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sid9221
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http://dl.dropbox.com/u/33103477/Linear%20oscilator.png
I am having trouble understand the question, what I have done its solve the equation using the substitution [tex]x=e^{rx}[/tex]
Then, I have the solution given by:
[tex] x(t)=c_1 e^{t(\sqrt{\gamma^2 - \omega^2 })} + c_2e^{-t(\sqrt{\gamma^2 - \omega^2)}} [/tex]
So at x(0)=0,
[tex] c_1 + c_2 = 0[/tex]
and, x'(0)=v
[tex] v=c_2(\sqrt{\gamma^2 - \omega^2} - \omega) + c_1(\sqrt{\gamma^2 - \omega^2}+\omega) [/tex]
Not quite sure how to proceed as if \omega is greater than \gamma I get imaginary values and the opposite gives real, but what does that mean ?
I am having trouble understand the question, what I have done its solve the equation using the substitution [tex]x=e^{rx}[/tex]
Then, I have the solution given by:
[tex] x(t)=c_1 e^{t(\sqrt{\gamma^2 - \omega^2 })} + c_2e^{-t(\sqrt{\gamma^2 - \omega^2)}} [/tex]
So at x(0)=0,
[tex] c_1 + c_2 = 0[/tex]
and, x'(0)=v
[tex] v=c_2(\sqrt{\gamma^2 - \omega^2} - \omega) + c_1(\sqrt{\gamma^2 - \omega^2}+\omega) [/tex]
Not quite sure how to proceed as if \omega is greater than \gamma I get imaginary values and the opposite gives real, but what does that mean ?
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