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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
2
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.
3
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.
4
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
5
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)
6