5Second-order differential equations
IA Differential Equations
5.5 Transients and damping
In many physical systems, there is some sort of restoring force and some damping,
e.g. car suspension system.
Consider a car of mass
M
with a vertical force
F
(
t
) acting on it (e.g. mouse
jumping on the car). We can consider the wheels to be springs (
F
=
kx
) with a
“shock absorber” (F = l ˙x). Then the equation of motion can be given by
M ¨x = F (t) − kx − l ˙x.
So we have
¨x +
l
M
˙x +
k
M
x =
1
M
F (t).
Note that if we don’t have the damping and the forcing, we end up with simple
harmonic motion of angular frequency
p
k/M
. Write
t
=
τ
p
M/k
, where
τ
is dimensionless. The timescale
p
M/k
is proportional to the period of the
undamped, unforced system (or 1 over its natural frequency). Then we obtain
¨x + 2κ ˙x + x = f(τ)
where, ˙x means
dx
dτ
, κ =
l
2
√
kM
and f =
F
k
.
By this substitution, we are now left with only one parameter
κ
instead of
the original three (M, l, k).
We will consider different possible cases.
Free (natural) response f = 0
¨x + 2κ ˙x + x = 0
We try x = e
λτ
λ
2
+ 2κλ + 1 = 0
λ = −κ ±
p
κ
2
− 1
= −λ
1
, −λ
2
where λ
1
and λ
2
have positive real parts.
Underdamping
If κ < 1, we have x = e
−κτ
(A sin
√
1 − κ
2
τ + B cos
√
1 − κ
2
τ).
The period is
2π
√
1−κ
2
and its amplitude decays in a characteristic of
O
(
1
κ
).
Note that the damping increases the period. As
κ →
1, the oscillation period
→ ∞.
τ
x
Critically damping
If κ = 1, then x = (A + Bτ )e
−κτ
.
The rise time and decay time are both
O
(
1
κ
) =
O
(1). So the dimensional rise
and decay times are O(
p
M/k).
τ
x
Overdamping
If
κ >
1, then
x
=
Ae
−λ
1
τ
+
Be
−λ
2
τ
with
λ
1
< λ
2
. Then the decay time is
O(1/λ
1
) and the rise time is O(1/λ
2
).
τ
x
Note that in all cases, it is possible to get a large initial increase in amplitude.
Forcing
In a forced system, the complementary functions typically determine the short-
time transient response, while the particular integral determines the long-time
(asymptotic) response. For example, if
f
(
τ
) =
sin τ
, then we can guess
x
p
=
C sin τ + D cos τ . In this case, it turns out that x
p
= −
1
2κ
cos τ.
The general solution is thus
x
=
Ae
−λ
1
τ
+
Be
−λτ
−
1
2κ
cos τ ∼ −
1
2κ
cos τ
as
τ → ∞ since Re(λ
1,2
) > 0.
It is important to note that the forcing response is out of phase with the
forcing.