Download Spring-Mass Problems An object has weight w (in pounds

Spring-Mass Problems
An object has weight w (in pounds, abbreviated lb). Weight w is mass times gravity, so that we have
.
Here, gravity is
32
since “down” in this scenario is considered positive, and weight is a force.
Hooke’s Law for springs states that the force
to extend a spring a distance L is proportional to L, or
. Thus, a mass attached to a spring has two forces acting on it: the downward weight (force) due
to gravity, and the upward force due to Hooke’s Law (i.e. the spring wants to contract opposite gravity).
Since the forces act opposite of one another, the overall force acting on the object is
. When the
mass is at rest, we have
0, or
.
Now assume
is the distance (in feet) after t seconds of the mass relative to the spring’s “rest” state.
Note that the spring is extended L units initially when the mass is attached, then another u units (up and
down) over time. Thus, Hooke’s Law here gets modified to include the bobbing nature of the spring:
.
The second derivative of
is the mass’s acceleration at time t. Using the formula F = ma, we have a
general force term
to describe the force of the bobbing mass.
A resistance force
is defined as some force (in pounds) acting against the bobbing mass when the mass
is at some speed, given by
. This resistive force is usually proportional to the speed, and acts against
the motion of the object. Thus, we can write
, where is the constant of proportionality.
The sum of all of the forces must equal
means that
0
Thus,
, per Newton’s law of motion (the F = ma rule). This
′
.Since
0
. Collecting the terms to the left side, we have
0.
The 0 on the right side indicates that there is no external force acting on the spring.
Here is the example from the text, Page 195: A mass weighing 4 lb stretches a spring 2 in (1/6 ft).
Suppose that the mass is given an additional 6 in of displacement in the positive direction and then
released. The mass is in a medium that exerts a viscous resistance of 6 lb when the mass has a velocity of
3 ft/s. Formulate the initial value problem that governs the motion of the mass.
Since there is no external force acting on the system, the general formula is
0.
The parameters are as follows:
, ormass
∙
. The units are
⁄
∙
, where the units are
⁄
.
.
, where the units are .
The function
is feet and t is in seconds. Thus
piecing this all together, we have the following:
is in
∙
0→
and
is in
∙
. Going by units alone,
ft .
All terms simplify into lb, which makes sense, since the whole equation is one of force. If you really want
to flesh out the general formula, you get
0.
The given problem then becomes
0, or
⁄
2
24
16
0
The initial conditions are
“released”.
0. Multiplying through by 8, we get
192
4 1 192
16
0
(6 inches, or half a foot), and
16
Using the quadratic formula on the auxiliary equation
16
0.
16
2
√ 512
2
192
16
0 since it was merely
0, we get
16 √2
2
8
8 √2.
Thus, the general solution is
cos 8√2
We then look at the initial condition
1, and the e factor becomes 1. Thus,
0
sin 8√2
.
. When t = 0, the sine term disappears, the cosine becomes
.
We now have
1
cos 8√2
2
sin 8√2
.
Taking the derivative is a chore, but when evaluated at 0, many terms will simplify to 0 or 1:
4√2 sin 8√2
8√2 cos 8√2
8
1
cos 8√2
2
sin 8√2
.
But now when t = 0, we get:
0
8√2
4.
Thus,
√2
,
4
The specific solution is
1
cos 8√2
2
√2
sin 8√2
4
.
Here is the graph of the bobbing of the spring over a 1-second interval of time:
As you can see, it bobs a little, passing through the “rest” state (where u = 0) once or twice before
essentially stopping due to the resistance of the surrounding medium.
Suppose there is no resistive force. Then
0 and the differential equation is
1
8
24
With the same initial conditions. The solution is
Here, the spring bobs forever, never losing amplitude.
0,
cos 8√3 . Its graph is
Now suppose the resistive force is twice what it originally was. The differential equation becomes
32
The auxiliary equation is
general solution
32
192
192
0.
0, which factors to
24
8
0. Thus, we have a
.
Using the same initial conditions as before, the specific solution is
1
4
3
4
.
Its graph is
The spring never bobs past the resting line. The medium is so viscous that the mass just slowly moves
toward the rest state, approaching it asymptotically. This system is over-damped.
The solutions of the general form
0 will be one of the three cases:
1. Complex conjugate solutions, which will result in a periodic motion, with or without an
“envelope” factor (the
term) that damps the periodic motion to zero. If it lacks this e factor,
then the system is undamped and it bobs up and down forever. If it has this e factor, then the
system is damped and it will slowly bob up and down, asymptotically to zero.
2. Two real but different solutions, both negative, so that the two terms are e raised to negative
coefficients of t. This results in an overdamped system. The mass won’t bob at all. It will just
move slowly back to the rest state.
3. One repeated real solution. In this case, the system is critically damped. Below this, the mass will
bob up and down, past this, it won’t bob at all.
In the case of one repeated real solution, we can make this happen by solving the auxiliary polynomial of
0, using the quadratic formula. Thus, we have
the generic differential equation
4
2
.
If there is to be just the one repeated root, then the discriminant (the portion under the radical) must be
2 1/8 24
zero. Thus,
2√ . In our example above, we have m = 1/8 and k = 24, so that
2√3.
Let’s set
8√6 and see what happens. We have
1
8
2√3
24
0.
Solving (with the same initial conditions as before), we get
√
1
.
2
4√3
The graph is
It does not look too much different than the one before it, except that the mass moves to the rest state a
little faster than the one before this one. This makes sense since the previous one was in a more viscous
medium, so that the mass would move slower. In fact, if we increase the value of , the above graph will
just spread out more, trending to 0 slower.
An External Forcing Function.
Suppose the original problem included an external forcing function. This forcing function would impart
its energy onto the mass in some way.
Suppose this forcing function is sin
. This is a periodic forcing function, its period being 4 seconds,
or a little over 12 seconds. Thus, the original problem is now
16
192
sin
1
2
,
0
1
,
2
0
0.
The solution is
0.5002172 cos 8√2
0.3535 sin 8√2
0.005206 sin
1
2
0.0002172 cos
1
2
.
Here is the graph over the first half second or so:
The spring starts at 0.5 foot away, bobs past the midline, then starts to slow due to the viscosity of the
medium.
Note what happens when we look at the first 25 seconds:
The forcing function is now guiding the bobbing of the mass. Whatever effects of the initial few partial
seconds is now completely overwhelmed by the forcing function. The amplitude is small, about 0.005
units.
Prepared by Scott Surgent ([email protected]). Gosh, I hope there are no errors, but if there are, email me.