Coupled Oscillators (Electrical Circuit)

[roam]

Homework Statement

http://img849.imageshack.us/img849/6315/63685525.jpg

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.

 

[tiny-tim]

hi roam!

Are these correct so far?

yes (except you forgot q₂/C₂)

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

 

[roam]

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

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?

 

[tiny-tim]

d²(aq₁ + bq₂)/dt² + C(aq₁ + bq₂) = …

 

[roam]

d²(aq₁ + bq₂)/dt² + C(aq₁ + bq₂) = …

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 V_s(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 V_s(t).

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

 

[tiny-tim]

hi roam

(just got up :zzz:)

erm …

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

 

[roam]

hi roam

(just got up :zzz:)

erm …

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

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?

 

[tiny-tim]

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

 

[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.

 

[tiny-tim]

isn't that the same?

 

[ehild]

isn't that the same?

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

 

[tiny-tim]

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

 

[ehild]

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

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

 

[tiny-tim]

yes, i meant C a constant

 

[ehild]

yes, i meant C a constant

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

ehild

 

[tiny-tim]

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

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 …

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 V_s(t)? …

 

[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

 

[tiny-tim]

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 …

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 )​

 

[ehild]

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. :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

 

[tiny-tim]

I meant question c

but there isn't a question c

 

[roam]

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.

 

[ehild]

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

 

[roam]

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.

Thanks. I tried to follow your post #9:

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

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

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

 

[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

 

[roam]

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)?

 

[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?

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

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

 

[roam]

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?

 

[ehild]

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

 

[roam]

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