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Unformatted text preview: 4. Review of Forced Damped SHO
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(a) Given : x + γ x + ω0 = a0 ω0 cos(ωt)
¨
˙
• And two initial conditions (like x(0) = xi and x(0) = vi
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• We ﬁnd a solution x(t) = xc (t) + X (t)
• Where the complementary solutions are I γ < 2ω0
Underdamped xc (t) = C1 e−γt/2 cos(ω t + C2 )
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ω = ω0 − γ 2 /4 II γ = 2ω0
Critically damped xc (t) = (C1 + C2 t)e−γt/2 γ > 2ω0
xc (t) = C1 eα1 t + C2 e−α2 t
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Overdamped
α1,2 = −γ/2 ± γ 2 /4 − ω0
• In all three cases C1,2 are determined by xi and vi
• Independent of the drive frequency ω or its amplitude a0 these contributions die away like e−γt/2 (or faster).
• The steady state solution has an amplitude and phase independent of time
III X (t) = A(ω ) cos(ωt − δ )
• Where:
A(ω ) = 2
a0 ω0
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(ω0 − ω 2 )2 + (γω )2 tan δ = 2
(ω0 ωγ
− ω2 ) • This will be the solution for any γ
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• But for γ 2 > 2ω0 there is no resonance • This solution is independent of initial conditions • The steady state solution can be decomposed into its two phases
X (t) = G cos(ωt) + H sin(ωt)
= A(ω )[cos δ cos(ωt) + sin δ sin(ωt)] ...
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 Fall '07
 PierreSavaria
 mechanics, Force, Statistical Mechanics

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