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 .

For the other two springs, the potential energy is given by $\frac{1}{2}\left(6k\right){\left(x\right)}^2$and$\frac{1}{2}\left(3k\right){\left(y\right)}^2$ .

So, the total potential energy is $V=\frac{1}{2}\left(6k\right){\left(x\right)}^2+\frac{1}{2}\left(2k\right){\left(x-y\right)}^2+\frac{1}{2}\left(3k\right){\left(y\right)}^2$.

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 $\frac{-鈭�V}{鈭�x}=-k\left(8x-2y\right)$and $\frac{-鈭�V}{鈭�y}=-k\left(-2x+5y\right)$.

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 $-m{蝇}^{2}x=-k\left(8x-2y\right)$and $-m{蝇}^{2}y=-k\left(-2x+5y\right)$.

Now, substitute $位=\frac{m{蝇}^2}{k}$. We get $位x=8x-2y$and $位y=\left(-2x+5y\right)$.

Now, write the above equation in the matrix form as $\left(\begin{array}{cc}8& -2\\ -2& 5\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=位\left(\begin{array}{c}x\\ y\end{array}\right)$. So, the characteristic equation becomes$\left|\begin{array}{cc}8-位& -2\\ -2& 5-位\end{array}\right|=0$. Solve this equation for $位$:

localid="1664374045167" $\begin{array}{rcl}\left(8-位\right)\left(5-位\right)-\left(-2\right)\left(-2\right)& =& 0\\ 40-13位+{位}^2-4& =& 0\\ {位}^2-13位+36& =& 0\\ \left(位-4\right)\left(位-9\right)& =& 0\end{array}$

So, the eigenvalues are 4 and 9.

Now, if localid="1664374052831" $\begin{array}{rcl}位& =& 4\end{array}$, then:

localid="1664374061150" $\begin{array}{rcl}{蝇}_1& =& \sqrt{\frac{k位}{m}}\\ & =& \sqrt{\frac{4k}{m}}\end{array}$

Now, if localid="1664374068591" $\begin{array}{rcl}位& =& 9\end{array}$, then:

$\begin{array}{rcl}{蝇}_2& =& \sqrt{\frac{k位}{m}}\\ & =& \sqrt{\frac{9k}{m}}\end{array}$

Thus, the characteristic frequencies are

$\begin{array}{rcl}{蝇}_1& =& \sqrt{\frac{4位}{m}}\\ {蝇}_2& =& \sqrt{\frac{9k}{m}}\end{array}$

Now, on substituting$\begin{array}{rcl}位& =& 1\end{array}$in$\begin{array}{rcl}位x& =& 8x-2y\end{array}$, we get$\begin{array}{rcl}2x& =& y\end{array}$. Substituting $\begin{array}{rcl}位& =& 9\end{array}$in$\begin{array}{rcl}位y& =& \left(-2x+5y\right)\end{array}$, we get$\begin{array}{rcl}x& =& 2y\end{array}$.

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$\begin{array}{rcl}& 鈫�& 鈫�\end{array}$.