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Boundary conditions don't apply in the equation's region of validity

  1. Apr 21, 2014 #1
    1. The problem statement, all variables and given/known data
    A tight string lies along the positive x-axis when unperturbed. Its displacement from the x-axis is denoted by [itex]y(x, t)[/itex]. It is attached to a boundary at [itex]x = 0[/itex]. The condition at the boundary is
    [tex]y+\alpha \frac{\partial y}{\partial x} =0[/tex] where [itex]\alpha[/itex] is a constant.

    Write the displacement as the sum of an incident wave and reflected wave,
    [tex]y(x, t) = e^{−ikx−i\omega t} + re^{ikx−i\omega t},\qquad x > 0,[/tex] and compute the reflection coefficient, [itex]r[/itex]. Writing [itex]r = |r|e^{i\phi}[/itex], show that [itex]|r| = 1[/itex] and find [itex]\phi[/itex].

    2. Relevant equations



    3. The attempt at a solution
    Since the boundary condition applies at [itex]x=0[/itex] and the equation given is only valid for [itex]x>0[/itex] I can't use that, so what equation should I use?


    (If you just apply the condition that the incident and reflected wave are equal at [itex]x=0[/itex], since there is no transmission, you get what I believe is the desired result, but how would one go about this problem with the method it wants)
     
  2. jcsd
  3. Apr 21, 2014 #2

    haruspex

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    The equation is only given for x > 0 because the derivative is not defined at x= 0. If we take the derivative at 0 as meaning the one sided derivative then the equation must also be valid at 0; if not there would be a discontinuity in the string. Just take it as valid at x = 0 and see where it leads.
     
  4. Apr 21, 2014 #3
    Am I doing this correctly.

    Just assuming it works at [itex]x=0[/itex] gives [tex]e^{−ikx−i\omega t} + |r|e^{ikx−i\omega t +i\phi} = \alpha ike^{−ikx−i\omega t} - \alpha ikre^{ikx−i\omega t +i\phi}[/tex] so [tex]e^{−i\omega t} + |r|e^{−i\omega t +i\phi} = \alpha ike^{−i\omega t} - \alpha ik|r|e^{−i\omega t +i\phi}[/tex] and dividing by [itex]e^{-i \omega t}[/itex] (or letting [itex]t=0[/itex], since the equation must hold then as well as any time) [tex]1+|r|e^{i\phi} = \alpha ik - \alpha ik|r|e^{i\phi} \\ |r|e^{i\phi}=\frac{\alpha ik -1}{\alpha ik +1}[/tex] and so [tex]|r|=|r||e^{i\phi}|=\left| \frac{\alpha ik -1}{\alpha ik +1} \right|=\frac{|\alpha ik -1|}{|\alpha ik +1|} = 1. [/tex] Now since [tex]\frac{\alpha ik -1}{\alpha ik +1} = \frac{\alpha k +i}{\alpha k -i} = \frac{\alpha^2 k^2 +2\alpha ki -1}{\alpha^2 k^2 +1} = \frac{\alpha^2 k^2 -1}{\alpha^2k^2 +1}+\frac{2\alpha ki}{\alpha^2k^2 +1}[/tex] we have [tex]\phi = \tan^{-1} \left\{ \frac{2\alpha k}{\alpha^2 k^2 -1} \right\}.[/tex] Is that right?
     
  5. Apr 21, 2014 #4
    I'm pretty sure that last part isn't right at all
     
  6. Apr 21, 2014 #5

    haruspex

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    That all looks good to me.
     
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