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## Coupled Oscillators (Electrical Circuit)

 Quote by ehild Sorry, that was Roam. You said " LHS parallel to the RHS " left hand side and right hand side of what? Of the circuit?
no, of the (combined) equation
 Quote by tiny-tim now look for a linear combination of the two equations that makes the LHS parallel to the RHS
… another way of saying, look for for the eigenvalues of the matrix
 What do you mean about the question c? Is it enough to write up an equation as you suggested d2(aq1 + bq2)/dt2 + C(aq1 + bq2)=0 or is it needed to write the equations in detail with the parameters L1,L2,C1, C2, C(capacitance) given?
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 )

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 Quote by tiny-tim 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?
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

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 Quote by ehild I meant question c
but there isn't a question c

 Quote by tiny-tim but there isn't a question c
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:

$\frac{d^2(aq_1+bq_2)}{dt^2} + k(aq_1+bq_2) = 0$

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.

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 Quote by 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: $\frac{d^2(aq_1+bq_2)}{dt^2} + k(aq_1+bq_2) = 0$ 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.
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

 Quote by ehild 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.

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

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

And the second equation:

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

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

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
 Recognitions: Homework Help You can divide the whole equations with sin(ωt), getting a system of linear equations for A and B. $$-L_1 \omega^2 A + \frac{A }{C_1} + \frac{(A-B) }{C} = 0$$ $$-L_2 \omega^2 B + \frac{B }{C_2} + \frac{(B-A) }{C} = 0$$ Collecting like terms, $$(-L_1 \omega^2 + \frac{1 }{C_1} + \frac{1 }{C})A-\frac{1}{C}B = 0$$ $$(-L_2 \omega^2 + \frac{1 }{C_2} + \frac{1 }{C})B-\frac{1}{C}A = 0$$ 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

 Quote by ehild You can divide the whole equations with sin(ωt), getting a system of linear equations for A and B. $$-L_1 \omega^2 A + \frac{A }{C_1} + \frac{(A-B) }{C} = 0$$ $$-L_2 \omega^2 B + \frac{B }{C_2} + \frac{(B-A) }{C} = 0$$ Collecting like terms, $$(-L_1 \omega^2 + \frac{1 }{C_1} + \frac{1 }{C})A-\frac{1}{C}B = 0$$ $$(-L_2 \omega^2 + \frac{1 }{C_2} + \frac{1 }{C})B-\frac{1}{C}A = 0$$ 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
Thank you so much, but the ratio A/B I get from the two equations are not equal. From the first equation I get

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

And according to the second equation it is:

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

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)?

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 Quote by 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 $$\frac{A}{B} = \frac{1}{\left( -L_1 \omega^2 + \frac{1 }{C_1} + \frac{1}{c} \right) C}$$ And according to the second equation it is: $$\frac{A}{B} = \left( -L_2 \omega^2 + \frac{1 }{C_2} + \frac{1 }{C}) \right) C$$ Shouldn't they be equal?
Yes, they have to be equal. Find the value of ω which makes them equal, or at least write the equation for ω.

 Quote by roam 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)?
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

 Quote by ehild Yes, they have to be equal. Find the value of ω which makes them equal, or at least write the equation for ω.
I tried to solve for an expression for ω by equating the two equations:

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

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?

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 Quote by roam I tried to solve for an expression for ω by equating the two equations: $$\frac{1}{\left( -L_1 \omega^2 + \frac{1 }{C_1} + \frac{1}{c} \right) C}=\left( -L_2 \omega^2 + \frac{1 }{C_2} + \frac{1 }{C} \right) C$$ 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?
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
 Okay. Thank you very much for all your help, that was very helpful. Cheers.