Find the characteristic frequencies and the characteristic modes of vibration for systems of masses and springs as in Figure 12.1 and Examples 3,4 and 6 for the following arrays.

k,m,2k,m,k

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$\mathit{m}\stackrel{\mathbf{,}\mathbf{,}}{\mathbf{x}}\mathbf{=}\mathbf{-}\frac{\mathbf{\partial }\mathbf{v}}{\mathbf{\partial }\mathbf{x}}$

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

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

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

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

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

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

solve further

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

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

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

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

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 $$\begin{gathered} A-\lambda I=\left(\begin{array}{cc}3& -2\\ 2& -1\end{array}\right)-\lambda \left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right) \\ =\left(\begin{array}{cc}3-\lambda & -2\\ 2& -1-\lambda \end{array}\right) \end{gathered}$$.

Hence, the characteristic equation of matrix A is

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

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

The eigenvector corresponding to eigenvalue $\lambda =1$is given by

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

Interpret the above results as follows:

$\lambda =1$

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

When

$$\begin{gathered} \lambda =5 \\ \Rightarrow \frac{m{\omega }_2^2}{k}=5 \\ \Rightarrow {\omega }_2=\sqrt{\frac{5k}{m}} \end{gathered}$$

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