91影视

The masses$\left(m_1=m_1{m}_2=m\right)$and spring system$\left(k_1=5k,k_2=2k,k_3=2k\right)$is as shown in figure here.

The potential energy of the spring at compression or extension x(say) is given by$\mathit{V}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{k}{\mathit{x}}^{\mathbf{2}}$( k is spring constant). The equation of motion of mass mattached to the spring and displacement from the origin is given by$\mathit{m}\stackrel{\mathbf{,}\mathbf{,}}{\mathbf{x}}\mathbf{=}\mathbf{-}\frac{\mathbf{鈭�}\mathbf{V}}{\mathbf{鈭�}\mathbf{x}}$.

A mass- springs system of masses $m_1=m_{,}{m}_2=m,k_1=k\mathrm{and}{k}_3=k$. Suppose at any time x and y are the displacements (extension) from the respective mean positions. The extension or compression of the middle spring is (x-y). Hence, the total potential energy of the masses- springs system is

$$\begin{gathered} V=\frac{1}{2}{k}_1{x}^2+\frac{1}{2}{k}_2{\left(x-y\right)}^2+\frac{1}{2}{k}_3{y}^2 \\ =\frac{1}{2}5k{x}^2+\frac{1}{2}2k{\left(x-y\right)}^2+\frac{1}{2}2k{y}^2 \\ Or \\ V=\frac{1}{2}k\left\{5x^2+2{\left(x-y\right)}^2+2y^2\right\} \\ =\frac{1}{2}k\left\{5x^2+2x^{2}x-2y^2-4xy+2y^2\right\} \\ =\frac{1}{2}k\left\{7x^2+4y^2-4xy\right\}...\left(1\right) \\ 鈬�\frac{鈭�V}{鈭�x}=\frac{k}{2}\left\{14x-4y\right\} \\ \mathrm{Solve}\mathrm{further} \\ =7\mathrm{kx}-2\mathrm{ky}. \\ \frac{鈭�\mathrm{V}}{鈭�\mathrm{x}}=\frac{\mathrm{k}}{2}\left\{8\mathrm{y}-4\mathrm{x}\right\} \\ =-2\mathrm{kx}+4\mathrm{ky} \end{gathered}$$

Therefore, equation of motion for masses is$m_1=m$

$$\begin{gathered} m_1\stackrel{,,}{x}=-\frac{鈭�V}{鈭�x} \\ =-\left\{7kx-2ky\right\} \\ -m{蝇}^{2}x=-\left\{7kx-2ky\right\} \\ 鈭�\stackrel{,,}{x}=-{蝇}^{2}x \end{gathered}$$

(For SHM of mass m)

role="math" localid="1659178645408" $\left(\frac{m{蝇}^2}{k}\right)x=7x-2y...\left(2\right)$

The equation of motion for masses$m_2=m$is

$$\begin{gathered} m_2\stackrel{,,}{y}=-\frac{鈭�V}{鈭�y} \\ =-\left\{-2kx+4ky\right\} \\ -m{蝇}^{2}y=-\left\{-2kx+4ky\right\} \\ 鈭�\stackrel{,,}{x}=-{蝇}^{2}x \end{gathered}$$

(For SHM of mass m)

$\left(\frac{m{蝇}^2}{k}\right)y=-2x-4y...\left(3\right)$

In matrix form combined equation (combination of equation (2) and (3)) of masses is

$\left(\frac{m{蝇}^2}{k}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{cc}7& -2\\ -2& 4\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)$

If$\frac{m{蝇}^2}{k}=位$ is the eigenvalue of coefficient matrix, then$位\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{cc}7& -2\\ -2& 4\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)...\left(4\right)$

Now, determine eigenvalues and eigenvectors of coefficient matrix$A=\left(\begin{array}{cc}7& -2\\ -2& 4\end{array}\right)$

The characteristic matrix of matrix is

$$\begin{gathered} A-位I=\left(\begin{array}{cc}7& -2\\ -2& 4\end{array}\right)-位\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right) \\ =\left(\begin{array}{cc}7-位& -2\\ -2& 4-位\end{array}\right) \end{gathered}$$

Hence, the characteristic equation of matrix A is

$$\begin{gathered} \left|A-位I\right|=0鈬�\left|\begin{array}{cc}7-位& -2\\ -2& 4-位\end{array}\right| \\ ={位}^2-11位+24 \\ =\left({位}^2-3\right)\left({位}^2-8\right)=0 \\ 鈬�位=3,位=8 \end{gathered}$$width="281">A-位I=0鈬�7-位-2-24-位=位2-11位+24=位2-3位-8=0鈬�位=3,位=8A-位I=0鈬�7-位-2-24-位=位2-11位+24=位2-3位-8=0鈬�位=3,位=8

Thus, the eigenvalues of matrix A are $位=3$and $位=8$.

The eigenvector corresponding to eigenvalue$位=3$is given by

localid="1659179997644" $$\begin{gathered} \left(A-3I\right)X_1=O\left(\mathrm{null}\mathrm{matrix}\right) \\ 鈬�\left(\begin{array}{cc}7& -2\\ -2& 4\end{array}\right)-3\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right) \\ =\left(\begin{array}{cc}4& -2\\ -2& 1\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right) \\ =\left(\begin{array}{cc}4x& -2y\\ -2x& y\end{array}\right) \\ =\left(\begin{array}{c}0\\ 0\end{array}\right) \\ 鈬�4x-2y=0 \\ 鈬�2x=y \end{gathered}$$

The eigenvector corresponding to eigenvalue$位=8$is given by role="math" localid="1659180315585" $\left(A-8I\right)X_1=O$(null matrix)

$$\begin{gathered} \left(\begin{array}{cc}7& -2\\ -2& 4\end{array}\right)-8\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{cc}-1& -2\\ -2& -4\end{array}\right)=\left(\begin{array}{c}x\\ y\end{array}\right) \\ \left(\begin{array}{cc}-x& -2y\\ -2x& -4y\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right) \\ 鈬�-x-2y=0 \\ 鈬�x=-2y, \end{gathered}$$

Interpret the above results as follows:

When

$$\begin{gathered} 位=3 \\ 鈬�\frac{m{蝇}_1^2}{k}=3 \\ 鈬�{蝇}_1=\sqrt{\frac{3k}{m}} \end{gathered}$$

and eigenvector$2x=y$ that means both the masses will oscillate with characteristic frequency ${蝇}_1=\sqrt{\frac{3k}{m}}$, back and forth together in same direction (as 2x=y ) like$鈫�鈫�$ and $鈫�鈫�$.

When

$$\begin{gathered} 位=8 \\ 鈬�\frac{m{蝇}_2^2}{k}=8 \\ 鈬�{蝇}_2=\sqrt{\frac{8k}{m}} \end{gathered}$$

and eigenvector x=-2y that means both the masses will oscillate with characteristic frequency ${蝇}_2=\sqrt{\frac{8k}{m}}$, back and forth together in opposite directions ( as x=-2y like $鈫�鈫�$and $鈫�鈫�$.