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

[dipset24]

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.

 

[CompuChip]

First of all, you can write down the amplitude of I_w(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)

 

[nlightner]

One approach to proving this statement is to use calculus and the first and second derivatives of the steady periodic solution. We can start by taking the first derivative of the solution with respect to \omega:

\frac{dI}{d\omega} = -\frac{E_0}{\sqrt{R^2+(wL-(1/wC))^2}} \cdot \frac{d}{d\omega} \left(R^2+(wL-(1/wC))^2 \right)^{-1/2}

Using the chain rule, we can simplify this to:

\frac{dI}{d\omega} = -\frac{E_0}{\sqrt{R^2+(wL-(1/wC))^2}} \cdot \left(-\frac{1}{2}\right) \cdot \frac{1}{\left(R^2+(wL-(1/wC))^2 \right)^{3/2}} \cdot \frac{d}{d\omega} \left(R^2+(wL-(1/wC))^2 \right)

Simplifying further, we get:

\frac{dI}{d\omega} = \frac{E_0}{2\left(R^2+(wL-(1/wC))^2 \right)^{3/2}} \cdot \frac{d}{d\omega} \left(R^2+(wL-(1/wC))^2 \right)

Next, we can take the second derivative of the solution with respect to \omega:

\frac{d^2I}{d\omega^2} = \frac{E_0}{2} \cdot \frac{d}{d\omega} \left(\frac{1}{\left(R^2+(wL-(1/wC))^2 \right)^{3/2}} \cdot \frac{d}{d\omega} \left(R^2+(wL-(1/wC))^2 \right) \right)

Using the chain rule and simplifying, we get:

\frac{d^2I}{d\omega^2} = \frac{E_0}{2} \cdot \frac{d}{d\omega} \left(\frac{1}{\left(R^2+(wL-(1/wC))^2 \right)^{3/2}} \cdot \left(2(wL-(1/w