Using Newton's 2nd Law for a damped oscillator:(adsbygoogle = window.adsbygoogle || []).push({});

[tex] ma = -kx - \alpha x [/tex]

which is a second order linear DE. To solve it we use the trial integrating factor [itex] e^{\lambda x} [/tex] to come to the root equation

[tex] mx^2 + \alpha x + k = 0[/tex] where we can find our two solutions to be

[tex] r_{1} \ and \ r_{2} \frac{-\alpha}{2m} +/- \frac{1}{2m} \sqrt{\alpha^2 - 4km} [/tex]

And the function [itex] x(t) [/itex] is supposed to be represented by

[tex] x(t) = Ae^{r_1 t } + Be^{r_2 t} [/tex] which can be solved with th given initial conditions [itex] x(0) = 0 \ and \ v(0) = v_0[/itex].

I'm supposed to end up with the product of two exponentials which dissolve to a cosine and sin function.

Four lines up is where I'm lost.. I'm not sure how to jump to the two exponential functions. What I have is:

[tex] x(t) = Ae^{r_1 t}(\cos(x(\alpha^2 - 4km)) +/- i \sin(x(\alpha^2 - 4km)) [/tex] which just smells really wrong. I can solve the particular equations once I can get to the general solution, but I'm stuck there. I know how the damping is determined and everything.. I just cant get the general equation :(

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# Homework Help: Deriving SHM equations

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