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Week 9: Oscillations |
satisfy arbitrary initial conditions, the inhomogeneous solution inherits these constants of integration and freedom, and the mathematical gods are thus pleased.
The solution to the homogeneous equation, however, is exponentially damped. If we wait long enough, no matter how we start the system o initially, the homogeneous solution will get as small as we like – basically, it will disappear and we’ll be left only with the particular solution. For that reason, in the context of driven harmonic oscillation the homogeneous solution is called the transient solution and depends on initial conditions, the particular solution is called the steady state solution and does not depend on the initial conditions!
We can then search for a steady state solution that does not contain any free parameters but is completely determined by the givens of the problem independent of the initial conditions!
Although it is not immediately obvious, given a harmonic driving force at a particular frequency ω, the steady state solution must also be a harmonic function at the same frequency ω. This is part of that deep math mojo that comes out of fourier transforming not only the driving force, but the entire ODE. The latter e ectively reduces the entire ODE to an algebraic problem. It is by far easiest to see how this goes by using a complex exponential driving force and solution and then taking the real part at the end, but we will skip doing this for now. Instead I will point out some true facts about the solution one obtains when one does it all right and completely. Interested parties can always look ahead into the companion volume for this course, Intro Physics 2, in the chapter on AC circuits to see how it really should be done.
a) The steady state solution to the driven SHO is:
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xp(t) = A cos(ωt − δ) |
(858) |
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where: |
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F0/m |
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A = |
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(859) |
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and the phase angle δ is: |
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p(ω2 − ω02)2 + (2ζω0ω)2 |
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µ ω2 − ω02 |
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δ = tan−1 |
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2ζω0ω |
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(860) |
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b)The average power delivered to the mass by the driving force is a quantity of great interest. This is the rate that work is being done on the mass by the driving force, and (because it is in steady state) is equal to the rate that the damping force removes the energy that is being added. Evaluating the instantaneous power is straightforward:
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(861) |
P (t) = F · ~v = F v = −F0Aω cos(ωt) sin(ωt − δ) |
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We use a trig identity to rewrite this as: |
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P (t) = |
−F0Aω cos(ωt)(sin(ωt) sin(δ) − cos(ωt) cos(δ)) |
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= |
F0Aω ¡cos2(ωt) sin(δ) − cos(ωt) sin(ωt) cos(δ)¢ |
(862) |
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The time average of cos2(ωt) is 1/2. The time average of cos(ωt) sin(ωt) is 0. Hence: |
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Pavg = |
F0Aω |
sin(δ) |
(863) |
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2 |
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c) If one sketches out a right triangle corresponding to: |
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tan(δ) = |
2ζω0ω |
(864) |
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ω2 − ω02 |
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one can see that: |
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2ζω0ω |
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sin(δ) = |
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(865) |
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p(ω2 − ω02)2 + (2ζω0ω)2 |
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Week 9: Oscillations |
413 |
Finally, substituting this equation and the equation for A above into the expression for the average power, we get:
Pavg(ω) = |
F02ω2ζω0 |
(866) |
m ((ω2 − ω02)2 + (2ζω0ω)2) |
This equation is maximum when ω = ω0, at resonance. At that point the peak average power
can be written: |
F 2 |
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2 |
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F |
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Pavg(ω0) = Pmax = |
0 |
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0 |
(867) |
4mζω0 |
2b |
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d)This shape of Pavg(ω) is very important to semi-quantitatively understand. It is (for weak damping) a sharply peaked curve with the peak centered on ω0, the natural frequency of the undamped oscillator. You can see from the expression given for Pavg why this would be so – at ω = ω0 the quantity in the denominator is minimum. This function is plotted in figure 125 below.
RESONANCE OF A DRIVEN HARMONIC OSCILLATOR
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Q = 3 |
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Q = 10 |
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10 |
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Q = 20 |
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POWER |
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0 |
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0.5 |
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ω |
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Figure 125: This plots the average power P (ω) for three resonances. In these figures, F0 = 1, k = 1, m = 1 (so ω0 = 1) and hence b is the only variable. Since Q = ω0/ = mω0/b we plot Q = 3, 10, 20 by selecting b = 0.333, 0.1, 0.05. Thus Pmax = F02/2b = 3/2, 5, 10 respectively (all in suitable units for the quantities involved). Note that the full width of e.g. the Q = 20 curve (with the sharpest/highest peak) is 1/20 at P = Pmax/2 = 5. P is the average power added to the oscillator by the driving force, which in turn in steady state motion must equal the average power lost to to dissipative drag forces.
e)The sharpness of the resonance is controlled by the dimensionless Q-factor, or “quality factor”, of the resonance. Q for the driven SHO is defined to be:
Q = |
ω0 |
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(868) |
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ω |
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where ω is the full width of Pavg(ω) at half-maximum! To find |
ω, one must use: |
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Pavg(ω) = |
Pmax |
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2 |
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and solve for the two roots where the equality is satisfied. The di erence between them is the full width.