2.6 Forced Oscillations and Resonance 1 Oscillator equation with external force F(t): basic case assumes F periodic, mx 00 + cx 0 + kx = F 0 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 = ω 0 = q k m (driving frequency 6 = natural frequency). Already found complementary function x C (t)= c 1 cos ω 0 t + c 2 sin ω 0 t. Particular integral: guess x P (t)= a cos ωt + b sin ωt. Then mx 00 P + kx P =(-maω 2 + ka) cos ωt +(-mbω 2 + kb) sin ωt = F 0 cos ωt ⇐⇒ a = F 0 k - mω 2 = F 0 m(ω 2 0 - ω 2 ) , b = 0 = ⇒ x P (t)= F 0 m(ω 2 0 - ω 2 ) cos ωt = ⇒ x(t)= x P (t)+ x C (t)= F 0 m(ω 2 0 - ω 2 ) cos ωt + c 1 cos ω 0 t + c 2 sin ω 0 t • Sum of distinct periodic motions. • Larger F 0 = ⇒ more motion. • ω close to ω 0 = ⇒ more motion. Suppose have initial conditions x(0)= 0 = x 0 (0) (periodic force applied to resting spring). Quickly obtain x(t)= F 0 m(ω 2 0 - ω 2 ) ( cos ωt - cos ω 0 t ) = 2F 0 m(ω 2 0 - ω 2 ) sin ω 0 - ω 2 t sin ω 0 + ω 2 t using a trigonometric identity. If ω 0 , ω close in value, then ω 0 - ω ω 0 + ω Amplitude beats at ω 0 -ω 2 rad/s. x 2 4 6 8 10 t ω 0 = 20 ω = 18 Graphics show x 00 + 400x = 38 cos 18t where ω 0 = 20. Solution x(t)= sin t sin 19t = A(t) sin 19t. High frequency vibration sin 19t with periodic amplitude A(t)= sin t. 1 This is an abstract summary. Study this open-book and pay attention to the numerical examples from lectures. . .
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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. . .
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.