Find the characteristics vibration frequencies of a system of masses and springs as in Figure 12.1 if the spring constants are $\mathit{k}\mathbf{,}\mathbf{3}\mathit{k}$ and$\mathit{k}$ .

For a matrix A , all possible values of $\mathit{\lambda }$which satisfies the equation $\left|A-\lambda I\right|\mathbf{=}\mathbf{0}$ , where I is the identity matrix, are considered as the eigenvalues of a matrix.

The figure with system of masses and springs is shown below:

Now, x-y is the compression or extension of the middle spring. So, the potential energy of the middle spring is given by $\frac{1}{2}\left(3k\right){\left(x-y\right)}^2$.

For the other two springs, the potential energy is given by $\frac{1}{2}\left(k\right){\left(x\right)}^2$and $\frac{1}{2}\left(k\right){\left(y\right)}^2$.

So, the total potential energy is $V=\frac{1}{2}k{x}^2+\frac{1}{2}\left(3k\right){\left(x-y\right)}^2+\frac{1}{2}k{y}^2$.

Here, V is a bi-variable function, so, the forces applied on two masses are $-\frac{\partial V}{\partial x}$and $-\frac{\partial V}{\partial y}$.

Substitute $x\text{'}\text{'}=\frac{d^{2}x}{d{t}^2}$and $x\text{'}=\frac{dx}{dt}$. Then the motions represented by $-\frac{\partial V}{\partial x}=-k\left(4x-3y\right)$and $-\frac{\partial V}{\partial y}=-k\left(-3x+4y\right)$.

Let $\omega$be the frequency for$x$and$y$. Then$x\text{'}\text{'}=-{\omega }^{2}x,y\text{'}\text{'}=-{\omega }^{2}y$. So, from above equations, we have$-m{\omega }^{2}x=-k\left(4x-3y\right)$and$-m{\omega }^{2}y=-k\left(-3x+4y\right)$.

Now, substitute $\lambda =\frac{-m{\omega }^2}{k}$. We get$\lambda x=4x-3y$and $\lambda y=\left(-3x+4y\right)$.

Now, write the above equation in the matrix form as $\left(\begin{array}{cc}4& -3\\ -3& 4\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\lambda \left(\begin{array}{c}x\\ y\end{array}\right)$. So, the characteristic equation becomes $\left|\begin{array}{cc}4-\lambda & -3\\ -3& 4-\lambda \end{array}\right|=0$. Solve this equation for :

$\begin{array}{rcl}\left(4-\lambda \right)\left(4-\lambda \right)-\left(-3\right)\left(-3\right)& =& 0\\ 16-8\lambda +{\lambda }^2-9& =& 0\\ {\lambda }^2-8\lambda +7& =& 0\\ \left(\lambda -7\right)\left(\lambda -1\right)& =& 0\end{array}$

So, the eigenvalues are 1 and 7.

Now, if $\begin{array}{rcl}\lambda & =& 1\end{array}$, then $\begin{array}{rcl}& & {\omega }_1\end{array}$is:

$\begin{array}{rcl}{\omega }_1& =& \sqrt{\frac{k\lambda }{m}}\\ & =& \sqrt{\frac{k}{m}}\end{array}$

Now, if $\begin{array}{rcl}\lambda & =& 7\end{array}$, then $\begin{array}{rcl}& & {\omega }_2\end{array}$is:

localid="1664371959358" $\begin{array}{rcl}{\omega }_2& =& \sqrt{\frac{k\lambda }{m}}\\ & =& \sqrt{\frac{7k}{m}}\end{array}$

Thus, the characteristic frequencies are$\begin{array}{rcl}{\omega }_1& =& \sqrt{\frac{k}{m}}\end{array}$and $\begin{array}{rcl}{\omega }_2& =& \sqrt{\frac{7k}{m}}\end{array}$.

Now, on substituting $\begin{array}{rcl}\lambda & =& 1\end{array}$in$\begin{array}{rcl}\lambda x& =& 4x-3y\end{array}$, we get$\begin{array}{rcl}x& =& y\end{array}$. Substituting $\begin{array}{rcl}\lambda & =& 7\end{array}$in $\begin{array}{rcl}\lambda y& =& \left(-3x+4y\right)\end{array}$, we get$\begin{array}{rcl}x& =& -y\end{array}$.

So, at frequency $\begin{array}{rcl}& & {\omega }_1\end{array}$, the two masses oscillate back and forth together like$\begin{array}{rcl}& \to & \to \end{array}$and then $\begin{array}{rcl}& \leftarrow & \leftarrow \end{array}$.

At frequency$\begin{array}{rcl}& & {\omega }_2\end{array}$, the two masses oscillate in opposite directions like $\begin{array}{rcl}& \leftarrow & \to \end{array}$ and then like$\begin{array}{rcl}& \to & \leftarrow \end{array}$.