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Why is the Flexural Wavelength Dependent on Density?

  1. Oct 5, 2012 #1
    I'm familiar with the derivation of flexural wavelength (λ=(4*D/gΔρ)^.25), but have no intuition for why λ's dependent on Δρ. My intuition tells me if you put a mass on a plate and it bends a little, then you increase the mass, the amplitude of the bend will change, but not the wavelength (assuming you're not stretching far enough to change the material properties). Anyone have any explanation to help with my false intuition? Thanks
  2. jcsd
  3. Oct 6, 2012 #2

    Intersting topic, tectonics isn't it ? I have found the explanation in http://web.ics.purdue.edu/~ecalais/teaching/eas450/Gravity5.pdf. The model seems to consider only the balance between vertical load and resistance within a stack of two layers having different densities. For me it is quite intuitive that both the deflection depth and width depend on density difference. May be a useful analogy would be that of a bimetal strip whose deformation depends on the difference between the linear thermal expansion coefficients of the two metals, the only thing is that in this case horizontal forces should be balanced.
  4. Oct 6, 2012 #3
    The answer is really pretty simple.

    When generating the relevant flexural wave equation the force = mass times acceleration

    and the mass = density times volume of the layer concerned.

    The spatial term is a fourth degree term due to the flexural action thus

    [tex]YI\frac{{{\partial ^4}\varphi }}{{\partial {X^4}}} = - \rho V\frac{{{\partial ^2}\varphi }}{{\partial {t^2}}}[/tex]

    The right hand side is the mass x acceleration term, [itex]\varphi [/itex] is the displacement variable.
    Last edited: Oct 6, 2012
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