The damped harmonic oscillator
D. Jaksch1
Goals:
• Understand the behaviour of this paradigm exactly solvable physics model that appears in numerous
applications.
• Understand the connection between the response to a sinusoidal driving force and intrinsic oscillator
properties.
• Understand the connection between the Q factor, width of this response and energy dissipation.
The damped harmonic oscillator
1. A damped harmonic oscillator is displaced by a distance x0 and released at time t = 0. Show that the
subsequent motion is described by the differential equation
m
dx
d2 x
+ mγ
+ mω02 x = 0 ,
dt2
dt
or equivalently
mẍ + mγ ẋ + mω02 x = 0 ,
with x = x0 and ẋ = 0 at t = 0, explaining the physical meaning of the parameters m, γ and ω0 .
• Find and sketch solutions for (i) overdamping, (ii) critical damping, and (iii) underdamping. (iv)
What happens for γ = 0?
• For a lightly damped oscillator the quality factor, or Q-factor, is defined as
Q=
energy stored
.
energy lost per radian of oscillation
Show that Q = ω0 /γ.
We now add a driving force F cos(ωt) to the harmonic oscillator so that its equation becomes
mẍ + mγ ẋ + mω02 x = F cos(ωt) .
• Explain what is meant by the steady state solution of this equation, and calculate the steady state
solution for the displacement x(t) and the velocity ẋ(t).
• Sketch the amplitude and phase of x(t) and ẋ(t) as a function of ω.
• Determine the resonant frequency for both the displacement and the velocity.
• Defining ∆ω as the full width at half maximum of the resonance peak calculate ∆ω/ω0 to leading
order in γ/ω0 .
• For a lightly damped, driven oscillator near resonance, calculate the energy stored and the power
supplied to the system. Hence confirm that Q = ω0 /γ as shown above. How is Q related to the
width of the resonance peak?
Solution: The forces on the mass m are Fs = −kx = −mω02 x due to the spring and Ff = −mγ ẋ due to
friction γ. The equation follows from Newton’s law mẍ = Fs + Ff .
The characteristic polynomial for ansatz x(t) = eλt is λ2 + γλ + ω02 = 0 leading to eigenfrequencies
r
γ2
γ
− ω02 .
λ1,2 = − ±
2
4
1 These
problems are based on problem sets by Prof J. Yeomans.
We get (i) overdamping when γ > 2ω0 and hence solutions do not oscillate, (ii) critical damping for
γ = 2ω0 and (iii) underdamping for γ < 2ω0 . Different solutions are shown in Fig. 1. The general
solution is given by
x(t) = < A1 eλ1 t + A2 eλ2 t .
p
and can be simplified for the different situations (writing α = |ω02 − γ 2 /4|) for the three cases
(i) x(t) = e−γt/2 [A cosh(αt) + B sinh(αt)] or equivalently x(t) = e−γt/2 (Ceαt + De−αt )
(ii) x(t) = e−γt/2 (A + Bt)
(iii) x(t) = e−γt/2 [A cos(αt) + B sin(αt)]
using the standard procedure for degenerate roots of the characteristic polynomial in (ii).
By matching the initial conditions we find for the different cases
(i) A = x0 and B = x0 γ/(2α) or equivalently C = x0 (α + γ/2)/(2α) and D = x0 (α − γ/2)/(2α)
(ii) A = x0 and B = x0 γ/2
(iii) A = x0 and B = x0 γ/(2α)
overdamped
critical
xHtL
1
underdamped
xHtL
1
xHtL
1
2
0
2
4
w0 t
0
2
4
w0 t
4
w0 t
-1
Figure 1: Oscillator displacement for different dampings.
The energy stored in the harmonic oscillator is the sum of kinetic and elastic energy
2
E(t) =
mẋ(t)
mω02 x(t)2
+
.
2
2
In order to proceed for the lightly damped case it is easiest to write x(t) = A cos(αt − φ)e−γt/2 and thus
ẋ(t) = −Aα sin(αt − φ)e−γt/2 − γx(t)/2. Since lightly damped means γ ω0 we may neglect the second
term in ẋ(t) and approximate α ≈ ω0 . Then the expression for the energy simplifies to
E(t) =
mω02 2 −γt
A e
.
2
A radian corresponds to the time difference τ = 1/ω0 and so we find the energy lost per radian
EL = E(0) − E(1/ω0 ) =
mω02 2
mω0 γ 2
A (1 − e−γ/ω0 ) ≈
A .
