Driven Oscillators Sect. 3.5
• Consider a 1d Driven Oscillator with damping, & a time dependent driving force Fd(t).
Newton’s 2nd Law Equation of Motion:F = ma = m(d2x/dt2) = - kx - bv + Fd(t)
• Can be solved in closed form for many special cases of Fd(t).• The text immediately goes to the very important special case of a
sinusoidal driving force (at frequency ω):
Fd(t) = F0 cos(ωt)• Treating this is equivalent to treating one Fourier component of a more
general t dependent force! This is because N’s 2nd Law is a linear differential equation!
Sinusoidal Driving Forces• Eqtn of Motion: F = m(d2x/dt2) = - kx -bv + Fd(t) (1)
• Assume Fd(t) = F0 cos(ωt)ω = Angular frequency of driving force(1) becomes:
mx + bx + kx = F0 cos(ωt)• Defining: β [b/(2m)]; A (F0/m), ω0
2 (k/m)
x + 2βx + ω02x = A cos(ωt)
– A linear, 2nd order, inhomogeneous, time dependent differential eqtn! Differential eqtn theory shows: (App. C): The general solution has 2 parts: x(t) = xc(t) + xp(t) Where xc(t) “Complimentary solution”xp(t) “Particular solution”
x + 2βx + ω02x = A cos(ωt)
x(t) = xc(t) + xp(t) • xc(t) “Complimentary solution”= the solution to the homogeneous eqtn (the
solution with with A = 0 or the solution for the damped oscillator just discussed!)
xc(t) = e-βt[A1 eαt + A2 e-αt] with α [β2 - ω02]½
• xp(t) “Particular solution” = a specific solution to the inhomogeneous eqtn (the solution with A 0)
• For the particular solution, try xp(t) = D cos(ωt - δ) Note: δ here is NOT the same δ as in the discussion of the ordinary SHO!
x + 2βx + ω02x = A cos(ωt) x(t) = xc(t) + xp(t)
• For the “Particular solution” xp(t) try xp(t) = D cos(ωt - δ). Substitution into the differential equation gives D & δ (Student exercise!):
{A-D[(ω02 -ω2)cosδ +(2ωβ)sinδ]}cos(ωt)
-{D[(ω02 - ω2)sinδ – (2ωβ)cosδ]}sin(ωt) = 0
• cos(ωt) & sin(ωt) are linearly independent Coefficients of these functions are separately = 0!
{A-D[(ω02 -ω2)cosδ +(2ωβ)sinδ]}cos(ωt)
-{D[(ω02 - ω2)sinδ – (2ωβ)cosδ]}sin(ωt) = 0
• Algebra: sin(ωt) term gives: tanδ = 2(ωβ)/(ω02 - ω2)
sinδ = 2(ωβ)/[(ω02 - ω2)2 +4(ωβ)2]½
cosδ =(ω02 - ω2)/[(ω0
2 - ω2)2 +4(ωβ)2]½
• Algebra: cos(ωt) term gives:D = (A)/[(ω0
2 - ω2)cosδ + 2(ωβ)sinδ]Or D = (A)/[(ω0
2 - ω2)2 +4(ωβ)2]½ xp(t) = D cos(ωt - δ)
xp(t) = [A cos(ωt - δ)]/[(ω02 - ω2)2+4(ωβ)2]½
And δ = tan-1[2(ωβ)/(ω02 - ω2)]
• Physics: δ = the phase difference between the driving force Fd(t) & the response. xp(t)
• Summary: x(t) = xc(t) + xp(t) wherexc(t) = e-βt [A1 eαt + A2 e-αt] with α = [β2 - ω0
2]½ Clearly a transient solution! Goes to zero after times t >> 1/β
xp(t) = [A cos(ωt - δ)]/[(ω02 - ω2)2+4(ωβ)2]½
with δ = tan-1[2(ωβ)/(ω02 - ω2)]
A steady state solution! Dominates at long times t >> 1/β
x(t >> 1/β) xp(t) • Motion details before the transient xc(t) dies to zero (t 1/β):
– Depend strongly on the conditions at time the force is first applied. – Depend clearly also on the relative magnitudes of the driving frequency ω and the
frequency with damping: ω1 [ω0
2- β2]½
• To understand this, its helpful to solve Problems 3-24 & 3-25 numerically (on a computer!) Results similar to figure:
If ω < ω1 = [ω02- β 2]½
the oscillator transient xc(t) greatly distorts the sinusoidal shape of the forcing function just after t = 0
• Figures for another case.
If ω > ω1 = [ω02- β 2]½
the oscillator transient xc(t) modulates the sinusoidal shape of the forcing function just after t = 0
Resonance• Focus on the steady state solution. Write it as: xp(t) = D cos(ωt - δ) with D = (A)/[(ω0
2 - ω2)2 +4(ωβ)2]½ A= (F0/m) δ = tan-1[2(ωβ)/(ω0
2 - ω2)]
• Plotting D vs. ω clearly gives a function with a peak! • Resonance frequency ωR the frequency at which the
amplitude D(ω) is a maximum. (dD/dω) = 0 Solving for ω = ωR gives: ωR = [ω0
2 - 2β2]½
• Clearly the resonance frequency ωR < ω0
where ω0 = (k/m)½ is the natural” frequency of oscillator! How much less obviously depends on the size of the damping constant β!
