- #1
Valeron21
- 8
- 0
The det. of the following matrix:
$$
\begin{matrix}
2k-ω^{2}m_{1} & -k\\ -k & k-ω^{2}m_{2}\\
\end{matrix}
$$
must be equal to 0 for there to be a non-trivial solution to the equation: $$(k - ω^{2}m)x =0$$
Where m is the mass matrix:
$$
\begin{matrix}
m_{1} & 0\\ 0& m_{2}\\
\end{matrix}
$$
k is the stiffness matrix:
$$
\begin{matrix}
2k& -k\\ -k & k\\
\end{matrix}
$$
and ω^2 is the eigenvalue.
I worked out the determinant as $$ω^{4}m_{1}m{2}-k(m_{1}+2m_{2})ω^{2}+k^{2}=0$$ and I could probably solve to find the two positive roots of this, but the roots are going to be pretty horrible and they are going to make finding the corresponding eigenvectors pretty difficult, no?
Am I doing something wrong? Is there an easier way to do this? Perhaps a substitution/alteration to the matrix?
$$
\begin{matrix}
2k-ω^{2}m_{1} & -k\\ -k & k-ω^{2}m_{2}\\
\end{matrix}
$$
must be equal to 0 for there to be a non-trivial solution to the equation: $$(k - ω^{2}m)x =0$$
Where m is the mass matrix:
$$
\begin{matrix}
m_{1} & 0\\ 0& m_{2}\\
\end{matrix}
$$
k is the stiffness matrix:
$$
\begin{matrix}
2k& -k\\ -k & k\\
\end{matrix}
$$
and ω^2 is the eigenvalue.
I worked out the determinant as $$ω^{4}m_{1}m{2}-k(m_{1}+2m_{2})ω^{2}+k^{2}=0$$ and I could probably solve to find the two positive roots of this, but the roots are going to be pretty horrible and they are going to make finding the corresponding eigenvectors pretty difficult, no?
Am I doing something wrong? Is there an easier way to do this? Perhaps a substitution/alteration to the matrix?