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Coupled Oscillators (Electrical Circuit)

  1. May 26, 2012 #1
    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:

    [itex]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[/itex]

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

    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. jcsd
  3. May 26, 2012 #2

    tiny-tim

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

    now look for a linear combination of the two equations that makes the LHS parallel to the RHS :wink:
     
  4. May 26, 2012 #3
    Hi Tiny-tim!

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

    [itex]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)[/itex]

    So, I'm a bit confused about (b) what should a linear combination of the two equations equal to?
     
  5. May 27, 2012 #4

    tiny-tim

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    d2(aq1 + bq2)/dt2 + C(aq1 + bq2) = … :wink:
     
  6. May 27, 2012 #5
    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):

    [itex]\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}[/itex]

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

    If it doesn't have to be equal to the source voltage, what could it be equal to?
     
    Last edited: May 27, 2012
  7. May 28, 2012 #6

    tiny-tim

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    hi roam :smile:

    (just got up :zzz:)

    erm :redface:

    you have noticed that (b) says that R1 R2 and V are all zero? :wink:
     
  8. May 28, 2012 #7
    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?
     
  9. May 29, 2012 #8

    tiny-tim

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    that equation is a normal mode, isn't it? :smile:
     
  10. May 29, 2012 #9

    ehild

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    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
    [tex]L_1 \frac{d^2q_1}{dt^2} + \frac{q_1}{c_1}+ \left( \frac{q_1-q_2}{C} \right) = 0[/tex]

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

    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.
     
  11. May 29, 2012 #10

    tiny-tim

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    isn't that the same? :confused:
     
  12. May 29, 2012 #11

    ehild

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    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
     
  13. May 29, 2012 #12

    tiny-tim

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    the same as finding a and b so that d2(aq1 + bq2)/dt2 + C(aq1 + bq2)=0 ?
     
  14. May 29, 2012 #13

    ehild

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    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
     
  15. May 29, 2012 #14

    tiny-tim

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    yes, i meant C a constant
     
  16. May 29, 2012 #15

    ehild

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

    ehild
     
  17. May 29, 2012 #16

    tiny-tim

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    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 :wink:
     
  18. May 29, 2012 #17

    ehild

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    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
     
  19. May 29, 2012 #18

    tiny-tim

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

    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 :rolleyes:)​
     
  20. May 29, 2012 #19

    ehild

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

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

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

    ehild
     
  21. May 29, 2012 #20

    tiny-tim

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    but there isn't a question c :confused:
     
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