Amplitude and phase shift as functions of ω
Consider the differential equation
1
u00 + u0 + 4u = 3 cos(ωt).
4
This DE models the motion of a spring-mass system where m = 1, γ = 41 , k = 4, and there
is an external force of 3 cos(ωt) Newtons.
Remember that we really care about the steady state part of the solution (called U (t))
to this DE, since it does not depend on the initial conditions and since the actual motion
u(t) is basically indistinguishable from the steady state after only a short amount of time
anyway.
Remember that the steady state part of the solution is the particular solution to the
nonhomogeneous equation. In this case it has the form
U (t) = A cos(ωt) + B sin(ωt).
As you saw in section 3.7, you can always rewrite this in the form
R cos(ωt − δ),
where R is the amplitude and δ is the phase shift.
Let’s see how R and δ change as ω changes. Here’s a table. Each row represents a different
choice for ω. The second column, U (t), is found by using the method of undetermined
coefficients (but I have already done the work for you). Then the numbers R and δ are
found by the procedure we used in 3.7 to combine sine and cosine terms.
Remember that everything in the system is kept completely identical, except for the
frequency of the external force function. The spring, mass, damping, and so on are always
the same.
ω
0
0.1
1
1.5
1.95
1.99
2
3
U (t), the steady state
0.75 = 0.75 cos(0t)
0.75 cos(0.1t) + 0.0047 sin(0.1t)
0.99 cos(t) + 0.08 sin(t)
1.64 cos(1.5t) + 0.35 sin(1.5t)
2.08 cos(1.95t) + 5.34 sin(1.95t)
0.04 cos(1.99t) + 5.99 sin(1.99t)
6 sin(2t)
−0.59 cos(3t) + 0.088 sin(3t)
R
δ
0.75
0
≈ 0.75 0.006
0.99
0.08
1.68
0.21
5.73 1.199
6.009 1.49
π
6
2
0.597 2.99
Observations:
1. Surprisingly, altering ω by a little bit can result in a radically different steady state
response! Usually the amplitude is rather small, but notice that there’s a window
where ω is about 1.95–2 where the amplitude is very large.
2. When ω is very close to zero, the steady state is mostly a cosine function, with very
little sine. On the other hand, when ω is equal to 2, the steady state is all sine. And
when ω gets bigger than 2 (say, 3), the steady state is mostly cosine again, but with a
negative coefficient of cosine.
3. The phase shift δ goes from 0 to about 3. If I added even more rows to this table, with
larger ωs, you would see that δ would approach π but never pass it, even as ω gets
arbitrarily large.
4. Interesting things (see items (1) and (2) above) seem to happen when ω is close to 2.
Also, 2 is the natural frequency of the system! This is not a coincidence.
Here’s a summary of what’s going on. The resonant frequency ωmax is close to the natural
frequency of 2, but slightly smaller (because γ is small but not zero). You can find a formula
in the book. This means that the amplitude of the steady state response is greatest when ω
is just below the natural frequency. As ω increases past the natural frequency, the amplitude
R drops off dramatically. This effect is largest when the damping constant γ is very small
compared to the other numbers in the equation.
Also, when ω = 2, the natural frequency, the phase shift δ is π2 . (This is the same thing as
saying that U (t) is a sine function.) This means that the graph of the steady state response
is a quarter-period out of phase with the forcing function. (Graph a sine function and a
cosine function on the same plane to see this in action.)
On the other hand, as ω gets large, δ approaches π. This means that the steady state
response and the forcing function are very close to being exactly out of phase (so that one
is positive while the other is negative). To see this in action, graph cos(t) and − cos(t) on
the same plane.
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