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Simply Armonic Movement and k constant

  1. Jan 16, 2006 #1
    I have a question on "k" of an armonic simple movement.

    If we take the equation of a wave, this is:

    [tex] \frac {\partial \psi (x,t)}{\partial x^2} = \frac 1 v^2 \frac {\partial^2 \psi (x,t)}{\partial t^2} [/tex]

    And this, if I'm not wrong, must to satisfy the independent of time Helmholtz equation:

    [tex] \frac {d^2 A(x)}{dt} + k A(x) = 0 [/tex]

    I'm ok since here?

    If it's ok, the solution of the first equation could be:

    [tex] \psi (x,t) = A e^{kx-wt} [/tex]

    Is this "k" the same that the "k" in Helmholzt equation? Is there any mistake in this?

    Last edited by a moderator: Jan 16, 2006
  2. jcsd
  3. Jan 17, 2006 #2

    Nobody can help me a little?
  4. Jan 23, 2006 #3

    Tom Mattson

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    Not quite. It should be:

    [tex] \frac {d^2 A(x)}{dx^2} + k^2 A(x) = 0 [/tex]

    If you use the corrected version of the equation that I posted, then yes the [itex]k[/itex]'s are the same.
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