anonymous
  • anonymous
Is \(y=Ae^{-x-t^2}\) a wave? I'm testing it by using the wave equation:\[\frac{\partial ^2}{\partial x^2}=\frac{1}{u^2} \frac{\partial ^2}{\partial t^2}\]So I currently have\[\frac{\partial ^2}{\partial x^2}=Ae^{-x-t^2}\]and\[\frac{\partial ^2}{\partial t^2}=Ae^{-x-t^2} [-2+4t^2]\]Normally, a wave has the form \(kx-\omega t\), and the velocity can be calculated from that, but this one has a \(t^2\), which confuses me. Any thoughts?
Physics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Those derivatives should all be with respect to y also.
anonymous
  • anonymous
They are derivatives of y, but not with respect to y. In the wave equation, u is a constant, i.e. the speed of propagation of the wave. Do your results suggest that your function satisfies the wave equation?
anonymous
  • anonymous
Aye, I meant derivatives of y. Early morning head I suppose. If the two sides were equal, the velocity squared would need to be \(-2+4t^2\), which is not constant, so I suppose it cannot be a wave.

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anonymous
  • anonymous
I suppose it would also have an imaginary velocity.
anonymous
  • anonymous
Mhmm, it's not a wave as defined by the wave equation.
anonymous
  • anonymous
I shouldn't say "as defined by the wave equation".... it suffices to say that the function you are given does not satisfy the one-dimensional wave equation :)
anonymous
  • anonymous
Thanks Jemurray3!
anonymous
  • anonymous
No problem :)

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