Template:Frequency domain equations for a floating plate: Difference between revisions
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If we make the assumption of [[Frequency Domain Problem]] that everything is proportional to | If we make the assumption of [[Frequency Domain Problem]] that everything is proportional to | ||
<math>\exp (\mathrm{i}\omega t)\,</math> the equations become | <math>\exp (-\mathrm{i}\omega t)\,</math> the equations become | ||
<center><math> | <center><math> | ||
\mathrm{i}\omega\zeta = \partial_z\phi , | \begin{align} | ||
-\mathrm{i}\omega\zeta &= \partial_z\phi , &z=0 \\ | |||
= D \partial_x^4 \eta -\omega^2 \rho_i h \zeta, | \rho g\zeta - \mathrm{i}\omega\rho \phi &= D \partial_x^4 \eta -\omega^2 \rho_i h \zeta, &z=0 \\ | ||
\Delta \phi &= 0, &-h<z<0 \\ | |||
\Delta \phi = 0, | \partial_z \phi &= 0, &z=-h, | ||
\end{align} | |||
\partial_z \phi = 0, | |||
</math></center> | </math></center> | ||
where | where | ||
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<math>\beta = D/\rho g</math> and <math>\gamma = \rho_i h/\rho</math> to obtain | <math>\beta = D/\rho g</math> and <math>\gamma = \rho_i h/\rho</math> to obtain | ||
<center><math> | <center><math> | ||
\Delta \phi = 0, | \begin{align} | ||
\Delta \phi &= 0, &-h < z \leq 0 \\ | |||
\partial_z \phi &= 0, &z = - h \\ | |||
\partial_z \phi = 0, | \beta \partial_x^4 \zeta + \left( 1 - \gamma\alpha \right) \zeta &= -\mathrm{i} \sqrt{\alpha}\phi, &z = 0 \\ | ||
-\mathrm{i}\omega\zeta &= \partial_z\phi , &z=0 . | |||
\end{align} | |||
\beta \partial_x^4 \zeta | |||
+ \left( 1 - \gamma\alpha \right) \zeta = | |||
z = 0. | |||
</math></center> | </math></center> |
Latest revision as of 11:03, 6 November 2010
If we make the assumption of Frequency Domain Problem that everything is proportional to [math]\displaystyle{ \exp (-\mathrm{i}\omega t)\, }[/math] the equations become
where [math]\displaystyle{ \zeta }[/math] is the surface displacement and [math]\displaystyle{ \phi }[/math] is the velocity potential in the frequency domain.
These equations can be simplified by defining [math]\displaystyle{ \alpha = \omega^2/g }[/math], [math]\displaystyle{ \beta = D/\rho g }[/math] and [math]\displaystyle{ \gamma = \rho_i h/\rho }[/math] to obtain