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Driven Oscillators Sect. 3.5

Driven Oscillators Sect. 3.5. Consider a 1d Driven Oscillator with damping, & a time dependent driving force F d (t). Newton’s 2 nd Law Equation of Motion: F = ma = m(d 2 x/dt 2 ) = - kx - bv + F d (t) Can be solved in closed form for many special cases of F d (t).

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Driven Oscillators Sect. 3.5

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  1. Driven OscillatorsSect. 3.5 • Consider a 1d Driven Oscillator with damping, & a time dependent driving forceFd(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!

  2. 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), ω02 (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”

  3. 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 thesolution for the damped oscillator just discussed!)  xc(t) = e-βt[A1 eαt + A2 e-αt]withα  [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!

  4. 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!

  5. {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)/[(ω02 - ω2)2 +4(ωβ)2]½ • Algebra:cos(ωt) term gives: D = (A)/[(ω02 - ω2)cosδ + 2(ωβ)sinδ] Or D = (A)/[(ω02 - 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 Andδ = tan-1[2(ωβ)/(ω02 - ω2)] • Physics:δ = the phase difference between the driving force Fd(t) & the response. xp(t)

  6. Summary: x(t) = xc(t) + xp(t)where xc(t) = e-βt [A1 eαt + A2 e-αt] withα = [β2 - ω02]½Clearly a transient solution! Goes to zero after times t >> 1/β xp(t) = [A cos(ωt - δ)]/[(ω02 - ω2)2+4(ωβ)2]½ withδ = 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 ωand the frequency with damping: ω1  [ω02- β2]½

  7. To understand this, its helpful to solve Problems 3-24 & 3-25 numerically (on a computer!) Results similar to figure: If ω < ω1 = [ω02- β2]½ the oscillator transient xc(t) greatly distorts the sinusoidal shape of the forcing function just after t = 0

  8. If ω > ω1 = [ω02- β2]½ the oscillator transient xc(t) modulatesthe sinusoidal shape of the forcing function just after t = 0 • Figures for another case.

  9. Resonance • Focus on the steady state solution. Write it as: xp(t) = D cos(ωt - δ)with D = (A)/[(ω02 - ω2)2 +4(ωβ)2]½ A= (F0/m)δ = tan-1[2(ωβ)/(ω02 - ω2)] • Plotting D vs. ω clearly gives a function with a peak! • Resonance frequencyωR  the frequency at which the amplitude D(ω) is a maximum.  (dD/dω) = 0 Solving for ω=ωR gives: ωR = [ω02 -2β2]½ • Clearly the resonance frequency ωR < ω0 where ω0 = (k/m)½is thenatural” frequency of oscillator! How much less obviously depends on the size of the damping constant β!

  10. 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 frequencyωR = [ω02 -2β2]½ • If ω02 <2β2, ωR is imaginary.  In this case, there is no resonance! D simplydecreases as ω increases. • Comparison of the fundamental oscillation frequencies for the driven oscillator: • Free oscillations: ω02 = (k/m) • Free oscillations + damping:ω12 = ω02 - β2 • Driven oscillations + damping:ωR2 = ω02 - 2β2 • Clearly: ω0 > ω1 > ωR

  11. 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 damping2β << 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!

  12. 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 

  13. δ = tan-1[2(ωβ)/(ω02 - ω2)] vs ω for different Q values (different relative sizes of ω0&β)

  14. 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 2ω’s where D(ω) = Dm/(2)½  0.707 Dm  Δω A measure of the “linewidth” (or, simply, the “width”) of the resonance.

  15. (D/Dm) vs ω showing relative positions of the 3 frequencies ω0 , ω1 &ωR

  16. Real physical oscillators: Values of Q vary greatly! • Mechanical systems (e.g., loudspeakers): Q  1 to a few100 • 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

  17. 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),

  18. Start with: xp(t) = D cos(ωt - δ), vp(t) = ωD sin(ωt - δ), D  D(ω) = (A)/[(ω02 - ω2)2 +4(ωβ)2]½ • The Potential Energyis: 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 ωfor which (d‹U›/dω) = 0  Occurs at the same ωas the amplitude resonance: ωR= [ω02 -2β2]½ . So: Potential Energy Resonance is the same as amplitude resonance!

  19. xp(t) = D cos(ωt - δ), vp(t) = ωD sin(ωt - δ), D  D(ω) = (A)/[(ω02 - ω2)2 +4(ωβ)2]½ • The Kinetic Energy is T = 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 ωfor which (d‹T›/dω) = 0  Occurs at a different ωthan the amplitude resonance: ωE= ω0So: Kinetic Energy Resonance is different than amplitude resonance! It occurs at the natural frequency!

  20. SUMMARY • Potential energy resonance occurs at ωR= [ω02 -2β2]½ • Kinetic energy resonance occurs atωE = 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.

  21. 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 ωfor which (d‹E›/dω) = 0

  22. “Lorentzian” Resonance Curves L(ω)  [D(ω)]2 from Energy Resonance  β = ω0 β = 2ω0  β = 3ω0

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