Finding the Resonant frequency in an LC circuit

In summary: Just because you can't see the analogy doesn't mean it isn't there. :)In summary, the resonant frequency of the free oscillator is 1/LC. The equation for SHM gives the same result. The attempt at a solution was off, but the final equation was still correct. Charge q is the same everywhere around the circuit and flow thru L to give the full differential equation.
  • #1
bsr25
7
0

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) + QTotal/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
 

Attachments

  • circuit.png
    circuit.png
    40.3 KB · Views: 555
Physics news on Phys.org
  • #3
bsr25 said:

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.
 
  • #4
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?
 
  • #5
bsr25 said:
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! :smile:
 
  • #6
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 ... ?
 
  • #7
bsr25 said:
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)∫0ti(t')dt' - Ldi/dt - V(C2)0 - (1/C2)∫0ti(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.
 

Related to Finding the Resonant frequency in an LC circuit

1. What is a resonant frequency in an LC circuit?

The resonant frequency in an LC circuit is the frequency at which the circuit reaches its maximum energy and the voltage across the capacitor and the current through the inductor are at their peak levels.

2. How do you calculate the resonant frequency in an LC circuit?

The resonant frequency can be calculated using the formula: f = 1 / (2π√LC), where f is the resonant frequency, L is the inductance in henries, and C is the capacitance in farads.

3. What factors affect the resonant frequency in an LC circuit?

The resonant frequency in an LC circuit can be affected by the values of the inductance and capacitance, the materials used for the components, and the circuit's resistance.

4. Why is finding the resonant frequency important in an LC circuit?

Finding the resonant frequency is important in an LC circuit because it allows us to determine the ideal frequency for the circuit to operate at its maximum efficiency and power. It is also essential in designing and tuning electronic circuits.

5. How do you experimentally determine the resonant frequency in an LC circuit?

To experimentally determine the resonant frequency in an LC circuit, you can use an oscilloscope and a function generator. Adjust the frequency of the function generator until the amplitude of the current through the circuit is at its maximum, which indicates the resonant frequency. You can also use a multimeter to measure the voltage across the capacitor and the current through the inductor and find the frequency at which they are at their peak values.

Similar threads

  • Introductory Physics Homework Help
Replies
5
Views
375
  • Introductory Physics Homework Help
Replies
6
Views
232
  • Introductory Physics Homework Help
Replies
3
Views
256
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
10
Views
597
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Electrical Engineering
Replies
17
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
2K
Back
Top