I'm working on a problem and am having trouble, here's my work so far:
The motion of a straight, slender beam with a constant cross-section is governed by the partial differential equation:
$\displaystyle EI \frac{\partial^4v}{\partial x ^4} + \rho A \frac{\partial^2v}{\partial t^2} = 0$
where $E$, $I$, $\rho$, and $A$ are constants, and $v(x, t)$ is the displacement of the beam away from equilibrium at location $x$ and time $t$.
$v(x, t) = V (x) sin( \omega t + f)$, for some constants $\omega$ and $f$. I want the ODE satisfied by $V(x)$.
I think this is
$\displaystyle \frac{\partial ^4 V(x)}{\partial x^4} sin (\omega t + f) - \rho A(\omega^2) sin(\omega t+f) = 0$
So we have, $\displaystyle \frac{\partial^4 V(x)}{\partial x^4} = \rho A \omega^2$
Supposing the beam has length $L$, and is pinned in place but able to rotate freely at both ends. So,
$\displaystyle v(0, t) = v(L, t) = 0~\forall t$, and
$\displaystyle \frac{\partial^2v}{\partial x^2}(0, t) = \frac{\partial^2v}{\partial x^2}(L, t) = 0~\forall t$
I need to express these as boundary conditions and solve the boundary value problem. I need to get all the values of $\omega$ for which there are solutions