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Coupled ODEs - Applied to EM

  1. Jul 29, 2013 #1
    1. The problem statement, all variables and given/known data
    Solve the equations of motion ##\ddot{y}= \omega \dot{z}## and ##\ddot{z}= \omega (\frac{E}{B}-\dot{y})##


    2. Relevant equations



    3. The attempt at a solution

    Integrate the first equation to get ##\dot{y}=\omega z + c_1## and plug into equation 2: ##\ddot{z}=\omega (\frac{E}{B}-\omega z + c_1## simplify ##\ddot{z}= \omega \frac{E}{B} - \omega ^2 z + \omega c_1## and integrating again leaves ##\dot{z}= \omega \frac{E}{B}t - \omega ^2 zt + c_1 t +c_2##

    The next step is where everything seems to go off the rails. second integration of ##\dot{y}## gives ##\omega z t + c_1 t + c_3 = y(t)## and the second integral of ##\dot{z}## gives ##\omega \frac{E}{B}t-\omega ^2 z t + \omega c_1 t + \omega c_4##

    The solution the worked example gives are $$ y(t) = c_1 cos(\omega t) + c_2 sin(\omega t) + (E/Bt + c_3$$ and $$z(t) = c_2 cos(\omega t) - c_1 sin(\omega t) + c_4$$

    Where did the trig functions come from?!
     
  2. jcsd
  3. Jul 29, 2013 #2

    WannabeNewton

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    You started off right. ##\dot{y} = \omega z + c_1## so ##\ddot{z} + \omega^{2} z - \omega(\frac{E}{B} + c_{1}) = 0##. The next step is where you messed up (you forgot that ##z = z(t)##). This is the equation for a driven oscillator. What's the solution for such an equation?
     
  4. Aug 1, 2013 #3
    Alright, so after doing some more reading, I'm still stuck.

    The general solution for a dampened oscillator of the form ##\ddot{z}+2 \beta \dot{x} + \omega ^2 x = f(t)## is ##Ae^{- \beta t}(C_1 e ^{\sqrt{ \beta ^2 - \omega ^2}}+C_2 e ^{-\sqrt{ \beta ^2 - \omega ^2}})##. If I understood correctly, the solution for a driven oscillator will be this expression PLUS the "particular" solution.

    In this case we have ##\ddot{z}-\omega ^2 z = \omega (E/B) + C_1 \omega## the solution, if this were a non-driven system, would be ##C_1 cos( \omega t) - C_2 sin(\omega t)##, but I know there needs to be a constant added here, I just don't know where it came from.

    Yeah, still confused by the extra force term. There is no damping force, only the natural frequency plus the driving force.
     
    Last edited: Aug 1, 2013
  5. Aug 1, 2013 #4

    ehild

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    If the driving force is constant you can find the particular solution by setting ##\ddot{z}=0##.

    ehild
     
  6. Aug 1, 2013 #5

    ehild

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    There are some typos in your post. I correct them.
    For a particular solution, see the attachment in https://www.physicsforums.com/showthread.php?t=641578, #17

    ehild
     
  7. Aug 1, 2013 #6
    Why would I set ##\ddot{z} = 0##? And even if I did, that gives ##\omega ^2 z = \omega (E/B) + C_1 \omega \rightarrow E/(\omega B) + C_1 / \omega ##
     
  8. Aug 1, 2013 #7

    ehild

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    That means z=const is a particular solution with the constant E/(ωB)+C1/ω. Check.

    The general solution of the equation ##\ddot{z}-\omega ^2 z = \omega (E/B) + C_1 \omega## is

    ##z=C_2\cos(ωt)+C_3\sin(ωt)+\frac{E}{ωB}+\frac{C_1}{ω}##

    ehild
     
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