2.6 Forced Oscillations and Resonance1

Oscillator equation with external force F(t): basic case assumes F periodic,

mx′′ + cx′ + kx = F0 cos ωt

Many real-life situations can be modelled with this equation, for example buildings in an earthquake.

There are three standard cases.

Case 1: Beating

Take c = 0 (no damping/friction) and ω 6= ω0 =√

km (driving frequency 6= natural frequency).

Already found complementary function xC(t) = c1 cos ω0t + c2 sin ω0t.Particular integral: guess xP(t) = a cos ωt + b sin ωt. Then

mx′′P + kxP = (−maω2 + ka) cos ωt + (−mbω2 + kb) sin ωt

= F0 cos ωt ⇐⇒ a =F0

k−mω2 =F0

m(ω20 −ω2)

, b = 0

=⇒ xP(t) =F0

m(ω20 −ω2)

cos ωt

=⇒ x(t) = xP(t) + xC(t) =F0

m(ω20 −ω2)

cos ωt + c1 cos ω0t + c2 sin ω0t

• Sum of distinct periodic motions.

• Larger F0 =⇒ more motion.

• ω close to ω0 =⇒ more motion.

Suppose have initial conditions x(0) = 0 = x′(0) (periodic force applied to resting spring). Quicklyobtain

x(t) =F0

m(ω20 −ω2)

(cos ωt− cos ω0t

)=

2F0

m(ω20 −ω2)

sinω0 −ω

2t sin

ω0 + ω

2t

using a trigonometric identity. If ω0, ω close in value, then ω0 −ω � ω0 + ωAmplitude beats at ω0−ω

2 rad/s.x

2 4 6 8 10t

ω0 = 20 ω = 18

Graphics show x′′ + 400x = 38 cos 18t where ω0 = 20. Solution x(t) = sin t sin 19t = A(t) sin 19t.High frequency vibration sin 19t with periodic amplitude A(t) = sin t.

1This is an abstract summary. Study this open-book and pay attention to the numerical examples from lectures. . .

Case 2: Resonance

No friction (c = 0) where ω = ω0 Driving frequency = natural frequency

Obvious guess xP(t) = a cos ω0t + b sin ω0t already solves homogeneous ODE, so try

xP(t) = at cos ω0t + bt sin ω0t

Substituting in the ODE and solving for a, b gives

xP(t) =F0

2mω0t sin ω0t

xP is also the solution corresponding to2 the initial conditions x(0) = x′(0) = 0.

• Sine wave with increasing amplitude A(t) =F0

2mω0t

• Spring tears itself apart!

Example Consider the IVP{

4x′′ + 64x = 4 cos ωtx(0) = x′(0) = 0

The natural frequency is ω0 =√

k/m = 4 rad/s. The four graphs show the solutions for fourdifferent driving frequencies: the last case is resonance.

0.005

0

x

2 4 6 8 10t

ω = 20 x(t) =1

192sin 8t sin 12t

0.1

0

x

2 4 6 8 10t

ω = 6 x(t) =110

sin t sin 5t

0.10

x

2 4 6 8 10t

ω = 5 x(t) =29

sint2

sin9t2

2

4

−4

−2

0

x

2 4 6 8 10t

ω = 4 x(t) =18

t sin 4t

2It is also what we obtain by taking limω→ω0 x(t) from the previous slide

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Summary of undamped-driven motion The initial value problem mx′′+ kx = F0 cos ωt with initialconditions x(0) = 0 = x′(0) has solution

x(t) =

2F0/mω2

0 −ω2sin

ω0 −ω

2t sin

ω0 + ω

2t if ω 6= ω0

F0

2mω0t sin ω0t if ω = ω0

where ω0 =√

k/m. As ω → ω0 low frequency beats of increasing amplitude occur.The clickable animation is generated with the same F0, m throughout

Case 3: Damped-driven motion (practical resonance)

c > 0 mx′′ + cx′ + kx = F0 cos ωt x′′ + 2px′ + ω20x =

F0

mcos ωt

Transient and steady-periodic solutions Three types:

Damping Condition Complementary Function xC(t)

Overdamping c2 > 4km e−pt(

c1e−√

p2−ω20 t + c2e

√p2−ω2

0 t)

Critical damping c2 = 4km (c1 + c2t)e−pt

Underdamping c2 < 4km e−pt(c1 cos ω1t + c2 sin ω1t)

where ω1 =√

ω20 − p2

• p > 0 =⇒ xC transient: limt→∞

xC(t) = 0

• Particular integral: standard guess xP(t) = a cos ωt + b sin ωt always works.

