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Superposition of 2 vectors

  1. Aug 23, 2006 #1
    I'm considering the superposition of 2 vectors.

    [tex]E_x=E_1cos(wt)
    E_y=E_2cos(wt+/pi/4)[/tex]

    Trying to eliminate t. Its easy when the phase shift is 0 or pi/2 but I'm not sure how to go about it in this case. I can get Ey to be a function of cos(wt)-sin(wt), or cos(wt)sin(wt) just using trigonometric formulas, but I don't know where to go from there.

    The end result should be an elipse with axis that are not aligned with the x-y axis. If its quite difficult then I won't worry about it. I just get the feeling that I've done this before and I should know it, but somethings not clicking. Thanks.


    edit: For some reason the LATEX graphics don't appear for me. Just in case I put the formulas in wrong they should be:

    Ex = E1 cos(wt)
    Ey = E2 cos(wt + pi/4)
     
    Last edited: Aug 23, 2006
  2. jcsd
  3. Aug 23, 2006 #2

    Andrew Mason

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    How about [tex]E = \sqrt{E_x^2 + E_y^2} [/tex]?

    As far as the latex problem, you have to use lower case for the tex and /tex commands.

    AM
     
  4. Aug 23, 2006 #3
    Thanks for the reply, I don't think I was very clear about what I'm trying to do though.

    If I have
    [tex]E_x=E_1cos(wt)[/tex]

    [tex]E_y=E_2cos(wt)[/tex]

    then [tex]E_1y=E_2x[/tex] gives me my equation of motion for the particle.

    If
    [tex]E_x=E_1cos(wt)[/tex]

    [tex]E_y=E_2cos(wt + \frac{\pi}{2})[/tex]

    then
    [tex]E_y=-E_2sin(wt)[/tex]

    [tex]\frac{x^2}{E_1}+\frac{y^2}{E_2}=1 [/tex]

    And the vector moves in an ellipse.

    I'm not sure how to get the same sort of equation (ie. no dependance on t), or if its possible, for

    [tex]E_x=E_1cos(wt)[/tex]

    [tex]E_y=E_2cos(wt+\frac{\pi}{4})[/tex]

    [tex]E_y=\frac{E_2}{\sqrt{2}}(cos(wt)-sin(wt))[/tex]

    [tex]E_y^2=-E_2^2(cos(wt)sin(wt))[/tex]

    Not sure where I can go from here
     
    Last edited: Aug 23, 2006
  5. Aug 23, 2006 #4

    robphy

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    Gold Member

    There's a rotation [transformation] involved.
    Complex numbers might simplify your calculation.
     
  6. Aug 23, 2006 #5
    I'm sorry but I'm still not sure how to get there.

    If I write it as,

    [tex]E_x=E_1e^{iwt}[/tex]
    [tex]E_y=E_2e^{iwt}e^{\frac{\pi}{4}}[/tex]

    I don't know where to go from here.

    I feel like its staring me right in the face. I know what the answer should look like. I know the transform should be along the lines of [cosA,-sinA;sinA,cosA] and I could probably work out A by playing with E1 and E2, but its just not falling into place for me. Maybe I'm just having a bad day, hopefully a good night's sleep will help.
     
  7. Aug 23, 2006 #6
    Just for clarification, I'll explain the context in which I ask this question. For a uGrad assignment I have been asked to propose an exam question and provide a solution (The hard part about this assignment is thinking of a question that hasn't already been on previous exams).

    Part of my question goes along the lines of:
    Consider 2 EM waves, prpagating along the z axis, with angular frequency w and wave number k. Wave 1 has its E-field aligned with the x axis, wave 2 with the y axis. There is a relative phase difference of [tex]\phi[/tex] between them. Describe how the polarisation of the superposition of waves 1 & 2 varies in time.

    For the case [tex]\phi=0[/tex] we have a plane wave with an E-field of magnitude [tex]\sqrt{E_1^2+E_2^2}[/tex]. For all [tex]0 < \phi < \pi[/tex] we should have an elliptically polarized plane wave, but I only know how to demonstrate this for the case where [tex]\phi=\frac{\pi}{2}.[/tex]
     
  8. Aug 24, 2006 #7
    Nevermind. Worked it out. Just wasn't thinking yesterday.
     
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