Oscillator Notes

# Oscillator Notes - Oscillator Notes Fall 2006 Physics 200a...

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Oscillator Notes. Fall 2006 Physics 200a R. Shankar I will discuss here only what was left out near the end of the lecture of November 6 We were considering the solution to the driven oscillator m d 2 x dt 2 + b dx dt + kx = F 0 cos( ωt ) ω 0 = s k m First note that even if F 0 = 0, there is a solution. (In class I referred to F 0 as simply F .) For the case γ < 2 ω 0 , which we will focus on, it is x c ( t ) = Ce - γt 2 cos( ω 0 t - φ 0 ) (1) where ω 0 = s ω 2 0 - γ 2 2 γ = b m (2) and the subscript c stands for complimentary solution , the solution with no driving force. I have referred to the phase as φ 0 since another phase φ will enter shortly. The parameters C and φ 0 are found by demanding that the initial position and velocity, x (0) and v (0), have some prescribed values at t = 0. Now consider the driven problem m d 2 x dt 2 + b dx dt + kx = F 0 cos( ωt ) (3) which we rewrite as d 2 x dt 2 + γ dx dt + ω 2 0 x = F 0 m cos( ωt ) . (4) Now introduce another problem where the driving force is F 0 sin( ωt ) and the solution is y ( t ): d 2 y dt 2 + γ dy dt + ω 2 0 y = F 0 m sin( ωt ) (5) Now if we form z ( t ) = x ( t ) + i y ( t ) (6) by adding the first equation 4 to i times the second, 5, we find d 2 z dt 2 + γ dz dt + ω 2 0 z = F 0 m e iωt . (7) Our plan is to solve this equation for z and take the real part, which is our x ( t ).

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