2.6 Forced Oscillations and Resonance1 Oscillator equation with external force F(t): basic case assumes F periodic,

mx00 + cx0 + 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 = ω = k (driving frequency = natural frequency). 6 0 m 6 Already found complementary function xC(t) = cq1 cos ω0t + c2 sin ω0t. Particular integral: guess xP(t) = a cos ωt + b sin ωt. Then 2 2 mx00 + kx = ( maω + ka) cos ωt + ( mbω + kb) sin ωt P P − − F0 F0 = F0 cos ωt a = = , b = 0 ⇐⇒ k mω2 m(ω2 ω2) − 0 − F0 = xP(t) = cos ωt ⇒ m(ω2 ω2) 0 − F0 = x(t) = xP(t) + xC(t) = cos ωt + c1 cos ω0t + c2 sin ω0t ⇒ m(ω2 ω2) 0 − • Sum of distinct periodic motions. • Larger F = more motion. 0 ⇒ • ω close to ω = more motion. 0 ⇒ Suppose have initial conditions x(0) = 0 = x0(0) (periodic force applied to resting spring). Quickly obtain F 2F ω ω ω + ω ( ) = 0 cos ω cos ω = 0 sin 0 sin 0 x t 2 2 t 0t 2 2 − t t m(ω0 ω ) − m(ω0 ω ) 2 2 −
− using a trigonometric identity. If ω0, ω close in value, then ω0 ω ω0 + ω ω0 ω −  Amplitude beats at 2− rad/s. x ω0 = 20 ω = 18

2 4 6 8 10 t

Graphics show x00 + 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

F0 xP(t) = t sin ω0t 2mω0

2 xP is also the solution corresponding to the initial conditions x(0) = x0(0) = 0. F • Sine wave with increasing amplitude A(t) = 0 t 2mω0 • Spring tears itself apart!

Example Consider the IVP

4x00 + 64x = 4 cos ωt (x(0) = x0(0) = 0

The natural frequency is ω0 = √k/m = 4 rad/s. The four graphs show the solutions for four different driving frequencies: the last case is resonance. x x

0.005 0.1

0 0 2468 10 2468 10 t t

1 1 ω = 20 x(t) = sin 8t sin 12t ω = 6 x(t) = sin t sin 5t x 192 x 10 4

0.1 2 0 0 2 4 6 8 10 t 2 4 6 8 10 2 t − 2 t 9t 1 ω = 5 x(t) = sin sin 4 ω = 4 x(t) = t sin 4t 9 2 2 − 8

2 It is also what we obtain by taking limω ω x(t) from the previous slide → 0

2 Summary of undamped-driven motion The initial value problem mx00 + kx = F0 cos ωt with initial conditions x(0) = 0 = x0(0) has solution

2F0/m ω0 ω ω0 + ω sin − t sin t if ω = ω0 ω2 ω2 2 2 6 x(t) =  0 −  F0  t sin ω0t if ω = ω0 2mω0  where ω = √k/m. As ω ω low frequency beats of increasing amplitude occur. 0 → 0 The clickable animation is generated with the same F0, m throughout

Case 3: Damped-driven motion (practical resonance)

2 F0 c > 0 mx00 + cx0 + kx = F0 cos ωt x00 + 2px0 + ω0 x = cos ωt m

Transient and steady-periodic solutions Three types:

Damping Condition Complementary Function xC(t)

2 pt √p2 ω2t √p2 ω2t Overdamping c > 4km e− c1e− − 0 + c2e − 0 2 pt Critical damping c = 4km (c1 +c2t)e− 

2 pt 2 2 Underdamping c < 4km e− c cos ω t + c sin ω t where ω = ω p 1 1 2 1 1 0 −
q • p > 0 = xC transient: lim xC(t) = 0 t ∞ ⇒ → • 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 x00 + 3x0 + 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 − − − ω2 2 3ω = a = − , b = ⇒ 9ω2 + (ω2 2)2 (ω2 2)2 + 9ω2 − − 1 1 = x (t) = (ω2 2) cos ωt + 3ω sin ωt = cos (ωt γ) ⇒ P (ω2 2)2 + 9ω2 − (ω2 2)2 + 9ω2 − −
− where γ is the phase angle.

3 General situation Long-term solution is

F0/m 2 2 xP(t) = (ω ω ) cos ωt + 2pω sin ωt (ω2 ω2)2 + (2pω)2 0 − 0 − F
2pω = 0 cos(ω γ) where tan γ = t 2 2 m (ω2 ω2)2 + (2pω)2 − ω0 ω 0 − − q • Amplitude is a function of frequency ω.

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

ω2 2p2 if 2p2 < ω2 (requires very light damping c2 < 2km) ω = 0 − 0 (0q if 2p2 ω2 ≥ 0 In the first situation this is known as practical resonance.

The animation shows the steady periodic solution for the equation x00 + 16x0 + 324x = cos ωt, namely 1 xP(t) = cos(ωt γ) (182 ω2)2 + (16ω)2 − − for different valuesp of driving frequency ω. Note that the practical resonant frequency (maximum amplitude) occurs when ω = 14.

4 Glass smashing! Model transverse motion of lip of a wine glass by3

1 1 10 x00 + 5 x0 + 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 x

t xC(t) = e− (c1 cos ω1t + c2 sin ω1t) where 0 ω ω ω 2 1 = 1 ω2 p2 = 1 k c = 503.2920959 Hz 123 2π 2π 0 − 2π m − 4m2 t q q ( 1 octave above middle C) Practical≈ resonance occurs when ω 1 f = = ω2 2p2 = 503.2920707 Hz 2π 2π 0 − q The animation shows the steady periodic solution for tiny variations of driven frequency f near the practical resonant frequency.

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

3Small mass, small damping, high spring constant

5 Application: Electrical circuits

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

dQ Current flow: I = dt where I(t) = current flow (amps) and Q(t) (coulombs) is the charge stored in the capacitor at time t

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

dI 1 d2 I dI 1 L + RI + Q = V(t) = L + R + I = V0(t) dt C ⇒ dt2 dt C Damped-driven spring equation in disguise, with

1 m L c R k C− F ωV ↔ ↔ ↔ 0 ↔ 0 V Electrical Resonance Amplitude of steady-periodic current I = 0 2 1 2 R + ωL ωC 2 − Maximal when ω LC = 1 applications in electronics. . . q
Audio Hum Background ‘noise’ often result of (practical) resonant current resonance 2 1 Solution: adjust capacitance C to be very different to ω− L− to minimize resonant current.

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

V I(t) = 0 cos(ωt γ) √R2 + ω2L2 −

2 1 Inductor reduces current flow: acts like extra resistance. Inserting capacitor C = ω− L− 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. 2 1 Variable capacitor tuned to C = ω− L− amplifies signal at frequency ω and diminishes current flow 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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