Mathematics for Physics 3: Dynamics and Differential Equations
10
58
Lecture 10: Forced (and damped) harmonic motion
In the previous lecture we introduced a new component into the harmonic oscillator, a dissipating force proportional
to the velocity. The equation of motion we obtained was more involved, but it was still homogeneous. Now we
consider in addition the application of an external force. This force, will be in general time-dependent, and it would
lead to non-homogeneous equations of motion.
Forced harmonic motion
Let us generalise the discussion of oscillators to include the effect of an external time-dependent force F (t). The
equation of motion becomes:
mẍ = −µẋ − kx + F (t) .
(168)
Let us move the elastic term to the l.h.s. and writing it as − dVdx(x) where V (x) = 12 kx2 . By multiplying the equation
by ẋ we obtain:
�
�
dE
d 1
2
=
mẋ + V (x) = −µẋ2 + ẋF (t) .
dt
dt 2
We now see that while the damping term is negative definite (it always dissipates the energy of the system) the
external force term, ẋF (t) can have any sign, and thus it can either pump energy into the system, or out of the
system.
Bringing the equation of motion into standard form we have:
�
k
1µ
1
2
ẍ + 2λẋ + ω x = F (t),
ω=
, λ=
(169)
m
m
2m
This is a linear inhomogeneous second order ODE. The inhomogeneous term is given by the external, timedependent force. As before we define the linear differential operator by
Ô =
d2
d
+ 2λ + ω 2 .
dt2
dt
While the fact that equation (169) is non-homogeneous renders it more difficult to solve, it is still linear, which
helps a lot.
The general solution of a linear inhomogeneous ODE of the form
Ô [x(t)] = f (t)
(170)
x(t) = xH (t) + xP (t)
(171)
is:
where xH (t) is the general solution of the corresponding homogeneous equation,
Ô [x(t)] = 0 ,
while xP (t) is a particular solution of the inhomogeneous equation (171). Note that the two coefficients that need
to be fixed by the initial conditions reside in xH (t). Note however that we can only fix these coefficients based on
the initial conditions using the general solution (172). Thus we must first determine xP (t).
Let us take a trivial example first. Assume that F (t)/m = a where a is a constant. Now our equation is
ẍ + 2λẋ + ω 2 x = a
(172)
We already know xH (t) from the previous discussion. A particular solution is xP (t) = a/ω 2 . Thus the general
solution is x(t) = xH (t) + a/ω 2 . This solution can be matched with the initial conditions to fix the so-far arbitrary
constants of integration in xH (t).
Linearity also helps when computing the particular solution. A simple illustration of this is the following. Assume
(for example) that the force on the r.h.s has two components, so the equation takes the form:
Ô [x(t)] = ag1 (t) + bg2 (t)
(173)
When looking for a solution of this equation, the principal of superposition takes the following form: if x1 (t) is a
solution to Ô [x(t)] = g1 (t) and x2 (t) is a solution to Ô [x(t)] = g2 (t) then the particular linear combination of the
two, ax1 (t) + bx2 (t), admits:
�
�
Ô ax1 (t) + bx2 (t) = ag1 (t) + bg2 (t) .
Note that here a and b are the same constants appearing on the r.h.s. of (170). These clearly depend only on the
force, not on the initial conditions.
Mathematics for Physics 3: Dynamics and Differential Equations
59
Periodic forced oscillations
Let us consider now applying an external force F (t) = f cos(nt) (where f and n are constants) so our equation
becomes
�
k
1µ
2
ẍ + 2λẋ + ω x = a cos(nt),
ω=
, λ=
, a = f /m
(174)
m
2m
We already know the homogeneous solutions xH (t). Thus we need to look for a particular solution. It is natural to
try an oscillatory solution with the same frequency as the inducing force,
xP (t) = A cos(nt − φ) .
(175)
Differentiating (175) we get:
ẋP (t) = −nA sin(nt − φ)
and
ẍP (t) = −n2 A cos(nt − φ)
Thus, substituting this solution into the equation yields:
��
�
�
A ω 2 − n2 cos(nt − φ) − 2λn sin(nt − φ) = a cos(nt)
Using the standard trigonometric relations for the cosine and sine of a sum of angles this equation becomes:
��
�
�
��
��
A ω 2 − n2 cos(nt) cos(φ) + sin(nt) sin(φ) − 2λn sin(nt) cos(φ) − cos(nt) sin(φ)
= a cos(nt)
The equality must be realised at any time t. We can therefore compare the coefficients of cos(nt) and sin(nt)
separately, obtaining two conditions:
��
�
�
A ω 2 − n2 cos(φ) + 2λn sin(φ) = a
(176)
��
�
�
A ω 2 − n2 sin(φ) − 2λn cos(φ) = 0
(177)
The second relation gives:
tan(φ) =
Recalling that
cos(φ) = �
the first relation gives:
1
1 + tan2 (φ)
2λn
− n2
ω2
tan(φ)
sin(φ) = �
1 + tan2 (φ)
,
�
a 1 + tan2 (φ)
a
�
A= � 2
= ��
�2
2
ω − n + 2λn tan(φ)
ω 2 − n2 + 4λ2 n2
(178)
so we conclude that indeed, a particular solution is given by the above ansatz, namely:
xP (t) = ��
a
�2
ω 2 − n2
+ 4λ2 n2
cos(nt − φ),
tan(φ) =
2λn
ω 2 − n2
To write the general solution to (174) we need to know whether the system is under-damped, over-damped or
critically-damped. Let us assume that λ < ω so the system is under-damped. The general solution is then:
x(t) = Ce−λt cos(t
�
ω 2 − λ2 + θ) + ��
a
�2
ω 2 − n2
+ 4λ2 n2
cos(nt − φ),
tan(φ) =
2λn
ω 2 − n2
(179)
where C and θ need to be determined by the initial conditions.
We observe that while the homogeneous piece vanishes as t → ∞ the particular solution does not. Here we
made a distinction between the transient term, which is a function of the initial conditions, and a steady-state
solution, which depends on the forcing term.
Mathematics for Physics 3: Dynamics and Differential Equations
60
Resonance
The steady-state solution’s amplitude (178) depends not only on the amplitude of the driving force, but also on
the relation between the frequency of the force n and the characteristic mode of the oscillator ω, as well as on the
damping parameter λ.
For n → 0 we get A → a/ω. For n → ∞ we obtain A → 0. In between these two extremes the amplitude may
reach a maximum, which we refer to as the resonance frequency. To find it let us look for the point where the
derivative of the denominator in (178) vanishes:
The non-trivial solution is
�
�2
�
�
d �� 2
ω − n2 + 4λ2 n2 = −4n ω 2 − n2 + 8λ2 n = 0
dn
n = nr =
�
ω 2 − 2λ2
This is the resonance frequency. Note that for weak damping λ → 0 the resonance frequency becomes close to
the characteristic frequency of the system√ω, and the peak is narrow. In contrast if λ becomes large the resonance
becomes wider, and eventually, for λ = ω/ 2 it disappears altogether, and the amplitude of the steady-state solution
becomes a monotonously-decreasing function of the forced-oscillation frequency.
Figure 12:
with ω = 1 and λ = 3/10 (left) and
�
√ The amplitude of the steady-state solution of a forced oscillator
λ = 1/ 2 (right). In the former case the resonance occurs for nr = 41/50 � 0.906; in the latter there is no
resonance.