MATH 2280-002
Lecture Notes: 2/11/2013
Math 2280 Section 002 [SPRING 2013]
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Forced Oscillations
Let’s return to mass-spring-dashpot systems. We’ve already considered the case where the motion
is free (in both the undamped and damped case); now it’s time to look at forced motion. Suppose
the external force exerted on the mass is simple harmonic:
F (t) = F0 cos(ωt).
That means x(t) (the distance from the mass to the equilibrium position) satisfies
mx00 + cx0 + kx = F0 cos(ωt).
The techniques we’ll discuss also work when F (t) = F1 sin(ωt) or F (t) = F0 cos(ωt) + F1 sin(ωt),
but I’ll that for you to work out in the homework.
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Undamped Forced Motion
Suppose there is no damping force (c = 0). The natural frequency of the mass-spring system is
r
k
ω0 =
.
m
This is what we called the undamped circular frequency earlier. We have three cases depending on
the relationship between ω and ω0 .
r
k
Case 1. ω 6= ω0 =
m
Our complementary function is xh (t) = c1 cos(ω0 t) + c2 sin(ω0 t). We’ll use the method of
undetermined coefficients to find a particular solution. Take
xp (t) = A cos(ωt) + B sin(ωt)
as a trial solution. I’ll leave solving for A and B as an exercise for you.
Exercise. Verify that
x(t) = c1 cos(ω0 t) + c2 sin(ω0 t) +
is the general solution to mx00 + kx = F0 cos(ωt).
Suppose we have the initial conditions
x(0) = x0 (0) = 0.
Then we can solve c1 and c2 in our general solution x(t).
1
F0
cos(ωt)
− ω2)
m(ω02
MATH 2280-002
Lecture Notes: 2/11/2013
Exercise. Verify that, in this case,
2F0
x(t) =
sin
m(ω02 − ω 2 )
ω0 + ω
ω0 − ω
t sin
t .
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2
Notice that as ω gets closeto ω0 , the amplitude “blows up” since, as ω → ω0 , ω02 − ω 2 goes
to 0 faster than sin ω02−ω t goes to 0. (Check this for yourself!)
Example. Find and graph the solution to the IVP
2x00 + 8x = 90 cos(7t),
x(0) = x0 (0) = 0.
I work this example out a Maple worksheet I’ve posted on the webpage, but here’s what the
solution looks like:
r
Case 2. ω ≈ ω0 =
k
m
When ω is relatively close but not equal to ω0 , we get beating – a rapid oscillation with a
(comparatively) slowly varying periodic amplitude. We get the same general solution as in the
previous case, but the graph of the solution displays certain distinctive behavior in this case.
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MATH 2280-002
Lecture Notes: 2/11/2013
Example. Find and graph the solution to the IVP
.4x00 + 1690x = 8 cos(68t),
x(0) = x0 (0) = 0.
Take a look at the Maple worksheet I posted. Here’s what the beating looks like:
r
Case 3. ω = ω0 =
k
m
Our complementary function is xh (t) = c1 cos(ω0 t) + c2 sin(ω0 t) (just like in the previous two
cases), but now for our trial solutionfor xp , we take
xp (t) = t[A cos(ωt) + B sin(ωt)].
Since ω = ω0 , we have to multiply by t to prevent duplication.
Exercise. Verify that
x(t) = c1 cos(ω0 t) + c2 sin(ω0 t) +
F0
t sin(ω0 t)
2mω0
is the general solution to mx00 + kx = F0 cos(ωt).
F0
t → ∞, no matter how small F0 6= 0 is! That means that each time the mass
2mω0
leaves the equilibrium position, the mass goes farther and farther away from the equilibrium
position. This phenomenon is called resonance.
As t → ∞,
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MATH 2280-002
Lecture Notes: 2/11/2013
Example. Find and graph the solution to the IVP
x00 + 36x = 3 cos(6t),
x(0) = x0 (0) = 0.
Again, take a look at the Maple worksheet I posted. The graph of x(t) is included below.
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