Oscillator Notes

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

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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 ).
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
The nice thing about z
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern