1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    That all looks good to me.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Threads - Boundary conditions don't Date
Boundary conditions in dielectric problems Jun 26, 2017
How electric potential boundary condition works Jun 11, 2017
Electric field-boundary conditions question Jan 16, 2017