LRC Circuit: Frequency of Oscillation determined by R

[jamesbiomed]

Homework Statement

How much resistance must be added to a pure LC circuit (L = 340 mH, C = 2200 pF) to change the oscillator's frequency by 0.10 percent?

Will the frequency be increased or decreased?

Homework Equations

I'm assuming critical damping: R^2=4L/C, w=((1/LC)-R^2/(4L^2))^1/2

With no R: w=(1/(LC))^1/2

The Attempt at a Solution

I've been at this for a few hours :/ I'm letting w stand in for frequency since w=2pif the percent change needed for f should be the same for w (I hope). I've tried several approaches:

First, I tried w=(1/LC)^1/2 to get w. Then multiplied w by .001 to get .1%. Then made an equation: w-((1/LC)-R^2/(4L^2))^1/2=.001w

That didn't work, but the answer was close to the given one.

I also tried simply setting .001w=(1/LC-R^2/2l^2)^1/2 and that answer was further off.

If I'm missing anything or you need information I'd be happy to give it I'm really enthusiastic to get this done. Thank you in advance!

 

[ehild]

The angular frequency of a damped oscillator is

\omega=\sqrt{\frac{1}{LC}-\left(\frac{R}{2L}\right)^2}

Your first approach is correct, but it would be easier to write ω=K√(1/LC). How much is K? less then 1 or greater then 1? Then you can square both sides. Do it symbolically, do not evaluate. ehild

 

[jamesbiomed]

Hey echild. The problem was I wasn't squaring k, just plugging it in after the fact.

K is less than one, so the frequency should be decreased.

For anyone with same problem:
w' is .1 % w.

w'=k(1/LC)^1/2=> (w')^2=k^2/LC=>(w')^2=(.001)^2/(LC) so omega is the square root of that.

Then w-(w^2+(R/2L)^2)^1/2=w'. Solve for R:

R=(((w-w')^2-w^2)4L^2)^1/2. That solved it.

Thank you!

 

[ehild]

I meant that the new angular frequency is 0.999 ω₀, where ω₀=1/√(LC).

So
\omega=\sqrt{\frac{1}{LC}-\left(\frac{R}{2L}\right)^2}=0.999\sqrt{\frac{1}{LC}}

Squaring both sides:\frac{1}{LC}-\left(\frac{R}{2L}\right)^2=0.998\frac{1}{LC}

0.002\frac{1}{LC}=\left(\frac{R}{2L}\right)^2

ehild