Why Is Amplitude Maximal at w = 1/sqrt(LC)?

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The discussion focuses on proving that the amplitude I_{0} of the steady periodic solution is maximal at w = 1/sqrt(LC). The equation governing the system is LI'' + RI' + (1/C)*I = wE_{0}cos(wt), with the amplitude expressed as A(ω) = E_0 / √(R² + (ωL - 1/(ωC))²). Observations indicate that the amplitude reaches its maximum at the resonant frequency w = 1/sqrt(LC) and approaches zero as ω approaches zero. To prove this mathematically, one can analyze the function A(ω) by finding its derivative A' and setting it to zero to identify critical points. Ultimately, this leads to confirming that the amplitude is indeed maximal at the specified frequency.
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Homework Statement



Prove that the amplitude I_{0} of the steady periodic solution is maximal at w =1/\sqrt{LC}

Homework Equations


LI''+RI'+(1/C)*I=wE_{0}cos(wt)

I(steady periodic)=E_{0}cos(wt-\alpha))/(\sqrt{R^2+(wL-(1/wC))^2}3. The Attempt at a Solution [/b
I can see by the graph that it reaches a maximum value at w =1/\sqrt{LC} and then approaches zero as \omega\rightarrow 0. But I have no idea where to start to try and prove this.
 
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First of all, you can write down the amplitude of Iw(t) as a function of omega, that is simply
A(\omega) = \frac{E_0}{\sqrt{R^2 + (\omega L -1/ (\omega C) )^2}}
(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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