xp(t) = D cos(ωt - δ), D = (A)/[(ω02 - ω2)2 +4(ωβ)2]½
δ = tan-1[2(ωβ)/(ω02 - ω2)]
• Consider the resonance frequency ωR = [ω02 - 2β2]½
• If ω02 < 2β2, ωR is imaginary. In this case, there is no resonance! D simply
decreases as ω increases.• Comparison of the fundamental oscillation frequencies for the driven oscillator:
– Free oscillations: ω02 = (k/m)
– Free oscillations + damping: ω12 = ω0
2 - β2 – Driven oscillations + damping: ωR
2 = ω02 - 2β2
– Clearly: ω0 > ω1 > ωR
Quality (Q) Factor of the Oscillator• For a driven oscillator, its useful to define the
QUALITY FACTOR: Q [ωR/(2β)]– Q is a measure of the damping strength of the oscillator. Also (as we’ll see next)
its a measure of how “sharp” the resonance is. ωR = [ω02 - 2β2]½
– For very small damping 2β << ω0, expand the square root in a Taylor’s series: ωR ω0[1 - (β/ω0)2 + ]
or ωR ω0 - something small. ωR ω0 as β 0 & at the same time Q [ω0
/(something small)]
That is Q very large ( ) as β 0– On the other hand, large damping Small Q & the destruction of the
resonance! See the figures!– Q clearly is a measure of the quality of the resonance!
D = (A)/[(ω02 - ω2)2 + 4(ωβ)2]½ vs ω for different Q values
(different relative sizes of ω0 & β). Δω = full width at half max
Δω
δ = tan-1[2(ωβ)/(ω02 - ω2)] vs ω for different Q values
(different relative sizes of ω0 & β)
• Problem 3-19 shows: For a “lightly damped” oscillator (ω02 >> β2):
Q [ωR/(2β)] [ω0/(Δω)]– Where Δω = full width at half max (from D(ω) plot). – This shows that Q is definitely a measure of the quality (sharpness) of the
resonance!Δω the interval between 2 points on the D(ω) curve on either side of the max, which have an amplitude 1/(2)½ 0.707 of the maximum amplitude. That is, if the maximum D(ωR) Dm, Δω = the interval between 2 ω’s where
D(ω) = Dm/(2)½ 0.707 Dm
Δω A measure of the “linewidth” (or, simply, the “width”) of the resonance.
(D/Dm) vs ω showing relative positions of the 3 frequencies ω0 , ω1 & ωR
• Real physical oscillators: Values of Q vary greatly!– Mechanical systems (e.g., loudspeakers):
Q 1 to a few 100– Quartz crystal oscillators & tuning forks: Q > 104 – Highly tuned electrical circuits:
Q 104 - 105 – Atomic systems: Electron oscillations in atoms Optical radiation.
Sharpness of spectral lines limited by energy loss due to radiation. Classical minimum linewidth: Δω 2 108 ω0 Q 5 107
– Largest known Q’s: Gas lasers: Q 1014
Energy Resonance• What we’ve talked about up to now should technically be called
“Amplitude Resonance” since the resonance occurs in the amplitude of
xp(t) = D cos(ωt - δ), D = (A)/[(ω02 - ω2)2 +4(ωβ)2]½
• A similar, related phenomenon, with a (slightly) different resonance frequency is “Energy Resonance”.
• First note that the velocity is vp(t) = (dxp(t)/dt) = ωD sin(ωt - δ),
• Start with: xp(t) = D cos(ωt - δ), vp(t) = ωD sin(ωt - δ), D D(ω) = (A)/[(ω0
2 - ω2)2 +4(ωβ)2]½
• The Potential Energy is:U = (½)k[xp(t)]2 = (½)kD2 cos2(ωt - δ)
• Consider the time average of U (over one period of the driving force: 0 < t < [2π/ω]): ‹U› (ω/2π)∫Udt
Note that ‹cos2(ωt - δ)› = (½) ‹U› (¼)kD2 (¼)k[D(ω)]2
That is, the resonance frequency of the potential energy = the ω for which (d‹U›/dω) = 0 Occurs at the same ω as the amplitude resonance: ωR= [ω0
2 - 2β2]½ . So: Potential Energy Resonance is the same as amplitude resonance!
xp(t) = D cos(ωt - δ), vp(t) = ωD sin(ωt - δ), D D(ω) = (A)/[(ω0
2 - ω2)2 +4(ωβ)2]½
• The Kinetic Energy isT = ½ m[vp(t)]2 = ½ m ω2D2sin2(ωt - δ)
• Consider the time average of T (over one period of the driving force: 0 < t < [2π/ω]): ‹T› (ω/2π)∫T dt Note that ‹cos2(ωt - δ)› = (½) ‹T› (¼)mω2D2
‹T› (¼) mω2[D(ω)]2 . That is the resonance frequency of the kinetic energy = the ω for which (d‹T›/dω) = 0 Occurs at a different ω than the amplitude resonance:ωE= ω0
So: Kinetic Energy Resonance is different than
amplitude resonance! It occurs at the natural frequency!
SUMMARY• Potential energy resonance occurs at
ωR= [ω02 - 2β2]½
• Kinetic energy resonance occurs at ωE = ω0
• PHYSICS: They occur at different frequencies because the driven, damped oscillator is not a conservative system!– Energy is continually exchanged between the (external) driving
mechanism & the oscillator. Energy is also continually lost to the damping medium.
Total Energy Resonance • Consider E = T + U for the driven, damped
oscillator.• Student exercise (as part of Ch. 3 homework!):
– Take time the average of E: (over one period of the driving force: 0 < t < [2π/ω]):
‹E› (ω/2π)∫E dt – Compute the resonance frequency of the total energy =
the ω for which (d‹E›/dω) = 0
“Lorentzian” Resonance Curves L(ω) [D(ω)]2
from Energy Resonance
β = ω0
β = 2ω0 β = 3ω0