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

  1. #1
    MHF Contributor arbolis's Avatar
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    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!
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by arbolis View Post
    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
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  3. #3
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    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!
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