The damped harmonic oscillatorD. Jaksch1

Goals:

• Understand the behaviour of this paradigm exactly solvable physics model that appears in numerousapplications.

• Understand the connection between the response to a sinusoidal driving force and intrinsic oscillatorproperties.

• 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 thesubsequent motion is described by the differential equation

md2x

dt2+mγ

dx

dt+mω2

0x = 0 ,

or equivalentlymx+mγx+mω2

0x = 0 ,

with x = x0 and x = 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

mx+mγx+mω20x = F cos(ωt) .

• Explain what is meant by the steady state solution of this equation, and calculate the steady statesolution for the displacement x(t) and the velocity x(t).

• Sketch the amplitude and phase of x(t) and x(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 leadingorder in γ/ω0.

• For a lightly damped, driven oscillator near resonance, calculate the energy stored and the powersupplied to the system. Hence confirm that Q = ω0/γ as shown above. How is Q related to thewidth of the resonance peak?

Solution: The forces on the mass m are Fs = −kx = −mω20x due to the spring and Ff = −mγx due to

friction γ. The equation follows from Newton’s law mx = Fs + Ff .

The characteristic polynomial for ansatz x(t) = eλt is λ2 + γλ+ ω20 = 0 leading to eigenfrequencies

λ1,2 = −γ2±√γ2

4− ω2

0 .

1These 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 forγ = 2ω0 and (iii) underdamping for γ < 2ω0. Different solutions are shown in Fig. 1. The generalsolution is given by

x(t) = <{A1eλ1t +A2eλ2t

}.

and can be simplified for the different situations (writing α =√|ω2

0 − γ2/4|) 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 = x0γ/(2α) or equivalently C = x0(α+ γ/2)/(2α) and D = x0(α− γ/2)/(2α)

(ii) A = x0 and B = x0γ/2

(iii) A = x0 and B = x0γ/(2α)

0 2 4w0 t

1

xHtLoverdamped

0 2 4w0 t

1

xHtLcritical

2 4w0 t

-1

1

xHtLunderdamped

Figure 1: Oscillator displacement for different dampings.

The energy stored in the harmonic oscillator is the sum of kinetic and elastic energy

E(t) =mx(t)

2

2+mω2

0x(t)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 thusx(t) = −Aα sin(αt−φ)e−γt/2− γx(t)/2. Since lightly damped means γ � ω0 we may neglect the secondterm in x(t) and approximate α ≈ ω0. Then the expression for the energy simplifies to

E(t) =mω2

0

2A2e−γt .

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) =mω2

0

2A2(1− e−γ/ω0) ≈ mω0γ

2A2 .

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 willdamp out on a time scale 1/γ. At times t � 1/γ only terms arising from the particular solution willremain. These terms describe the stationary state2. We work out a particular solution using the ansatzx(t) = <

{A(ω)eiωt

}and find

A(ω) =F

m(ω20 − ω2 + iγω)

= |A(ω)|e−iϕ ,

where

|A(ω)| = F

m√

((ω20 − ω2)2 + γ2ω2)

and ϕ =

arctan(

γωω2

0−ω2

)for ω ≤ ω0

π + arctan(

γωω2

0−ω2

)for ω > ω

.

2Since we can always move a term from the homogeneous solution to the particular solution it is strictly speaking not accurateto 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 x(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 x(t) compared to that ofx(t). Amplitude and phase of x(t) are shown in Fig. 2.

0 1 2wêw0

p

2

p

f

0 1 2wêw0

5

10

»A»êA0

1 2wêw0

p

f

0 1 2wêw0

5

10

w»A»êHA0w0L

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 resonancefrequency ω2

x = ω20 − γ2/2. For the maximum velocity amplitude we solve d|ωA(ω)|/dω = 0 and find the

resonance frequency ωx = ω0.

We write the full width half maximum as ∆ω = ω2 − ω1 with A(ωi=1,2) = A(ωx)/2. We take the squareof this expression and find

1

(ω20 − ω2

i )2 + γ2ω2i

=1

4

1

(ω20 − ω2

x)2 + γ2ω2x

This can be re-written as γ4 + γ2(ω2i − 4ω2

0) + (ω20 − ω2

i )2 = 0. This can in principle be solved for ωibut since we have assumed the oscillator to be lightly damped and have worked out quantities like Qonly to lowest order in γ/ω0 we instead only look for a solution valid to this order. We thus substituteωi ≈ ω0(1 + βγ) obtaining γ4 + γ2(βγω0 − 3ω2

0) + β2γ2ω20 = 0. We now ignore any terms of O(γ3) and

O(γ4) and thus get the approximate solution β = ±√

3 and thus ω20 − ω2

i ≈ ±√

3γω0 to lowest order. ATaylor series expansion in γ/ω0 yields

∆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

E =m

2x(t)2 +

mω20

2x(t)2 ≈ F 2

2mγ2.

The average supplied power is given by

P = F cos(ωt)x(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

2=

F 2

2mγ.

In the steady state the energy dissipated per radian must be equal to the energy supplied by the externalforce per radian EL = Pτ = P/ω0. Thus

Q =E

EL=ω0

γand

∆ω

ω0=

√3

Q.