Help with interpretation of question. (Differential equations)

  • Thread starter sid9221
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  • #1
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http://dl.dropbox.com/u/33103477/Linear%20oscilator.png [Broken]

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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Answers and Replies

  • #2
699
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Assuming everything is correct and I see it is correct up to taking the derivative and plugging in zero (I was too lazy to take the derivative). If omega > gamma, you would have [itex]e^{ait}[/itex] where I am using a = radical.

What is [itex]e^{ait}\mbox{?}[/itex] Can you rewrite e in this form as some other transcendental?

If they are real, you just have a DE with of the form Ae+Be.
Equal you just have a constant.

Also, I see you didn't solve for you constant. You set it up but that was it.
$$
\begin{bmatrix}1&1&0\\
\sqrt{\gamma^2-\omega^2}-\omega & \sqrt{\gamma^2-\omega^2}+\omega & v
\end{bmatrix}
$$
When you solve this system, what do you get for [itex]c_1,c_2[/itex]?
 
Last edited:
  • #3
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Okay so working out [tex] c_1 = -v/2\omega [/tex] and [tex] c_2 = v/2\omega [/tex] what does that mean for the formula I don't get what I'm suppose to interpret from these results ?
 

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