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Changing the wave equation

  1. Mar 27, 2006 #1
    What happens when you make the x varible in the wave egn to some power

    m(dx/dt) + k x^n =0

    What happens when n increases/decreases?
  2. jcsd
  3. Mar 27, 2006 #2
    Um, perhaps you might want to get the correct form of the wave equation first?
  4. Mar 27, 2006 #3


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    That diff. Eq. is seperable, and has the solution

    [tex]x(t)=\left( C_1+(n-1)\frac{k}{m}t\right) ^{-\frac{1}{n-1}}[/tex]

    which is vastly different from

    m(dx/dt) + k x =0

    which is also seperable, and has the solution


    but it is notable that the limit as n->1 of the former solution is the later solution if [tex]C_1=C_2=1[/tex].
    Last edited: Mar 27, 2006
  5. Mar 27, 2006 #4


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    I guess you meant C times (1+ (n-1).....)

    It's interesting. It's neat to see the exponential recovered as n ->1.
  6. Mar 27, 2006 #5


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    If that were a second derivative, then you would be closer to a wave equation.
  7. Mar 27, 2006 #6
    Yeah sorry I forgot to make the eqn: m(d^(2)x/dt^(2)) + k X^(n)=0

    this is the eqn of the wave cause by a simple harmonic Isolator.
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