Download Sect. 3.6 - TTU Physics

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), 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”
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)/[(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
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) where
xc(t) = e-βt [A1 eαt + A2 e-αt] with α = [β2 - ω02]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  [ω02- 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)/[(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 frequency ωR  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)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 = ω02 - β2
Driven oscillations + damping: ωR2 = 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)/[(ω02 - 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= [ω02 - 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)/[(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 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