the
oscillation that we typically associate with vibrations.
Let’s take a look at one more example before moving on the
next type of vibrations.
Example 4 Take
the spring and mass system from the first example and for this example let’s
attach a damper to it that will exert a force of 5 lbs when the velocity is 2
ft/s. Find the displacement at any
time t, u(t).
Solution
So, let’s get the damping coefficient.
So it’s under damping this time. That shouldn’t be too surprising given the
first two examples. The IVP for this example is,
In this case the roots of the characteristic equation are
They are complex as we expected to get since we are in the
under damped case. The general
solution and actual solution are
Let’s convert this to a single cosine as we did in the
undamped case.
As with the undamped case we can use the coefficients of
the cosine and the sine to determine which phase shift that we should
use. The coefficient of the cosine (c_{1}) is negative and so cosδ must
also be negative. Likewise, the
coefficient of the sine (c_{2})
is also negative and so sinδ
must also be negative. This
means that δ must be in the Quadrant III and so the
second angle is the one that we want.
The displacement is then
Here is a sketch of this displacement.

In this case we finally got what we usually consider to be a
true vibration. In fact that is the
point of critical damping. As we
increase the damping coefficient, the critical damping coefficient will be the
first one in which a true oscillation in the displacement will not occur. For all values of the damping coefficient
larger than this (i.e. over damping)
we will also not see a true oscillation in the displacement.
From a physical standpoint critical (and over) damping is
usually preferred to under damping.
Think of the shock absorbers in your car. When you hit a bump you don’t want to spend
the next few minutes bouncing up and down while the vibration set up by the
bump die out. You would like there to be
as little movement as possible. In other
words, you will want to set up the shock absorbers in your car so get at the
least critical damping so that you can avoid the oscillations that will arise
from an under damped case.
It’s now time to look at systems in which we allow other
external forces to act on the object in the system.
Undamped, Forced
Vibrations
We will first take a
look at the undamped case. The
differential equation in this case is
This is just a nonhomogeneous differential equation and we
know how to solve these. The general
solution will be
where the complementary solution is the solution to the
free, undamped vibration case. To get
the particular solution we can use either undetermined coefficients or
variation of parameters depending on which we find easier for a given forcing
function.
There is a particular type of forcing function that we
should take a look at since it leads to some interesting results. Let’s suppose that the forcing function is a
simple periodic function of the form
For the purposes of this discussion we’ll use the first
one. Using this, the IVP becomes,
The complementary solution, as pointed out above, is just
where is the natural frequency.
We will need to be careful in finding a particular
solution. The reason for this will be
clear if we use undetermined coefficients.
With undetermined coefficients our guess for the form of the particular
solution would be,
Now, this guess will be problems if . If this were to happen the guess for the
particular solution is exactly the complementary solution and so we’d need to
add in a t. Of course if we don’t have then there will be nothing wrong with the
guess.
So, we will need to look at this in two cases.

In this case our initial guess is
okay since it won’t be the complementary solution. Upon differentiating the guess and plugging
it into the differential equation and simplifying we get,
Setting coefficients equal gives
us,
The particular solution is then
Note that we rearranged things a
little. Depending on the form that you’d
like the displacement to be in we can have either of the following.
If we used the sine form of the
forcing function we could get a similar formula.

In this case we will need to add in
a t to the guess for the particular
solution.
Note that we went ahead and
acknowledge that in our guess.
Acknowledging this will help with some simplification that we’ll need to
do later on. Differentiating our guess,
plugging it into the differential equation and simplifying gives us the
following.
Before setting coefficients equal,
let’s remember the definition of the natural frequency and note that
So, the first two terms actually
drop out (which is a very good thing…) and this gives us,
Now let’s set coefficient equal.
In this case the particular will
be,
The displacement for this case is
then
depending on the form that you
prefer for the displacement.
So, what was the point of the two cases here? Well in the first case, our displacement function consists of two
cosines and is nice and well behaved for all time.
In contrast, the second case, will have some serious issues at t increases. The addition of the t in the particular solution will mean that we are going to see an
oscillation that grows in amplitude as t
increases. This case is called resonance and we would generally like
to avoid this at all costs.
In this case resonance arose by assuming that the forcing
function was,
We would also have the possibility of resonance if we
assumed a forcing function of the form.
We should also take care to not assume that a forcing
function will be in one of these two forms.
Forcing functions can come in a wide variety of forms. If we do run into a forcing function different
from the one that used here you will have to go through undetermined coefficients
or variation of parameters to determine the particular solution.
Example 5 A
3 kg object is attached to spring and will stretch the spring 392 mm by
itself. There is no damping in the
system and a forcing function of the form
is attached to the object and the system will experience
resonance. If the object is initially
displaced 20 cm downward from its equilibrium position and given a velocity
of 10 cm/sec upward find the displacement at any time t.
Solution
Since we are in the metric system we won’t need to find
mass as it’s been given to us. Also,
for all calculations we’ll be converting all lengths over to meters.
The first thing we need to do is find k.
Now, we are told that the system experiences resonance so
let’s go ahead and get the natural frequency so we can completely set up the
IVP.
The IVP for this is then
Solution wise there isn’t a whole lot to do here. The complementary solution is the free
undamped solution which is easy to get and for the particular solution we can
just use the formula that we derived above.
The general solution is then,
Applying the initial conditions gives the displacement at
any time t. We’ll leave the details to you to check.
The last thing that we’ll do is combine the first two
terms into a single cosine.
In this case the coefficient of the cosine is positive and
the coefficient of the sine is negative.
This forces cosδ
to be positive and sinδ
to be negative. This means that
the phase shift needs to be in Quadrant IV and so the first one is the
correct phase shift this time.
The displacement then becomes,
Here is a sketch of the displacement for this example.

It’s now time to look at the final vibration case.
Forced, Damped
Vibrations
This is the full blown case where we consider every last
possible force that can act upon the system.
The differential equation for this case is,
The displacement function this time will be,
where the complementary solution will be the solution to the
free, damped case and the particular solution will be found using undetermined
coefficients or variation of parameter, whichever is most convenient to use.
There are a couple of things to note here about this
case. First, from our work back in the
free, damped case we know that the complementary solution will approach zero as
t increases. Because of this the complementary solution is
often called the transient solution
in this case.
Also, because of this behavior the displacement will start
to look more and more like the particular solution as t increases and so the particular solution is often called the steady state solution or forced response.
Let’s work one final example before leaving this
section. As with the previous examples,
we’re going to leave most of the details out for you to check.
Example 6 Take
the system from the last example and add in a damper that will exert a force
of 45 Newtons
when then velocity is 50 cm/sec.
Solution
So, all we need to do is compute the damping coefficient
for this problem then pull everything else down from the previous
problem. The damping coefficient is
The IVP for this problem is.
The complementary solution for this example is
For the particular solution we the form will be,
Plugging this into the differential equation and
simplifying gives us,
Setting coefficient equal gives,
The general solution is then
Applying the initial condition gives
Here is a sketch of the displacement for this example.
