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Odd constraint problem: Reflected and Transmitted Power of String

  1. Mar 31, 2014 #1
    1. The problem statement, all variables and given/known data
    Given that a string is constrained such that dy/dx = 0 at x = 0 and unconstrained otherwise, what is the reflected and transmitted power?
    y is the deflection of the string from the x-axis. y_1 is incident wave, y_r is reflected and y_t is transmitted.

    2. Relevant equations

    Reflected power, transmitted power have already been derived in terms of impedances.
    [tex] Impedance Z = \frac{Driving Force}{string element velocity} [/tex]
    Continuity of y and dy/dx.

    3. The attempt at a solution
    Knowing that y and dy/dx are continuous, I wrote [itex] \frac{\partial y_1}{\partial x} +\frac{\partial y_r}{\partial x} = \frac{\partial y_t}{\partial x} =0 [/itex] at x = 0.

    Substituting in the general solution [itex] y_1 = e^{-ikx+iwt}, y_r = re^{+ikx+iwt}, y_t = te^{-ikx+iwt} [/itex];

    I got 1 + r = t and 1 - r = t = 0 at x = 0.

    The latter suggests all reflection, no transmission, which isn't correct because it doesn't satisfy the first equation.
     
    Last edited: Apr 1, 2014
  2. jcsd
  3. Apr 1, 2014 #2

    BvU

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    It does not satisfy which equation precisely ?
     
  4. Apr 1, 2014 #3
    The given constraint (and continuity of gradient) means that -ik(1-r) = -ik(t) = 0.
    Continuity of string means that 1 + r = t.

    If t = 0, then 1-r = 0 so r = 1. But then 1 + r = 2 =/= t = 0.
     
  5. Apr 1, 2014 #4

    BvU

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    Yes, so there must be something wrong at a more elementary level: even if you only impose continuity of y and dy/dx on these "general solutions" you end up with r = 0 ! This way you have three equations with only t and r as unknowns. Are you sure the y are general enough ?
     
  6. Apr 1, 2014 #5
    I know the string satisfies a wave equation. Evanescent waves would be my next guess, but beyond that I'm stuck.
     
  7. Apr 1, 2014 #6
    The solution for that problem requires incoming waves from both directions at the same time. (producing a standing wave with an anti-node at the origin)
     
    Last edited: Apr 1, 2014
  8. Apr 1, 2014 #7
    There'd be no reflected and transmitted power then, would there? Since the wave is standing?
     
  9. Apr 1, 2014 #8
    Good point. I'm not sure what to make of that. The question doesn't seem to make much sense since there is incoming energy from both ends.
     
  10. Apr 1, 2014 #9
    Earlier in the question, it did say to use a complex exponential for y (=A exp(ikx-iwt)) to prove that time-averaged power = 1/2 Zw^2 A^2. I assumed that carried forwards.

    I suppose I'll have to ask my supervisor for this, then. Perhaps the answer really is zero.
     
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