I noticed a long time ago the similarity between Schrodinger equation and Euler-Bernoulli beam equation. Namely, Euler-Bernoulli equation is equivalent to the system of Schrodinger equation for a free particle and its complex conjugate.
Euler-Bernoulli equation for vibrations of beams ($\Psi$ here - amplitude of bending, Y - Young modulus, etc):
$$ \frac{YI}{\rho A} \frac{\partial ^4 \Psi}{\partial x^4}+\frac{\partial ^2 \Psi}{\partial t^2}=0 $$
If we transform the beam parameters to a particle parameters appropriately:
$$ \frac{\hbar^2}{4m^2} \frac{\partial ^4 \Psi}{\partial x^4} + \frac{\partial ^2 \Psi}{\partial t^2}=0 $$
$$ \left( \frac{\hbar}{2m} \frac{\partial ^2 }{\partial x^2}+i\frac{\partial }{\partial t} \right) \left( \frac{\hbar}{2m} \frac{\partial ^2 }{\partial x^2}-i\frac{\partial}{\partial t} \right) \Psi =0 $$
Which is obviously a system of two Schrodinger equations ($\Psi$ becomes a wavefunction):
$$ \frac{\hbar}{2m} \frac{\partial ^2 \Psi}{\partial x^2}+i\frac{\partial \Psi}{\partial t}=0 $$ $$ \frac{\hbar}{2m} \frac{\partial ^2 \Psi}{\partial x^2}-i\frac{\partial \Psi}{\partial t}=0 $$
The main (mathematical) difference between the solution of this system and a single Schrodinger equation is of course that here
$$\Psi=\Psi^*$$
Which is not true in quantum mechanics, since complex conjugation is equivalent to time-reversal (and momentum-reversal for a free particle).
Also, the general solution of B-E contains $\cosh$ and $\sinh$ as well as $\cos$ and $\sin$ and thus requires four boundary conditions instead of two.
But the $\omega(k)$ spectrum is the same for both equations.
So, do you think there is any physical meaning for this mathematical similarity, or can it prove to be in any way useful, like in modeling the behavior of particles with acoustic experiments?