For a beam that is fixed at both ends, the boundary conditions are,

\[ \begin{equation} w = 0\hspace{0.5cm}\text{and}\hspace{0.5cm}\frac{\partial w}{\partial x}=0\hspace{0.5cm}\text{at}\hspace{0.5cm}x=0, \end{equation} \] \[ \begin{equation} w = 0\hspace{0.5cm}\text{and}\hspace{0.5cm}\frac{\partial w}{\partial x}=0\hspace{0.5cm}\text{at}\hspace{0.5cm}x=L_x. \end{equation} \]

In this case, the normal mode solutions to the Euler-Lagrange equation for an unloaded beam are,

\[ \begin{equation} w_n(x,t) = A_n\left(\cosh (k_nx)-\cos (k_nx)+\frac{\left(\cosh (k_nL_x)-\cos (k_nL_x)\right)\left(\sin (k_nx)-\sinh (k_nx)\right)}{\sinh (k_nL_x)-\sin (k_nL_x)}\right)\hspace{1cm}n=1,2,3,\cdots \end{equation} \]

where the allowed wavenumbers are determined by the condition, $\cosh(k_nL_x)\cos(k_nL_x) =1$. This nonlinear equation must be solved numerically. The first few solutions are $k_1L_x=4.730,\, k_2L_x = 7.853,\, k_3L_x = 10.996,\, k_4L_x = 14.137,\,\cdots$

The corresponding frequencies are given by the formula,

\[ \begin{equation} \omega = \sqrt{\frac{EI}{\mu}}k^2. \end{equation} \]

The figure below shows the form of the first four modes. These are small amplitude oscillations. Notice that the vertical scale is expanded compared to the horizontal scale.



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