91影视

The masses $\left(m_1=4m,m_2=2m\right)$and spring system $\left(k_1=2k,k_2=k\right)$is as shown in figure here.

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

A mass- springs system of masses $\left(m_1=4m,m_2=2m\right)$, and $\left(k_1=2k,k2=k\right)$. Suppose at any time x and yare the displacements (extension) of masses from their mean positions. Then extension or compression of the lowest spring is(x - y).

Hence, the total kinetic energy of the masses- springs system is

$\begin{array}{l}T=\frac{1}{2}{m}_1{\stackrel{藱}{x}}^2+\frac{1}{2}{m}_2{\stackrel{藱}{y}}^2\\ =\frac{1}{2}4m{\stackrel{藱}{x}}^{{}^2}+\frac{1}{2}2m{\stackrel{藱}{y}}^2\\ =\frac{1}{1}m\left(4{\stackrel{藱}{x}}^2+2{\stackrel{藱}{y}}^2\right).....\left(1\right)\end{array}$

The kinetic energy$T=\frac{1}{2}m\left(4\stackrel{藱}{x}+2{\stackrel{藱}{y}}^2\right)$ of the system may be written as

$T=\frac{1}{2}m{\stackrel{藱}{r}}^{T}T\stackrel{藱}{r}$,

Here

$T=\left(\begin{array}{cc}4& 0\\ 0& 2\end{array}\right),\stackrel{藱}{r}=\left(\begin{array}{c}x\\ y\end{array}\right)\mathrm{and}{\stackrel{藱}{r}}^T=\left(x\mathit{}\mathit{}\mathit{}\mathit{}y\right)$.

The matrix of the system is$T=\left(\begin{array}{cc}4& 0\\ 0& 2\end{array}\right)........\left(2\right)$

The total potential energy of the masses- springs system is

$$\begin{gathered} V=\frac{1}{2}{k}_1{x}^2+\frac{1}{2}{k}_2{\left(x-y\right)}^2 \\ =\frac{1}{2}2k{x}^2+\frac{1}{2}k{\left(x-y\right)}^2 \\ =\frac{k}{2}\left\{2x^2+{\left(x-y\right)}^2\right\}\mathrm{Or} \\ \mathrm{V}=\frac{\mathrm{k}}{2}\left\{3{\mathrm{x}}^2-2\mathrm{xy}+{\mathrm{y}}^2\right\} \end{gathered}$$

Solve further

$$\begin{gathered} \frac{鈭�V}{鈭�x}=k\{3x-y\},\mathrm{and} \\ \frac{鈭�\mathrm{V}}{鈭�\mathrm{y}}=\mathrm{k}\{-\mathrm{x}-\mathrm{y}\}......\left(3\right) \end{gathered}$$

Therefore, equation of motion for masses$m_1=3m$is

$$\begin{gathered} m_1\stackrel{篓}{x}=-\frac{鈭�V}{鈭�x} \\ =-k\{3x-y\} \\ \stackrel{篓}{x}=-{蝇}^{2}x\left(\mathrm{For}\mathrm{SMH}\mathrm{of}\mathrm{mass}\mathrm{m}\right) \\ -4m{蝇}^{2}x=-k\{3x-y\}.......\left(4\right) \end{gathered}$$

Solve further

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

$$\begin{gathered} m_2\stackrel{篓}{y}=-\frac{鈭�V}{鈭�y} \\ =-k\{-x+y\} \\ \stackrel{篓}{y}=-{蝇}^{2}y\left(\mathrm{For}\mathrm{SMH}\mathrm{of}\mathrm{mass}m\right) \\ -2m{蝇}^{2}y=-k\{-x+y\}......\left(5\right) \end{gathered}$$

Therefore, in matrix form of combined equation from equations (2), (4) and (5), we have

(As left had sides of equations (4) and (5) are kinetic energies of masses-springs system.)

$\left(\frac{m{蝇}^2}{k}\right)\left(\left.\begin{array}{c}\begin{array}{cc}4& 0\\ 0& 2\end{array}\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{cc}3& -1\\ -1& 1\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right.\right)$

,$位Tr=\mathrm{V}r$wherelocalid="1659147442006" $\frac{m{蝇}^2}{k}=位$

and$\left(\begin{array}{cc}3& -1\\ -1& 1\end{array}\right)=\mathrm{V}....\left(6\right)$

The equation (6) may be written as

$位r=\left({\mathrm{T}}^{-1}\mathrm{V}\right)r=\mathrm{M}r......\left(7\right)$

Clearly, $\frac{m{蝇}^2}{k}=位$are the eigenvalues of matrix M, we determine here.

The matrix

$$\begin{gathered} M={\left(\begin{array}{cc}4& 0\\ 0& 2\end{array}\right)}^{-1}\left(\begin{array}{cc}3& -1\\ -1& 1\end{array}\right) \\ =\frac{1}{8}\left(\begin{array}{cc}2& 0\\ 0& 4\end{array}\right)\left(\begin{array}{cc}3& -1\\ -1& 1\end{array}\right) \\ =\frac{1}{8}\left(\begin{array}{cc}6& -2\\ -4& 4\end{array}\right) \\ =\left(\begin{array}{ccc}3/4& & -1/4\\ -1/2& & 1/2\end{array}\right) \end{gathered}$$.

