Finding the Resonant frequency in an LC circuit

[bsr25]

Homework Statement

The figure opposite shows a free oscillator composed of an LC circuit. Derive an expression for the resonance frequency. Show all of your working.

Homework Equations

d²x/dt² + ω²x = 0

The Attempt at a Solution

First I tried using Kirchoffs Voltage Law to set up a differential equation defining the circuit.

L(d²q/dt²) + Q₁/C + Q₂/C = 0

I then combined the two voltages across the capcitor, and divided by L

(d²q/dt²) + Q_Total/LC = 0

Comparing this to the equation for SHM:

d²x/dt² + ω²x = 0

shows that ω²= 1/LC.

But I figure this cannot be the case as that is the result expected for a simple LC circuit with one inductor and one capcitor. As this is two capcitors and one inductor I figure the rsult must be different.

One thing I notice on the diagram is the Capcitors have both negative plates facing each other, I wasnt sure what effect if any this would have on the circuit.

Other than this attempt at a solution which I am fairly sure if way off, I am a bit lost. Any help towards reahcing the right answer would be greatly appreciated!

Thanks

 

[sankalpmittal]

Look here: http://books.google.co.in/books?id=...t lc circuit is a harmonic oscillator&f=false

Scroll down to look for resonant frequency in electronic circuit. That will tell you, where you are going wrong.

 

[rude man]

Homework Statement

The figure opposite shows a free oscillator composed of an LC circuit. Derive an expression for the resonance frequency. Show all of your working.

Homework Equations

d²x/dt² + ω²x = 0

The Attempt at a Solution

First I tried using Kirchoffs Voltage Law to set up a differential equation defining the circuit.

L(d²q/dt²) + Q₁/C + Q₂/C = 0

Q1 and Q2 are meant to be initial conditions.

Replace Q1 and Q2 with q and away you go.

P.S. yes there is significance to the polarities of Q1 and Q2. If Q1 = Q2 there would be no current unless an initial current were also specified. The two capacitors would buck each other's voltage out like two batteries connected back-to-back.

 

[bsr25]

So if isntead I said:

L(d²q/dt²) + q/C + q/C = 0

(d²q/dt²) + q*(2/LC) = 0

ω² = 2/LC

Would that be correct?

 

[rude man]

So if isntead I said:

L(d²q/dt²) + q/C + q/C = 0

(d²q/dt²) + q*(2/LC) = 0

ω² = 2/LC

Would that be correct?

It surely would!

 

[bsr25]

Sweet. Many thanks for your help!

My next query would be as to what q is exactly. For example in a mass on a spring, the equation of motion is d2x/dt2 + ω2x = 0 and x is the displacement of the mass. Is q the charge across the inductor or one fo the capcitors or ... ?

 

[rude man]

Sweet. Many thanks for your help!

My next query would be as to what q is exactly. For example in a mass on a spring, the equation of motion is d2x/dt2 + ω2x = 0 and x is the displacement of the mass. Is q the charge across the inductor or one fo the capcitors or ... ?

Charge q is defined by i = dq/dt = current. Current is the same everywhere around the loop.
And therefore in any given time interval t, total charge flowing past any point along the circuit is also the same.

Let's call the left-hand capacitor C1 and the right-hand one C2. Then we sum voltage drops around the loop = 0:

V(C1)₀ - (1/C1)∫₀^ti(t')dt' - Ldi/dt - V(C2)₀ - (1/C2)∫₀^ti(t')dt' = 0. Note that current is leaving C1 and entering C2 which explains the polarities I used. V(C1)₀ and V(C2)₀ are the initial voltages on the two capacitors.

So looking at this, what is the first intergal ∫^tidt'? It's just charge q leaving C1 in time t since current is flowing OUT OF C1. And the second integral is charge entering C2 in time t, which is the same charge q. And both those charge quantities must flow thru L, but the voltage across L is Ldi/dt = Ld²q/dt². So, in time t, q leaves C1, flows thru L, and enters C2 and we wind up with

-q/C1 - Ld²q/dt² -q/C2 = V(C2)₀ - V(C1)₀.

which is the full diff. eq. including the initial conditions on C1 and C2.

I can't give you a good picture why mass should be analogous to inductance, etc. etc. except to point out that the mathematical equations are similar.