2
2
by expanding e−γ/ω0 ≈ 1 − γ/ω0 for γ ω0 . Hence the result Q = E(0)/EL = ω0 /γ follows as required.
We now turn to the forced damped harmonic oscillator. The solutions to the homogeneous equation will
damp out on a time scale 1/γ. At times t 1/γ only terms arising from the particular solution will
2
remain. These
terms
describe the stationary state . We work out a particular solution using the ansatz
iωt
x(t) = < A(ω)e
and find
A(ω) =
F
= |A(ω)|e−iϕ ,
m(ω02 − ω 2 + iγω)
where
|A(ω)| =
F
p
m ((ω02 − ω 2 )2 + γ 2 ω 2 )
and ϕ =


arctan
γω
ω02 −ω 2
 π + arctan
γω
ω02 −ω 2
for ω ≤ ω0
.
for ω > ω
2 Since we can always move a term from the homogeneous solution to the particular solution it is strictly speaking not accurate
to say that the particular solution is the stationary state.
Magnitude
|A(ω)|
and phase ϕ are shown in Fig. 2 as a function of ω. The velocity is given by ẋ(t) =
< i|A(ω)|ωeiωt = −|A(ω)|ω cos(ωt − (ϕ + π/2)), i.e. there is an additional shift of π/2 compared to
the displacement. The additional factor of ω shifts the maximum amplitude of ẋ(t) compared to that of
x(t). Amplitude and phase of ẋ(t) are shown in Fig. 2.
ÈAÈ•A0
10
f
p
p
5
2
0
1
w•w0
2
f
0
1
w•w0
2
wÈAÈ•HA0w0L
10
5
p
1
2
w•w0
0
1
2
w•w0
Figure 2: Displacement and velocity response to periodic driving for γ = ω0 /10 and γ = ω0 /4.
The maximum of the displacement amplitude is found by solving d|A(ω)|/dω = 0 giving a resonance
frequency ωx2 = ω02 − γ 2 /2. For the maximum velocity amplitude we solve d|ωA(ω)|/dω = 0 and find the
resonance frequency ωẋ = ω0 .
We write the full width half maximum as ∆ω = ω2 − ω1 with A(ωi=1,2 ) = A(ωx )/2. We take the square
of this expression and find
1
1
1
=
(ω02 − ωi2 )2 + γ 2 ωi2
4 (ω02 − ωx2 )2 + γ 2 ωx2
This can be re-written as γ 4 + γ 2 (ωi2 − 4ω02 ) + (ω02 − ωi2 )2 = 0. This can in principle be solved for ωi
but since we have assumed the oscillator to be lightly damped and have worked out quantities like Q
only to lowest order in γ/ω0 we instead only look for a solution valid to this order. We thus substitute
2 2 2
3
ωi ≈ ω0 (1 + βγ) obtaining γ 4 + γ 2 (βγω0 − 3ω02 ) + β√
γ ω0 = 0. We now ignore
√any terms of O(γ ) and
O(γ 4 ) and thus get the approximate solution β = ± 3 and thus ω02 − ωi2 ≈ ± 3γω0 to lowest order. A
Taylor series expansion in γ/ω0 yields
q
q
√
√
√
∆ω = ω2 − ω1 = ω0
1 + 3γ/ω0 − 1 − 3γ/ω0 ≈ 3γ .
Hence ∆ω/ω0 =
√
3γ/ω0 .
Near resonance ω ≈ ω0 and we thus find for the energy of the oscillator
m
mω02
F2
ẋ(t)2 +
x(t)2 ≈
.
2
2
2mγ 2
E=
The average supplied power is given by
P = F cos(ωt)ẋ(t) = −F |A(ω)|ωcos(ωt) sin(ωt − ϕ) = −F |A(ω)|ωcos(ωt)(sin(ωt) cos(ϕ) − cos(ωt) sin(ϕ)) .
Near resonance we have ϕ ≈ π/2 and ω ≈ ω0 so that
P =
F |A(ω0 )|ω0
F2
=
.
2
2mγ
In the steady state the energy dissipated per radian must be equal to the energy supplied by the external
force per radian EL = P τ = P/ω0 . Thus
√
E
ω0
∆ω
3
Q=
=
and
=
.
EL
γ
ω0
Q