91影视

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

The potential energy of the spring at compression or extension x(say) is given by$\mathit{v}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{k}{\mathit{x}}^{\mathbf{2}}$(kis 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 masses-springs system of masses,$m_1=m,m_2=m,k_1=k,\mathrm{and}{k}_3=k$and . 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

$$\begin{gathered} V=\frac{1}{2}{k}_1{x}^2+\frac{1}{2}{k}_2{\left(x-y\right)}^2+\frac{1}{2}{k}_3{y}^2 \\ =\frac{1}{2}k{x}^2+\frac{1}{2}k{\left(x-y\right)}^2+\frac{1}{2}k{y}^2 \end{gathered}$$

Or

$$\begin{gathered} V=\frac{1}{2}k\left\{x^2+2{\left(x-y\right)}^2+y^2\right\} \\ \frac{鈭�v}{鈭�x}=\frac{k}{2}\left\{2x+4x-4y\right\} \\ =3kx-2ky, \\ \frac{鈭�v}{鈭�x}=\frac{k}{2}\left\{4x+4y-2y\right\} \\ \mathrm{Solve}\mathrm{further} \\ =2\mathrm{kx}=\mathrm{ky} \end{gathered}$$

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

localid="1659174372069" $$\begin{gathered} m_1\stackrel{,,}{x}=-\frac{鈭�v}{鈭�x} \\ =-\left\{3kx-2ky\right\} \\ 鈬�=m{蝇}^{2}x \\ =-\left\{3kx-2ky\right\}鈭�\stackrel{,,}{x}=-{蝇}^{2}x \\ \left(\frac{m_{{蝇}^2}}{k}\right)x=3x-2y...\left(1\right) \end{gathered}$$ (For SHM of mass )

The equation of motion for masses $m_1=m$and $m_2=m$are

localid="1659174630193" $$\begin{gathered} m_2\stackrel{,,}{y}=-\frac{鈭�V}{鈭�y} \\ =-\left\{2kx-ky\right\} \\ -m{蝇}^{2}y=-\left\{2kx-ky\right\} \\ 鈭�\stackrel{,,}{x}=-{蝇}^2脳\left(\mathrm{For}\mathrm{SHM}\mathrm{of}\mathrm{mass}\mathrm{m}\right) \end{gathered}$$

solve further

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

In matrix form combined equation (combination of equation (1) and (2)) of masses is

$(\frac{m{蝇}^2}{k}\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{cc}3& -2\\ 2& -1\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right).$

If $\frac{m{蝇}^2}{k}=位$is the eigenvalue of coefficient matrix, then$位\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{cc}3& -2\\ 2& -1\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)...\left(3\right)$

Now, determine eigenvalues and eigenvectors of coefficient matrix $A=\left(\begin{array}{cc}3& -2\\ 2& -1\end{array}\right)$.

The characteristic matrix of matrix A is $$\begin{gathered} A-位I=\left(\begin{array}{cc}3& -2\\ 2& -1\end{array}\right)-位\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right) \\ =\left(\begin{array}{cc}3-位& -2\\ 2& -1-位\end{array}\right) \end{gathered}$$.

Hence, the characteristic equation of matrix A is

$$\begin{gathered} \left|A-位I\right|=0 \\ \left|\begin{array}{cc}3-位& -2\\ -2& 3-位\end{array}\right|={位}^2-6位+5 \\ \left(位-1\right)\left(位-5\right)=0 \\ 位=1,位=5 \end{gathered}$$

Thus, the eigenvalues of matrix A are $位=1$and $位=5$.

The eigenvector corresponding to eigenvalue $位=1$is given by

localid="1659176461901" $$\begin{gathered} \left(\begin{array}{cc}3-位& -2\\ -2& 3-位\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{cc}3-1& -2\\ -2& 3-1\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right) \\ =\left(\begin{array}{cc}2& -2\\ -2& 2\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right) \\ =\left(\begin{array}{c}0\\ 0\end{array}\right) \\ \mathrm{Solve}\mathrm{further} \\ 鈬�2\mathrm{x}-2\mathrm{y}=0 \\ -2\mathrm{x}+2\mathrm{y}=0 \\ 鈬�\mathrm{x}=\mathrm{y} \\ \mathrm{The}\mathrm{eigenvector}\mathrm{corresponding}\mathrm{to}\mathrm{eigenvalue}\mathrm{位}=5\mathrm{is}\mathrm{given}\mathrm{by} \\ \left(\mathrm{A}-\mathrm{位滨}\right){\mathrm{x}}_1=\mathrm{o}(\mathrm{Null}\mathrm{matrix}) \\ \left(\begin{array}{cc}3-位& -2\\ -2& 3-位\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{cc}3-5& -2\\ -2& 3-5\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right) \\ \left(\begin{array}{cc}-2& -2\\ -2& -2\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right) \\ \begin{array}{cc}-2x& -2y=0\\ -2x& -2y=0\\ & \end{array} \\ \mathrm{Solve}\mathrm{further} \\ 鈬�\mathrm{x}=-\mathrm{y} \end{gathered}$$

Interpret the above results as follows:

$位=1$

When$鈬�\frac{m{蝇}_1^2}{k}=1$and eigenvector x=y that means both the masses will oscillate $鈬�{蝇}_1=\sqrt{\frac{k}{m}}$with characteristic frequency ${蝇}_1=\sqrt{\frac{k}{m}}$, back and forth together in same direction (as x=y) like$鈫�鈫�$ and $鈫�鈫�$.

When

$$\begin{gathered} 位=5 \\ 鈬�\frac{m{蝇}_2^2}{k}=5 \\ 鈬�{蝇}_2=\sqrt{\frac{5k}{m}} \end{gathered}$$

And eigenvector x = -y that means both the masses will oscillate with characteristic frequency, ${蝇}_2=\sqrt{\frac{5k}{m}}$back and forth together in opposite directions $($ as x = - y ) like $鈫�鈫�$and$鈫�鈫�$