phy293_l07.page3 - 4. Review of Forced Damped SHO 2 2 (a)...

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Unformatted text preview: 4. Review of Forced Damped SHO 2 2 (a) Given : x + γ x + ω0 = a0 ω0 cos(ωt) ¨ ˙ • And two initial conditions (like x(0) = xi and x(0) = vi ˙ • We find a solution x(t) = xc (t) + X (t) • Where the complementary solutions are I γ < 2ω0 Under-damped xc (t) = C1 e−γt/2 cos(ω t + C2 ) 2 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 2 Over-damped 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 2 (ω0 − ω 2 )2 + (γω )2 tan δ = 2 (ω0 ωγ − ω2 ) • This will be the solution for any γ 2 • But for γ 2 > 2ω0 there is no resonance • This solution is independent of initial conditions • The steady state solution can be de-composed 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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