• Regardless of initial conditions, x(t) = xP(t) for large t: steady-periodic solution.

Example Find the steady-periodic solution to x′′ + 3x′ + 2x = cos ωt.

Try xP(t) = a cos ωt + b sin ωt. Substitute in the ODE:

(−aω2 + 3bω + 2a) cos ωt + (−bω2 − 3aω + 2b) sin ωt = cos ωt

=⇒ a =ω2 − 2

9ω2 + (ω2 − 2)2 , b =3ω

(ω2 − 2)2 + 9ω2

=⇒ xP(t) =1

(ω2 − 2)2 + 9ω2

((ω2 − 2) cos ωt + 3ω sin ωt

)=

1(ω2 − 2)2 + 9ω2 cos (ωt− γ)

where γ is the phase angle.

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General situation Long-term solution is

xP(t) =F0/m

(ω20 −ω2)2 + (2pω)2

((ω2

0 −ω2) cos ωt + 2pω sin ωt)

=F0

m√(ω2

0 −ω2)2 + (2pω)2cos(ωt− γ) where tan γ =

2pω

ω20 −ω2

• Amplitude is a function of frequency ω.

• Maximum amplitude when denominator minimal: a little calculus shows this is when

ω =

{√ω2

0 − 2p2 if 2p2 < ω20 (requires very light damping c2 < 2km)

0 if 2p2 ≥ ω20

In the first situation this is known as practical resonance.

The animation shows the steady periodic solution for the equation x′′+ 16x′+ 324x = cos ωt, namely

xP(t) =1√

(182 −ω2)2 + (16ω)2cos(ωt− γ)

for different values of driving frequency ω. Note that the practical resonant frequency (maximumamplitude) occurs when ω = 14.

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Glass smashing!

Model transverse motion of lip of a wine glass by3

110 x′′ + 1

5 x′ + 1, 000, 000x = F0 cos ωt

F0 cos ωt models vibration of the air due to ambient sound.

Unforced motion (F0 = 0) of the glass is

xC(t) = e−t(c1 cos ω1t + c2 sin ω1t)

where

ω1

2π=

ω1

2π

√ω2

0 − p2 =ω1

2π

√km − c2

4m2 = 503.2920959Hz

(≈ 1 octave above middle C)

0

x

1 2 3t

Practical resonance occurs when

f =ω

2π=

12π

√ω2

0 − 2p2 = 503.2920707Hz

The animation shows the steady periodic solution for tiny variations of driven frequency f near thepractical resonant frequency.

A singer must be very loud + accurate to crack the glass...

3Small mass, small damping, high spring constant

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Application: Electrical circuits

An RLC circuit has a source voltage4 V(t) = V0 sin ωt, a resistor R ohms, a capacitor C farads, andan inductor L henries connected in series.

Current flow: I = dQdt where I(t) = current flow (amps) and Q(t) (coulombs) is the charge stored in

the capacitor at time t

Calculate voltage drop across each component: RI, 1C Q, L dI

dt respectively ODE

LdIdt

+ RI +1C

Q = V(t) =⇒ Ld2 Idt2 + R

dIdt

+1C

I = V ′(t)

Damped-driven spring equation in disguise, with

m↔ L c↔ R k↔ C−1 F0 ↔ ωV0

Electrical Resonance Amplitude of steady-periodic current I =V0√

R2 +(ωL− 1

ωC

)2

Maximal when ω2LC = 1 applications in electronics. . .

Audio Hum Background ‘noise’ often result of (practical) resonant current resonanceSolution: adjust capacitance C to be very different to ω−2L−1 to minimize resonant current.

Reducing power loss If C = 0, the current flow is

I(t) =V0√

R2 + ω2L2cos(ωt− γ)

Inductor reduces current flow: acts like extra resistance. Inserting capacitor C = ω−2L−1 in-creases current to V0/R: reduces power loss from natural inductance.

Tuning an (AM) radio Radio station frequency ω induces voltage V(t) = V0 sin ωt in a radio an-tenna, resulting in a current flow which (after amplification) powers a loudspeaker.

Variable capacitor tuned to C = ω−2L−1 amplifies signal at frequency ω and diminishes currentflow due to other radio frequencies.

Analogue radio tuning dials were variable capacitors: changing C changed the resonant fre-quency, and thus which radio station was amplified most.

4In US, mains electricity is V(t) = 110√

2 sin(120πt)

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