Now, determine eigenvalues and eigenvectors of coefficient matrix $M=\left(\begin{array}{cc}3/4& -1/4\\ -1/2& 1/2\end{array}\right)$.

The characteristic matrix of matrix M is

$$\begin{gathered} M-位=\left(\begin{array}{cc}3/4& -1/4\\ -1/2& 1/2\end{array}\right)-位\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right) \\ =\left(\begin{array}{cc}\frac{3}{4}-位& -\frac{1}{4}\\ -\frac{1}{2}& \frac{1}{2}-位\end{array}\right) \end{gathered}$$.

Hence, the characteristic equation of matrix A is

$$\begin{gathered} \left|\mathrm{M}-\mathrm{位濒}\right|=0 \\ 鈬�\left|\begin{array}{cc}\frac{3}{4}-\mathrm{位}& -\frac{1}{4}\\ -\frac{1}{2}& \frac{1}{2}-\mathrm{位}\end{array}\right| \\ =\frac{8{\mathrm{位}}^2-10\mathrm{位}+2}{8}=0 \\ 鈬�\left(\mathrm{位}-1\right)\left(8\mathrm{位}-2\right)=0 \end{gathered}$$

Solve further

$鈬�\mathrm{位}=1,位=\frac{1}{4}路$

Thus, the eigenvalues of matrix M are $位=1$and $位=\frac{1}{4}$.

The eigenvector X corresponding to eigenvalue $位=1$is given by$\left(M-位\mathrm{l}\right)X=0$

$$\begin{gathered} \left(\mathrm{M}-1\mathrm{l}\right)\mathrm{X}=0 \\ \left(\begin{array}{cc}\frac{3}{4}& -\frac{1}{4}\\ -\frac{1}{2}& \frac{1}{2}\end{array}\right)-1\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\left(\begin{array}{c}\mathrm{x}\\ \mathrm{y}\end{array}\right)=\left(\begin{array}{cc}-\frac{1}{4}& -\frac{1}{4}\\ -\frac{1}{2}& -\frac{1}{2}\end{array}\right)\left(\begin{array}{c}\mathrm{x}\\ \mathrm{y}\end{array}\right) \\ \left(\begin{array}{cc}-\frac{1}{4}\mathrm{x}& -\frac{1}{4}\mathrm{y}\\ -\frac{1}{2}\mathrm{x}& -\frac{1}{2}\mathrm{y}\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right) \\ -\frac{1}{4}\mathrm{x}-\frac{1}{4}\mathrm{y}=0 \end{gathered}$$

Solve further

$\begin{array}{l}x=-y,-\frac{1}{2}x-\frac{1}{2}y=0\\ x=-y,\end{array}$

The eigenvector corresponding to eigenvalue $位=\frac{1}{4}$is given by$\left(M-位l\right)X=0$

$$\begin{gathered} \left(\mathrm{M}-\frac{1}{4}\mathrm{l}\right)\mathrm{X}=0 \\ \left(\begin{array}{cc}\frac{3}{4}& -\frac{1}{4}\\ -\frac{1}{2}& \frac{1}{2}\end{array}\right)-\frac{1}{4}\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\left(\begin{array}{c}\mathrm{x}\\ \mathrm{y}\end{array}\right)=\left(\begin{array}{cc}\frac{1}{2}& -\frac{1}{4}\\ -\frac{1}{2}& \frac{1}{4}\end{array}\right)\left(\begin{array}{c}\mathrm{x}\\ \mathrm{y}\end{array}\right) \\ \left(\begin{array}{cc}\frac{1}{2}\mathrm{x}& -\frac{1}{4}\mathrm{y}\\ -\frac{1}{2}\mathrm{x}& +\frac{1}{4}\mathrm{y}\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right) \\ -\frac{1}{2}\mathrm{x}-\frac{1}{4}\mathrm{y}=0 \end{gathered}$$

Solve further

$\begin{array}{l}鈬�2x=y,-\frac{1}{2}x+\frac{1}{4}y=0\\ 鈬�2x=y.\end{array}$

Interpret the above results as follows:

When

$$\begin{gathered} 位=1 \\ \frac{m{蝇}_1^2}{k}=1 \\ {蝇}_1=\sqrt{\frac{k}{m}} \end{gathered}$$

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

When

$$\begin{gathered} 位=\frac{1}{4} \\ 鈬�\frac{m{蝇}_2^2}{k}=\frac{1}{4} \\ 鈬�{蝇}_2=\sqrt{\frac{k}{4m}} \end{gathered}$$

and eigenvector 2x = y that means both the masses will oscillate with characteristic frequency ${蝇}_2=\sqrt{\frac{k}{4m}}$, back and forth together in same directions (as 2x = y) like,$鈫�鈫�$and $鈫�鈫�$.