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  , \ z=0; </math></center>
\begin{align}
<center><math> \rho g\zeta  + \mathrm{i}\omega\rho \phi
-\mathrm{i}\omega\zeta &= \partial_z\phi  , &z=0  \\
= D \partial_x^4 \eta -\omega^2 \rho_i h \zeta, \ z=0; </math></center>
\rho g\zeta  - \mathrm{i}\omega\rho \phi &= D \partial_x^4 \eta -\omega^2 \rho_i h \zeta, &z=0 \\
<center><math>
\Delta \phi &= 0, &-h<z<0 \\
\Delta \phi = 0,\,\,-h<z<0
\partial_z \phi &= 0, &z=-h,  
</math></center>
\end{align}
<center><math>
\partial_z \phi = 0,\,\,z=-h,
</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, \;\;\; -h < z \leq 0,
\begin{align}
</math></center>
\Delta \phi &= 0, &-h < z \leq 0 \\
<center><math>
\partial_z \phi &= 0, &z = - h \\
\partial_z \phi = 0, \;\;\; z = - h,
\beta \partial_x^4 \zeta + \left( 1 - \gamma\alpha \right) \zeta &= -\mathrm{i} \sqrt{\alpha}\phi, &z = 0 \\
</math></center>
-\mathrm{i}\omega\zeta &= \partial_z\phi , &z=0 .
<center><math>
\end{align}
\beta \partial_x^4 \zeta
+ \left( 1 - \gamma\alpha \right) \zeta = \mathrm{i} \sqrt{\alpha}\phi, \;\;
  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

[math]\displaystyle{ \begin{align} -\mathrm{i}\omega\zeta &= \partial_z\phi , &z=0 \\ \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\lt z\lt 0 \\ \partial_z \phi &= 0, &z=-h, \end{align} }[/math]

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

[math]\displaystyle{ \begin{align} \Delta \phi &= 0, &-h \lt z \leq 0 \\ \partial_z \phi &= 0, &z = - h \\ \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} }[/math]