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Coordinates of a particle (coupled differential equations)

  1. Dec 14, 2011 #1
    1. The problem statement, all variables and given/known data
    I'm given two equations for coordinates of a certain particle in the xy plane, [itex]\dot{x}[/itex]+ωy=0
    and [itex]\dot{y}[/itex]-ωx=0.
    Then using the complex variable z=x+iy, find the differential for z, and solve it. Hence give x and y as functions of time.


    2. Relevant equations


    3. The attempt at a solution
    I'm not sure how this is gotten:
    [itex]\dot{z}[/itex]-iωz=0

    Any help would be highly appreciated. Thank you!
     
  2. jcsd
  3. Dec 14, 2011 #2

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    Hi Linday12! :smile:

    Can you write [itex]\dot z[/itex] in terms of x and y?
    And then substitute the equations you have?

    Can you also write iωz in terms of x and y?
     
  4. Dec 14, 2011 #3
    Hi. I'm not really sure what to do. As far as I know, taking [itex]\dot{z}[/itex]=[itex]\dot{x}[/itex]+i[itex]\dot{y}[/itex], and [itex]\dot{y}[/itex]=ωx and [itex]\dot{x}[/itex]=-ωy, which doesn't seem to get me anywhere. I have no idea what I am doing.
     
  5. Dec 14, 2011 #4

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    So substitute your 2nd and 3rd equation in the first?
     
  6. Dec 14, 2011 #5
    Yes. I then get [itex]\dot{z}[/itex]=-ωy+iωx=ω(-y+ix). So I can sort of see the relation here now, except my variables seem to be the wrong way, and I have no idea where the [itex]\dot{z}[/itex]-iωz=o comes from still, because i would be gone if the variables were the proper way and I subbed in z for the x-iy.
     
  7. Dec 14, 2011 #6

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    What is iz in terms of x and y?
     
  8. Dec 14, 2011 #7
    Wow, interesting. So [itex]\dot{z}[/itex]=iωz → [itex]\dot{z}[/itex]-iωz=0 because iz=ix-y. Thank you! Sorry, one more question. Now that I have that, when solving it I get z=[itex]z_{0}[/itex][itex]e^{ωt}[/itex], but the answer has an additional phase angle [itex]\phi[/itex] in it, so I was just wondering if there was a quick explanation for that.

    It's been a while since I did any differentials (I haven't had a class in them yet), I think perhaps I'm used to the general formula that includes the phase angle in the constant z_0, but if you were to solve it you would get ln(z)=iωt+c, which is where the phase angle came from, but then where does the z_0 come from? Because the answer is z=z_0[itex]e^{ωt+\phi}[/itex] Sorry, I hope that makes sense.
     
  9. Dec 15, 2011 #8

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    Your z0 is the integration constant that is itself a complex number.
    Write [itex]z_0=r_0e^{i\phi}[/itex] and you get:
    [tex]z=z_0e^{i\omega t}=r_0e^{i\phi}e^{i\omega t}=r_0e^{i(\omega t + \phi)}[/tex]

    Btw, note the extra [itex]i[/itex] and the use of the non-negative real number r0 instead of z0.
     
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