# 9 Lecture 9: Damped harmonic motion

```Mathematics for Physics 3: Dynamics and Diﬀerential Equations
9
56
Lecture 9: Damped harmonic motion
In the previous lectures we considered in some detail the example of a pendulum in the small angle approximation,
and then analysed periodic motion in a more general potential. What is common to all oscillation examples considered
so far is that the total mechanical energy is conserved. We now drop this assumption and analyse the case of an
oscillating system that is dissipating its energy. In the next lecture we will consider in addition a time dependent
force. Both these lectures serve as an introduction to the subject of linear second order diﬀerential equations.
Damped harmonic motion
Let us consider a one-dimensional oscillator, such as a spring with a Hooke’s constant k, Felastic = −kx, which also
has a linear friction Ffriction = −µẋ, which is always against the direction of motion. The equation of motion is:
mẍ = −kx − µẋ
Dividing by m we can bring the equation of motion into the following standard form:
�
k
1µ
2
ẍ + 2λẋ + ω x = 0,
ω=
, λ=
,
m
2m
(157)
(158)
where λ as well as ω are positive. A similar equation can be obtained for a pendulum at small angles and many
other systems near equilibrium, where now we do not neglect the damping force.
We observed that (158) is a second order linear ordinary diﬀerential equation with constant coeﬃcients. Moreover,
this equation is homogeneous: it can be written using a diﬀerential operator as follows:
Ô [x(t)] = 0
where
Ô =
d2
d
+ 2λ + ω 2 .
2
dt
dt
Linear homogeneous equations have the property of superposition, namely that if x1 (t) and x2 (t) are solutions,
namely Ô [x1 (t)] = 0 and Ô [x2 (t)] = 0, then also any linear combination of x1 and x2 is a solution:
�
�
Ô ax1 (t) + bx2 (t) = 0 .
Moreover second order equations have two linearly independent solutions (they involve two constants of integration),
so if we found two such linearly independent solutions, ax1 (t) + bx2 (t) is in fact the most general solution.
For the case of constant coeﬃcients we may find the independent solutions by using a trial function of the form:
x(t) = eα t
(159)
Inserting this into (158) we get the following characteristic equation:
α2 + 2λα + ω 2 = 0
whose roots are:
α1,2 = −λ ±
�
λ2 − ω 2
(160)
Thus, assuming that ω �= λ we obtain two distinct solutions, and our general solution is
x(t) = aeα1 t + beα2 t
According to (160) the motion is qualitatively diﬀerent in the following three situations:
• λ > ω: over-damping
• λ < ω: under-damping
• λ = ω: critical damping
We now analyse them one by one.
(161)
Mathematics for Physics 3: Dynamics and Diﬀerential Equations
57
Over-damped harmonic motion
If λ > ω we have two distinct real negative roots
�
α1 = −λ + λ2 − ω 2
α2 = −λ −
so the most general solution is
√
�
λ2 − ω 2
(162)
√
2
2
2
2
x(t) = ae−(λ− λ −ω )t + be−(λ+ λ −ω ) t
�
�
√
√
2
2
2
2
= e−λt ae+t λ −ω + be−t λ −ω
�
� �
�
� �
��
= e−λt A cosh t λ2 − ω 2 + B sinh t λ2 − ω 2
(163)
We note that both exponentials tend to zero at large times. There is no oscillation at all: the damping is suﬃciently
fast that the oscillatory motion is not observed. The constants of integration a and b, or alternatively A and B,
should be fixed based on the initial conditions.
Under-damped harmonic motion
If λ < ω we have two distinct complex roots, which are complex conjugates of each other:
�
�
α1 = −λ + i ω 2 − λ2
α2 = −λ − i ω 2 − λ2
(164)
so the most general solution is
√
√
2
2
2
2
x(t) = ae−(λ−i ω −λ )t + be−(λ+i ω −λ ) t
�
� �
�
� �
��
= e−λt A cos t ω 2 − λ2 + B sin t ω 2 − λ2 ,
(165)
where the second expression is more natural if x(t) (and thus A and B) is real, as is indeed the case in classical
mechanics. We now observe that oscillations are present. The damping exponential e−λt suppresses the amplitude
of the oscillations.
Figure 11: Examples of an over-damped and an under-damped harmonic motion. In the latter case we also show the
enveloping amplitude.
Critically-damped harmonic motion
Finally consider the case where the two roots become degenerate, i.e. λ = ω, so
α1 = α2 = −λ .
Now the two exponentials eα1 t = e−λt and eα2 t = e−λt are clearly not linearly independent, so we have just one of
the two functions in the general solution. The other is provided by t e−λt . In other words, our claim is the general
solution of
ẍ + 2λẋ + λ2 x = 0,
(166)
is
x(t) = (a + bt) e−λt
(167)
As can be readily verified by substitution. Similarly to the over-damped oscillator, no oscillations are observed, just
monotonous damping.
```
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