Nonlinear finite elements/Natural vibration

Special case : natural vibrations
Recall that the finite element system of equations has the form

\mathbf{M}~\ddot{\mathbf{u}} = \mathbf{f}_{\text{ext}} - \mathbf{f}_{\text{int}}~. $$ We could also have written this equation as

\mathbf{M}~\ddot{\mathbf{u}} + \mathbf{K}~\mathbf{u} = \mathbf{f} ~. $$ For natural vibrations, the forces and the displacements are assumed to be periodic in time, i.e.,

\mathbf{u} = \mathbf{u}^0~\exp(i\omega t) ~. $$ and

\mathbf{f} = \mathbf{f}^0~\exp(i\omega t) ~. $$ Then, the accelerations take the form

\ddot{\mathbf{u}} = (i\omega)^2~\mathbf{u}^0~\exp(i\omega t) = -\omega^2~\mathbf{u}^0~\exp(i\omega t) ~. $$ Plugging these into the FE system of equations, we get

[-\omega^2~\exp(i\omega t)]\mathbf{M}~\mathbf{u}^0 + \exp(i\omega t)~\mathbf{K}~\mathbf{u}^0 = \exp(i\omega t)~\mathbf{f}^0 ~. $$ After simplification, we get

{ \left(-\omega^2\mathbf{M} + \mathbf{K}\right)~\mathbf{u}^0 = \mathbf{f}^0 ~. } $$ If there is no forcing, the right hand side is zero, and we get the finite element system of equations for  free vibrations

{ \left(-\omega^2\mathbf{M} + \mathbf{K}\right)~\mathbf{u}^0 =0 ~. } $$ The above equation is similar to the  eigenvalue problem of the form

\mathbf{A}~\mathbf{x} = \lambda~\mathbf{x} \qquad \equiv \qquad \left(\mathbf{A} - \lambda\mathbf{I}\right)\mathbf{x} = 0 ~. $$ Since the right hand side is zero, the finite element system of equations has a solution only if

{ \det(\mathbf{K} -\omega^2\mathbf{M}) =0 ~. } $$ For a two noded element,

\mathbf{K} = \begin{bmatrix} K_{11} & K_{12} \\ K_{21} & K_{22} \end{bmatrix} ~\text{and}~ \mathbf{M} = \begin{bmatrix} M_{11} & M_{12} \\ M_{21} & M_{22} \end{bmatrix} ~. $$ Therefore,

\mathbf{K} - \omega^2~\mathbf{M} = \begin{bmatrix} K_{11} -\omega^2~M_{11} & K_{12} -\omega^2~M_{12} \\ K_{21} -\omega^2~M_{21} & K_{22} -\omega^2~M_{22} \end{bmatrix}~. $$ The determinant is

\det(\mathbf{K} - \omega^2~\mathbf{M}) = (K_{11} -\omega^2~M_{11})(K_{22} -\omega^2~M_{22}) - (K_{12} -\omega^2~M_{12})(K_{21} -\omega^2~M_{21})~. $$ This gives us a quadratic equation in $$\omega^2$$ which can be solved to find the natural frequencies of the element.