MATH 2280-002
Lecture Notes: 2/05/2013
Math 2280 Section 002 [SPRING 2013]
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Mass-Spring-Dashpot Systems (Continued)
Last time we looked at mass-spring-dashpot systems.
Remember that we can model this system using the following linear DE:
mx00 + cx0 + kx = F (t),
where
• m is the mass attached the spring;
• k is the spring constant;
• c is the damping constant; and
• F (t) is the external force.
We’ve handled the free, undamped case, so now it’s time to consider the free, damped case.
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Free Damped Motion
If we have a damping force (c > 0) but no external force (F (t) = 0),
mx00 + cx0 + kx = 0.
Divide by m.
c 0
k
x + x = 0.
m
m
Just like before, we’re going to define certain variables in a way that may seem arbitrary now but
make equations look nicer later. Also just like before, we’ll define the undamped circular frequency
r
k
ω0 =
.
m
x00 +
Set p =
c
2m .
That means our DE can be rewritten as
x00 + 2px0 + ω02 x = 0.
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MATH 2280-002
Lecture Notes: 2/05/2013
The characeristic polynomial r2 + 2pr + ω02 has roots
r1 , r2 = −p ± (p2 − ω02 )1/2 .
We have three possibilities depending on whether p2 − ω02 is positive, zero, or negative. Notice that
k
c2 − 4km
c2
−
=
,
4m2 m
4m2
√
√
√
so p2 − ω02 is positive, zero, or
√ negative if and only if c > 2 km, c = 2 km, or c < 2 km,
respectively. We’ll call ccr = 2 km the critical damping, and each of these cases depend on how
the damping constant c compares with the critical damping.
p2 − ω02 =
√
Case 1. c > ccr = 2 km
This is called the overdamped case, and we have that
p2 − ω02 > 0.
That means that our roots r1 , r2 = −p ± (p2 − ω02 )1/2 are real, distinct, and negative, so the general
solution is
x(t) = c1 er1 t + c2 er2 t .
Since r1 and r2 are negative,
x(t) → 0 as t → ∞.
We expected this since eventually the resistance will eventually stop the spring’s motion and the
mass will return to the equilibrium position (x = 0).
Example. The motion modeled by
3x00 (t) + 8x0 (t) + 4x(t) = 0
is overdamped. Some sample solution curves have been plotted below.
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MATH 2280-002
Lecture Notes: 2/05/2013
Case 2. c = ccr
This is called the critically damped case, and we have that
p2 − ω02 = 0.
That means that our roots r1 , r2 are both equal to the real, negative number −p, so the general
solution is
x(t) = c1 e−pt + c2 te−pt .
Again we observe that
x(t) → 0 as t → ∞.
(Because of l’Höpital’s rule, we know that e−pt decays much faster than t grows.)
Example. The motion modeled by
x00 (t) + 4x0 (t) + 4x(t) = 0
is critically damped. Some sample solution curves have been plotted below.
Case 3. c < ccr
This is called the underdamped case, and we have that
p2 − ω02 < 0.
That means that our roots r1 , r2 = −p ± (p2 − ω02 )1/2 = −p ± i(ω02 − p2 )1/2 are distinct complex
numbers. Let
ω1 = (ω02 − p2 )1/2 .
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MATH 2280-002
Lecture Notes: 2/05/2013
The general solution is
x(t) = e−pt [c1 cos(ω1 t) + c2 sin(ω1 t)].
Define new constants C and α by
C=
p
A2 + B 2 ,
cos(α) =
A
,
C
sin(α) =
B
.
C
The picture to have in mind when defining C and α is the right triangle below.
If this seems familiar, it’s because we used the same trick in the undamped case to rewrite our
solution.
Exercise: Verify that x(t) = Ce−pt cos(ω1 t − α).
In this form, it is easy to describe the behavior of the graph of x(t). The graph is bounded above
by Ce−pt and below by −Ce−pt . The cosine part is causing the graph to oscillate between negative
and positive values, but because Ce−pt and −Ce−pt approach 0 as t → ∞,
x(t) → 0 as t → ∞.
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MATH 2280-002
Lecture Notes: 2/05/2013
It’s easy to visualize the motion of the mass. For example, in the graph above, the spring starts
stretched (x > 0). The mass gets pulled toward the equilibrium position (x = 0) but overshoots
and the spring gets compressed (x < 0). Then the spring pushes out past the equilibrium position
but, due to resistance, not as far as it started. This continues; each time the spring stretches or
compresses less and less.
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