# Homework Help: Coupled Oscillators (Electrical Circuit)

1. May 26, 2012

### roam

1. The problem statement, all variables and given/known data

http://img849.imageshack.us/img849/6315/63685525.jpg [Broken]

3. The attempt at a solution

(a) I think since i=dq/dt we will have:

$L_1 \frac{d^2q_1}{dt^2} + R_1 \frac{dq_1}{dt}+ \frac{q_1}{c_1}+ \left( \frac{q_1-q_2}{C} \right) = 0$

$L_2 \frac{d^2q_2}{dt^2}+ R_2 \frac{dq_2}{dt} + \frac{q_2-q_1}{C} = V_s(t)$

Are these correct so far?

(b) I'm really confused about this part. I would appreciate any guidance on how to approach this part.

Last edited by a moderator: May 6, 2017
2. May 26, 2012

### tiny-tim

hi roam!
yes (except you forgot q2/C2)

now look for a linear combination of the two equations that makes the LHS parallel to the RHS

3. May 26, 2012

### roam

Hi Tiny-tim!

Thank you very much. That was a typo, I meant:

$L_2 \frac{d^2q_2}{dt^2}+ R_2 \frac{dq_2}{dt} + \frac{q_2}{C_2} + \frac{q_2-q_1}{C} = V_s(t)$

So, I'm a bit confused about (b) what should a linear combination of the two equations equal to?

4. May 27, 2012

### tiny-tim

d2(aq1 + bq2)/dt2 + C(aq1 + bq2) = …

5. May 27, 2012

### roam

Ah, this expression gives the voltage across the branch in the middle. We need to have the two loops to be parallel. So does this expression have to be equal to Vs(t)?

If so, since the sourve voltage is given by the second equation (that gives a normal mode of oscillation with a frequency determined by L2, R2 and C2):

$\frac{d^2(aq_1+bq_2)}{dt^2} + c(aq_1+bq_2) = L_2 \frac{d^2q_2}{dt^2}+ R_2 \frac{dq_2}{dt} + \frac{q_2}{C_2} + \frac{q_2-q_1}{C}$

But I tried different values for the constants a and b, there is no linear combination that makes the expression equal to Vs(t).

If it doesn't have to be equal to the source voltage, what could it be equal to?

Last edited: May 27, 2012
6. May 28, 2012

### tiny-tim

hi roam

(just got up :zzz:)

erm

you have noticed that (b) says that R1 R2 and V are all zero?

7. May 28, 2012

### roam

Hi Tiny-tim,

Ah, so that equation must equal to zero: d2(aq1 + bq2)/dt2 + C(aq1 + bq2)=0

So this has to be satisfied by the angular frequency of the normal mode of the system?

8. May 29, 2012

### tiny-tim

that equation is a normal mode, isn't it?

9. May 29, 2012

### ehild

Normal mod means that both charges q1 and q2 vary sinusoidally with the same frequency and in phase or out-of phase with respect to each other.

Substitute the trial solution q1=Asin(ωt) q2=Bsin(ωt) into the equations
$$L_1 \frac{d^2q_1}{dt^2} + \frac{q_1}{c_1}+ \left( \frac{q_1-q_2}{C} \right) = 0$$

$$L_2 \frac{d^2q_2}{dt^2} + \frac{q_2}{c_2}+ \left( \frac{q_2-q_1}{C} \right) = 0$$

You get two linear equations for the coefficients A, B, and ω.
The determinant of the coefficients has to be zero. You find the possible values of ω from this condition.

10. May 29, 2012

### tiny-tim

isn't that the same?

11. May 29, 2012

### ehild

I do not understand the question. What is the same as what? This system has got two normal modes. They are solutions of the original system of differential equations.

ehild

12. May 29, 2012

### tiny-tim

the same as finding a and b so that d2(aq1 + bq2)/dt2 + C(aq1 + bq2)=0 ?

13. May 29, 2012

### ehild

The normal mode is a solution, not the way of finding it.
What is C in your equation? If it is a constant, you are right, a normal mode obeys the equation q"+w2q=0, and q is a linear combination of q1 and q2. But C means the coupling capacitor in this problem. The notation confuses the OP.

ehild

14. May 29, 2012

### tiny-tim

yes, i meant C a constant

15. May 29, 2012

### ehild

Oh, well... To tell the truth, I do not understand your hints. What do you mean on "to make the two loops parallel"?

ehild

16. May 29, 2012

### tiny-tim

i didn't say that, roam did, i think he was confused because he'd forgotten that in part (b) the non-harmonic non-homgeneous constants were zero

17. May 29, 2012

### ehild

Sorry, that was Roam. You said " LHS parallel to the RHS " left hand side and right hand side of what? Of the circuit?

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?

ehild

18. May 29, 2012

### 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 )​

19. May 29, 2012

### ehild

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

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

20. May 29, 2012

### tiny-tim

but there isn't a question c

21. May 29, 2012

### 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.

22. May 29, 2012

### 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

23. May 29, 2012

### roam

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

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

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

24. May 30, 2012

### 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

25. May 31, 2012

### 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?

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)?