Vibration Analysis - 2

With $M = K = 1$ , the equations of motion for the four masses assuming their displacements from the original positions of unstretched springs are $x_1 , x_2 , x_3, x_4$ are:

$\ddot{x_1} = x_2 - x_1$

$\ddot{x_2} = -\dfrac{2}{3} (x_2 - x_1) + \dfrac{1}{3} (x_3 - x_2 )$

$\ddot{x_3} = -\dfrac{1}{5} (x_3 - x_2) + \dfrac{3}{5} (x_4 - x_3 )$

$\ddot{x_4} = -3 ( x_4 - x_3 )$

Thus if we define the state vector $\mathbf{x} = [ x_1, x_2, x_3, x_4 , \dot{x_1}, \dot{x_2}, \dot{x_3} , \dot{x_4 } ]$ , then the state equation is

$\mathbf{\dot{x}} = \mathbf{A x }$

where

$A = \begin{bmatrix} \mathbf{O_4} && \mathbf{I_4} \\ \mathbf{C} && \mathbf{O_4} \end{bmatrix}$

where $\mathbf{O_4}$ is the $4 \times 4$ zero matrix, and $\mathbf{I_4}$ is the $4 \times 4$ identity matrix, and

$\mathbf{C} = \displaystyle \begin{bmatrix} -1 && 1 && 0 && 0 \\ \frac{2}{3} && -1 && \frac{1}{3} && 0 \\ 0 && \frac{1}{5} && \frac{-4}{5} && \frac{3}{5} \\ 0 && 0 && 3 && -3 \end{bmatrix}$

The natural frequencies of this oscillatory system are the magnitudes of the imaginary eigenvalues of the above $8 \times 8$ matrix.

To compute the eigenvalues, we form the determinant equation $| \lambda \mathbf{I - A} | = 0$ and find its roots. Now

$| \lambda \mathbf{I - A} | = \begin{vmatrix} \lambda \mathbf{I_4} && \mathbf{- I_4} \\ \mathbf{ - C} && \lambda \mathbf{I_4} \end{vmatrix}$

Now we need the following theorem that $\begin{vmatrix} \mathbf{A_1} && \mathbf{A_2} \\ \mathbf{A_3} && \mathbf{A_4} \end{vmatrix} = | A_1 | | A_4 - A_2 A_1^{-1} A_3 |$

Applying this to our matrix, $| \lambda \mathbf{I - A} | = \lambda^4 | \lambda \mathbf{I_4} + \dfrac{1}{\lambda} (\mathbf{- C} ) | = | \lambda^2 \mathbf{I_4 - C} |$

So the eigenvalues are the square roots of the eigenvalues of matrix $C$ . At this point, we can compute the eigenvalues of matrix $\mathbf{C}$ using a suitable computer application such as wolframalpha.com , and they turn out to be:

$0,-3.640078328, -1.829582612, -0.330339059$

So they are all real and negative (except for the first one), which means the eigenvalues of the system will be either 0 or imaginary. Their magnitudes squared are the absolute value of the eigenvalues of C. The sum of these values is the negative of trace of matrix $\mathbf{C}$ . This comes to $\boxed{5.8}$ .