Does this equation makes sense? If so, how to solve it?

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December 2nd 2009, 10:06 AM
arbolis

Does this equation makes sense? If so, how to solve it?

I'm wondering if the following equation makes sense, and how to solve for $q(t)$ .

$-\frac{dt^2}{LC}=\frac {d^2q}{q}$ .

I found this equation when trying to solve the following problem : http://www.physicshelpforum.com/phys...pacitance.html.
I wanted to find $V(t)$ . I realize the solution should be sinusoidal, so something like $V(t)=V_{max} \sin (\omega t + \phi)$ .
Regarding the problem, when the capacitor halved its energy, I've found out that $V(0.0021s)=\frac{V_{max}}{\sqrt 2}$ . So I've found the boundary values and if I find $V(t)$ I'm done.
If I find $q(t)$ , I have that $V(t)=\frac{q(t)}{C}$ and so I'm done.
I used Kirchhoff's law of voltage to find $-\frac{dt^2}{LC}=\frac {d^2q}{q}$ . I might have done errors though.

Thank you for any help!
December 2nd 2009, 01:46 PM
CaptainBlack

Quote:

Originally Posted by arbolis

I'm wondering if the following equation makes sense, and how to solve for $q(t)$ .

$-\frac{dt^2}{LC}=\frac {d^2q}{q}$ .

I found this equation when trying to solve the following problem : http://www.physicshelpforum.com/phys...pacitance.html.
I wanted to find $V(t)$ . I realize the solution should be sinusoidal, so something like $V(t)=V_{max} \sin (\omega t + \phi)$ .
Regarding the problem, when the capacitor halved its energy, I've found out that $V(0.0021s)=\frac{V_{max}}{\sqrt 2}$ . So I've found the boundary values and if I find $V(t)$ I'm done.
If I find $q(t)$ , I have that $V(t)=\frac{q(t)}{C}$ and so I'm done.
I used Kirchhoff's law of voltage to find $-\frac{dt^2}{LC}=\frac {d^2q}{q}$ . I might have done errors though.

Thank you for any help!

It looks like you have tried separating variables for what is essentially a second order homogeneous constant coefficient linear ODE. That is not the way to solve this, you normally use a trial solution of the form $q(t)=e^{\lambda t}$ which will give a quadratic for $\lambda$ . Alternatively recognise the ODE as that of SHM and just write down the solution.

CB
December 2nd 2009, 01:47 PM
arbolis

Quote:

Originally Posted by CaptainBlack

It looks like you have tried separating variables for what is essentially a second order homogeneous constant coefficient linear ODE. That is not the way to solve this, you normally use a trial solution of the form $q(t)=e^{\lambda t}$ which will give a quadratic for $\lambda$ . Alternatively recognise the ODE as that of SHM and just write down the solution.

CB

Thank you very much!