Coupled Oscillators (Electrical Circuit)

tiny-tim

no, of the (combined) equation …
… another way of saying, look for for the eigenvalues of the matrix
not really following you

a b and C are constants that have to be found

(or do you mean, is my C the same as the C in the question?

no, that was my mistake, i didn't notice C was already in use )

ehild

I meant question c, the third c already in this problem.

Well, I have to improve my English, I see.

Any hint how to find a,b,C? My old-fashion method gives rather ugly expressions for the eigenvalues.

ehild

tiny-tim

but there isn't a question c

roam

I think he meant the third capacitance, denoted "C" (the other two being C₁ and C₂). But we need to work out the constants in:

[itex]\frac{d^2(aq_1+bq_2)}{dt^2} + k(aq_1+bq_2) = 0[/itex]

I'm really confused right now. How do I exactly find the equation that must be satisfied by the ω of a normal mode of this system? I'm not sure which way to go...

We eventually have to find explicit expressions for the normal modes of oscillation, but now we are not given any numerical values for the inductances and capacitances.

ehild

Sorry, I remembered wrong. So it was question b.

You can write the normal mode frequencies in terms of the parameters, L1,L2,C1,C2,C. Read my post #9. Find the equation for ω. You get a rather ugly expression. I do not think there is a simpler one.

ehild

roam

Thanks. I tried to follow your post #9:

[itex]L_1 \frac{d^2(A \sin (\omega t))}{dt^2}+ \frac{A \sin (\omega t)}{C_1} + \left( \frac{A \sin(\omega t)-B \sin (\omega t)}{C} \right) = 0[/itex]

[itex]-L_1 \omega^2 A \sin (\omega t) + \frac{A \sin (\omega t)}{C_1} + \frac{(A-B) \sin(\omega t)}{C} = 0[/itex]

And the second equation:

[itex]L_2 \frac{d^2 (B \sin (\omega t))}{dt^2} + \frac{B \sin (\omega t)}{C_2} + \left( \frac{B \sin (\omega t)-A \sin(\omega t)}{C} \right)=0[/itex]

[itex]-L_2 \omega^2 B \sin(\omega t) + \frac{B \sin(\omega t)}{C_2} + \frac{(B-A) \sin(\omega t)}{C} = 0[/itex]

Is this what you meant? How exactly do I have to solve for ω in each case? The omegas in the sine arguments are a bit confusing

ehild

You can divide the whole equations with sin(ωt), getting a system of linear equations for A and B.

[tex]-L_1 \omega^2 A + \frac{A }{C_1} + \frac{(A-B) }{C} = 0 [/tex]
[tex]-L_2 \omega^2 B + \frac{B }{C_2} + \frac{(B-A) }{C} = 0[/tex]

Collecting like terms,

[tex](-L_1 \omega^2 + \frac{1 }{C_1} + \frac{1 }{C})A-\frac{1}{C}B = 0 [/tex]
[tex](-L_2 \omega^2 + \frac{1 }{C_2} + \frac{1 }{C})B-\frac{1}{C}A = 0[/tex]

A=0, B=0 is solution (trivial solution) but you want at least one from A and B different from 0. You can determine only the ratio of A and B, one of them is arbitrary, different from zero.

If you studied linear equations, you have to know that for nonzero solution, the determinant of the coefficients must be zero.

If you do not know it yet, find A/B from both equations. They must be equal, which is true only for special values of ω. These are the angular frequencies of the normal modes. Find that equation for ω.

ehild

roam

Thank you so much, but the ratio A/B I get from the two equations are not equal. From the first equation I get

[tex]\frac{A}{B} = \frac{1}{\left( -L_1 \omega^2 + \frac{1 }{C_1} + \frac{1}{c} \right) C}[/tex]

And according to the second equation it is:

[tex]\frac{A}{B} = \left( -L_2 \omega^2 + \frac{1 }{C_2} + \frac{1 }{C}) \right) C[/tex]

Shouldn't they be equal?

I also had another relevant question, once we found the the equation that must be satisfied by the angular frequency of a normal mode of this system, how do we find the actual frequency (the frequency of the system itself at which energy is transferred back and forth between the two loops)?

ehild

Yes, they have to be equal. Find the value of ω which makes them equal, or at least write the equation for ω.

The system will oscillate with one of those normal-mode frequencies or both.

It is useful to try a numerical solution. Assume that C=1μF, C1=C2=2μF, L1=L2=0.5H.

ehild

roam

I tried to solve for an expression for ω by equating the two equations:

[tex]\frac{1}{\left( -L_1 \omega^2 + \frac{1 }{C_1} + \frac{1}{c} \right) C}=\left( -L_2 \omega^2 + \frac{1 }{C_2} + \frac{1 }{C} \right) C[/tex]

But things turn out to be too messy, because I keep getting a few omega terms and one ω⁴ and I don't see how it can be simplified in order to have a single equation for ω... So, I'm wondering if this is the correct approach?

At this stage we don't know how many millihenries and microfarads are on the inductors and capacitors, that's why I can't find a suitable ω by inspection. Is there another way to solve this?

ehild

Multiplying the whole equation with the denominator and arranging everything on one side, expanding and collecting like terms, you get a quadratic equation for ω² in the form pω⁴+qω²+r=0. I think the problem asks only that equation, not the solution.

Coupled oscillators are not easy things.
You would get a more clear picture of the whole thing by solving a numerical example.

Try C=1μF, C1=C2=2μF, L1=L2=0.5H

ehild

roam

Okay. Thank you very much for all your help, that was very helpful. Cheers.