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Circular Polarization of Wave

  1. Feb 15, 2016 #1

    cpburris

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    1. The problem statement, all variables and given/known data
    The horizontal component of a wave is in the y-direction, and the vertical component is in the x-direction. If the horizontal and vertical components of a wave on a string have the same amplitude and are 90 degrees out of phase (say δv=0 and δh= 90°). At a fixed point z, show that the string moves in a circle around the z-axis.
    2. Relevant equations
    ƒ
    i(z,t)=A Cos(kz-ωt+δi)

    3. The attempt at a solution
    First of all, don't tell me ƒv2h2=A2 therefore it is a circle. I know that, and I don't like it. If that is the ONLY way to solve the problem so be it, but I think I can do it a different way.

    I think I should be able to express ƒ in cylindrical coordinates, ƒ=s(t)s+φ(t)φ where s and φ are the s and φ unit vectors, and show that s(t) = constant and that φ(t)≠constant and φ(t) has range s.t. φ(t)max≥2π (i.e. will go around a full circle, not do some silly oscillation between 0 and π or something. That wouldn't really make any physical sense I don't think, but I think it is necessary for the proof.).

    So here we go with my attempt(forgive my notation, hopefully you can make sense of it, if not let me know and ill do everything proper) :

    f_v=A Cos(kz-wt) x_unitvect.
    f_h=A Cos(kz-wt+π/2) y_unitvect. = - A Sin(kz-wt) y_unitvect.

    f=f_v+f_h= A [Cos(kz-wt) x_unitvect. - Sin(kz-wt) y_unitvect.]

    Converting to cylindrical:

    f= A [Cos(kz-wt) Cos(φ) - Sin(kz-wt) Sin(φ)] s_unitvect. - A[Cos(kz-wt)Sin(φ)+Sin(kz-wt)Cost(φ)] φ_unitvect.
    Using Cos/Sin (A±B) Trig Identities:
    f= A Cos[kz-wt+φ] s_unitvect. - A Sin[kz-wt+φ] φ_unitvect.
    If φ= wt (?), Then reduces to
    f= A Cos[kz] s_unitvect. - A Sin[kz] φ_unitvect.
    which would make both time independent. Yikes.
     
  2. jcsd
  3. Feb 16, 2016 #2

    blue_leaf77

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    First of all, I will tell you ƒv2h2=A2 therefore it is a circle.
    Using our intuition, the field at any instant of time is radial, i.e. for all times the resultant E field must only has ##\hat{s}## component and ##\hat{\phi}##, only that the direction is changing. But your last equation shows otherwise, it has ##\hat{\phi}## component.
    What you did above is actually finding the field vector at a point different from the origin and the coordinate of this point is given by the angle ##\phi##. This means, ##\phi## must be fixed, it cannot be a function of time as you have assumed.
    However, if you want to describe the time dependent of the field vector at the origin, you have to be careful because the angle ##\phi## associated with the origin is not defined. In this case, the phase of the wave itself becomes ##\phi##, namely ##\phi = kz-\omega t##.
     
    Last edited: Feb 16, 2016
  4. Feb 16, 2016 #3

    cpburris

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    Hm. I don't understand your point about the field. This is a wave on a string, created by a driving force (someone just moving the end of the string). What is the field?

    P.S. I think I may sound slightly combative, especially with the whole "Don't tell me" thing, I was just very frustrated and in my mood I was perhaps overly...blunt. I want to apologize for that.
     
  5. Feb 16, 2016 #4

    blue_leaf77

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    Ah I completely forgot that your problem concerns a string, then in this case the last two lines of my previous post applies, namely ##\phi=kz-\omega t##. Taking this into account, your steps:
    become unnecessary (in fact, it's misleading).
    Instead, you should directly substitute the above relation for ##\phi## in
     
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