91影视

For a matrix A , all possible values of $\mathit{位}$which satisfies the equation $\left|A-位I\right|\mathbf{=}\mathbf{0}$ , 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 $-\frac{鈭�V}{鈭�x}$and $-\frac{鈭�V}{鈭�y}$.

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

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

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

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

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

So, the eigenvalues are 1 and 7.

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

$\begin{array}{rcl}{蝇}_1& =& \sqrt{\frac{k位}{m}}\\ & =& \sqrt{\frac{k}{m}}\end{array}$

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

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

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

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

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

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