Example 3 Write
down the system of differential equations for the spring and mass system
above.
Solution
To help us out let’s first take a quick look at a
situation in which both of the masses have been moved. This is shown below.
Before
proceeding let’s note that this is only a representation of a typical case,
but most definitely not all possible cases.
In this case
we’re assuming that both and are positive and that ,
or in other words, both masses have been moved to the right of their
respective equilibrium points and that has been moved farther than . So, under these assumption on and we know that the spring on the left (with
spring constant ) has been stretched past it’s
natural length while the middle spring (spring constant ) and the right spring (spring
constant ) are both under compression.
Also, we’ve
shown the external forces, and ,
as present and acting in the positive direction. They do not, in practice, need to be
present in every situation in which case we will assume that and/or . Likewise, if the forces are in fact acting
in the negative direction we will then assume that and/or .
Before
proceeding we need to talk a little bit about how the middle spring will
behave as the masses move. Here are
all the possibilities that we can have and the affect each will have on . Note that in each case the amount of
compression/stretch in the spring is given by although we won’t be using the absolute
value bars when we set up the differential equations.
1. If both mass move the same amount
in the same direction then the middle spring will not have changed length and
we’ll have .
2. If both masses move in the
positive direction then the sign of will tell us which has moved more. If moves more than then the spring will be in compression and . Likewise, if moves more than then the spring will have been stretched and
.
3. If both masses move in the
negative direction we’ll have pretty much the opposite behavior as #2. If moves more than then the spring will have been stretched and
. Likewise, if moves more than then the spring will be in compression and .
4. If moves in the positive direction and moves in the negative direction then the
spring will be in compression and .
5. Finally, if moves in the negative direction and moves in the positive direction then the
spring will have been stretched and .
Now, we’ll use
the figure above to help us develop the differential equations (the figure
corresponds to case 2 above…) and then make sure that they will also hold for
the other cases as well.
Let’s start off
by getting the differential equation for the forces acting on . Here is a quick sketch of the forces acting
on for the figure above.
In this case and so the first spring has been stretched
and so will exert a negative (i.e.
to the left) force on the mass. The
force from the first spring is then and the “” is needed because the force is
negative but both and are positive.
Next, because
we’re assuming that has moved more than and both have moved in the positive
direction we also know that . Because has moved more than we know that the second spring will be under
compression and so the force should be acting in the negative direction on and so the force will be . Note that because is positive and is negative this force will have the correct
sign (i.e. negative).
The
differential equation for is then,
Note that this
will also hold for all the other cases.
If has been moved in the negative direction the
force form the spring on the right that acts on the mass will be positive and
will be a positive quantity in this
case. Next, if the middle is has been
stretched (i.e. ) then the force from this spring on will be in the positive direction and will be a positive quantity in this
case. Therefore, this differential
equation holds for all cases not just the one we illustrated at the start of
this problem.
Let’s now write
down the differential equation for all the forces that are acting on . Here is a sketch of the forces acting on
this mass for the situation sketched out in the figure above.
In this case is positive and so the spring on the right
is under compression and will exert a negative force on and so this force should be ,
where the “” is required because both and are positive. Also, the middle spring is still under
compression but the force that it exerts on this mass is now a positive
force, unlike in the case of ,
and so is given by . The “” on this force is required because is negative and the force must be positive.
The
differential equation for is then,
We’ll leave it
to you to verify that this differential equation does in fact hold for all
the other cases.
Putting all of
this together and doing a little rewriting will then give the following
system of differential equations for this situation.
This is a
system to two linear second order differential equations that may or may not
be nonhomogeneous depending whether there are any external forces, and ,
acting on the masses.
