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Proving Amplitude is maximal

  1. Jun 22, 2009 #1
    1. The problem statement, all variables and given/known data

    Prove that the amplitude I[tex]_{0}[/tex] of the steady periodic solution is maximal at w =1/[tex]\sqrt{LC}[/tex]


    2. Relevant equations
    LI''+RI'+(1/C)*I=wE[tex]_{0}[/tex]cos(wt)

    I(steady periodic)=E[tex]_{0}[/tex]cos(wt-[tex]\alpha[/tex]))/([tex]\sqrt{R^2+(wL-(1/wC))^2}[/tex]


    3. The attempt at a solution[/b
    I can see by the graph that it reaches a maximum value at w =1/[tex]\sqrt{LC}[/tex] and then approaches zero as [tex]\omega[/tex][tex]\rightarrow[/tex] 0. But I have no idea where to start to try and prove this.
     
  2. jcsd
  3. Jun 23, 2009 #2

    CompuChip

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    Homework Helper

    First of all, you can write down the amplitude of Iw(t) as a function of omega, that is simply
    [tex]A(\omega) = \frac{E_0}{\sqrt{R^2 + (\omega L -1/ (\omega C) )^2}}[/tex]
    (why?)

    Now you can argue why it should be maximal at 1/sqrt(LC) by mathematical arguments such as: A is maximal when the denominator is [maximal/minimal], which happens when the square is [maximal/minimal], etc.

    If you want a really mathematical looking argument, you should calculate the derivative A' and set it to zero, then solve for omega. (Which looks, by the way, horrible when calculating, but in the end isn't half so bad)
     
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