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Unformatted text preview: 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 cos( ωt ) ω = s k m First note that even if F = 0, there is a solution. (In class I referred to F as simply F .) For the case γ < 2 ω , which we will focus on, it is x c ( t ) = Ce- γt 2 cos( ω t- φ ) (1) where ω = s ω 2- 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 since another phase φ will enter shortly. The parameters C and φ 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 cos( ωt ) (3) which we rewrite as d 2 x dt 2 + γ dx dt + ω 2 x = F m cos( ωt ) . (4) Now introduce another problem where the driving force is F sin( ωt ) and the solution is y ( t ): d 2 y dt 2 + γ dy dt + ω 2 y = F 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 z = F m e iωt . (7) Our plan is to solve this equation for...
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This note was uploaded on 07/13/2008 for the course PHYS 200 taught by Professor Ramamurtishankar during the Fall '08 term at Yale.
- Fall '08