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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

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You lost $EI$ off (or incorporated them in the definition of $V$), but you can just define $k=\frac{\rho A}{EI}$ and save typing as the constants only come in this combination. $f$ is also not contributing, as nothing in the problem cares about the phase. –  Ross Millikan Jul 14 '11 at 15:54

1 Answer 1

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V(0,t)=0 V(L,t)=0 where V(x,t)=V(x)sin(wx+f) <- inside sin there is no x so conditions for V(x) is V(0)=0 and V(L)=0

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