Question

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.

3 k, 3 m, 2 k, 4 m, 2 k

Short answer

The characteristic frequencies and the characteristic modes of vibration for system of masses and springs are:

Characteristic frequency${\omega }_1=\sqrt{\frac{2k}{m}}$ and mode of vibrations x=-2y or $\leftarrow \to$, and $\to \leftarrow$.

Characteristic frequency ${\omega }_2=\sqrt{\frac{2k}{3m}}$and mode of vibrations 3x=2y or $\to \to$, and. $\leftarrow \leftarrow$.

Step-by-step solution

Given information

The masses$\left(m_1=3m,m_2=4m\right)$and spring system$\left(k_1=3k,k_1=2k,k_1=2k\right)$is as shown in figure here.

Potential energy of the spring

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 m attached to the spring and displacement from the origin is given by $\mathit{m}\stackrel{\mathbf{,}\mathbf{,}}{\mathbf{x}}\mathbf{=}\mathbf{-}\frac{\mathbf{\partial }\mathbf{V}}{\mathbf{\partial }\mathbf{x}}$.

Total potential energy of the masses- springs system

A mass- springs system of masses $m_1=3m_1{m}_2=4m,k_1=3k,k_2=2k$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

$$\begin{gathered} V=\frac{1}{2}{k}_1{x}^2+\frac{1}{2}{k}_2{\left(x-y\right)}^2+\frac{1}{2}{k}_3{y}^2 \\ \mathrm{Or} \\ \mathrm{V}=\frac{1}{2}3{\mathrm{kx}}^2+\frac{1}{2}2\mathrm{k}{\left(\mathrm{x}-\mathrm{y}\right)}^2+\frac{1}{2}2{\mathrm{ky}}^2 \\ =\mathrm{k}\left\{\frac{3}{2}{\mathrm{x}}^2+{\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{xy}+{\mathrm{y}}^2\right\} \\ \mathrm{V}=\mathrm{k}\left\{\frac{5}{2}{\mathrm{x}}^2-2\mathrm{xy}+{\mathrm{y}}^2\right\} \\ \mathrm{Solve}\mathrm{further} \\ \Rightarrow \frac{\partial \mathrm{V}}{\partial \mathrm{x}}=\mathrm{k}\left\{5\mathrm{x}-2\mathrm{y}\right\},\mathrm{and}\frac{\partial \mathrm{V}}{\partial \mathrm{x}}=\mathrm{k}\left\{-2\mathrm{x}+4\mathrm{y}\right\}......\left(1\right) \end{gathered}$$

Equation of motion for masses

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

$$\begin{gathered} m_1\stackrel{,,}{x}=-\frac{\partial V}{\partial x} \\ =-k\left\{5x-2y\right\} \\ \Rightarrow -3m{\omega }^{2}x \\ =-k\left\{5x-2y\right\} \\ \mathrm{Solve}\mathrm{further} \\ \because \stackrel{,,}{\mathrm{x}}=-{\mathrm{\omega }}^2\mathrm{x}(\mathrm{For}\mathrm{SHM}\mathrm{of}\mathrm{mass}) \\ \left(\frac{3{\mathrm{m\omega }}^2}{\mathrm{k}}\right)\mathrm{x}=5\mathrm{x}-2\mathrm{y} \\ \left(\frac{{\mathrm{m\omega }}^2}{\mathrm{k}}\right)\mathrm{x}=\frac{5}{3}\times =\frac{2}{3}\mathrm{y}......\left(2\right) \\ \mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{motion}\mathrm{for}\mathrm{masses}{\mathrm{m}}_2=4\mathrm{m}\mathrm{ism} \\ {\mathrm{m}}_2\stackrel{,,}{\mathrm{y}}=-\frac{\partial \mathrm{V}}{\partial \mathrm{y}} \\ =-\mathrm{k}\left\{-2\mathrm{x}+4\mathrm{y}\right\}(\mathrm{For}\mathrm{SHM}\mathrm{of}\mathrm{mass}\mathrm{m}) \\ -4{\mathrm{m\omega }}^2\mathrm{y}=-\mathrm{k}\left\{-2\mathrm{x}+4\mathrm{y}\right\} \\ \because \stackrel{,,}{\mathrm{y}}=-{\mathrm{\omega }}^2\mathrm{y} \\ \mathrm{Solve}\mathrm{further} \\ \left(\frac{4{\mathrm{m\omega }}^2}{\mathrm{k}}\right)\mathrm{y}=-2\mathrm{x}+4\mathrm{y} \\ \left(\frac{{\mathrm{m\omega }}^2}{\mathrm{k}}\right)\mathrm{y}=-\frac{1}{2}\mathrm{x}+\mathrm{y}......\left(3\right) \end{gathered}$$

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

$\left(\frac{m{\omega }^2}{k}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{cc}5/3& -2/3\\ -1/2& 1\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)$

If $\frac{m{\omega }^2}{k}=\lambda$is the eigenvalue of coefficient matrix, then

$\lambda \left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{cc}5/3& -2/3\\ -1/2& 1\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)...\left(4\right)$

Eigen values and eigen vectors of matrix

Now, determine eigenvalues and eigenvectors of coefficient matrix

$A=\left(\begin{array}{cc}5/3& -2/3\\ -1/2& 1\end{array}\right)$

The characteristic matrix of matrix A is

$$\begin{gathered} A-\lambda I=\left(\begin{array}{cc}5/3& -2/3\\ -1/2& 1\end{array}\right)-\lambda \left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right) \\ =\left(\begin{array}{cc}\frac{5}{3}-\lambda & -\frac{2}{3}\\ -\frac{1}{2}& 1-\lambda \end{array}\right) \end{gathered}$$

Hence, the characteristic equation of matrix Ais

$$\begin{gathered} \left|A-\lambda I\right|=0 \\ \left(\begin{array}{cc}\frac{5}{3}-\lambda & -\frac{2}{3}\\ -\frac{1}{2}& 1-\lambda \end{array}\right)=3{\lambda }^2-8\lambda +4 \\ \left(\lambda -2\right)\left(3\lambda -2\right)=0 \\ \lambda -2,\lambda =\frac{2}{3} \end{gathered}$$

Thus, the eigenvalues of matrix A are $\lambda -2,\mathrm{and}\lambda =\frac{2}{3}$

The eigenvector corresponding to eigenvalue $\lambda =2$is given by $\left(A-\lambda I\right)X=0$

$$\begin{gathered} \left(\mathrm{A}-2\mathrm{I}\right)X=O(\mathrm{Null}\mathrm{matrix}) \\ \left(\begin{array}{cc}5/3& -2/3\\ -1/2& 1\end{array}\right)-2\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\left(\begin{array}{c}\mathrm{x}\\ \mathrm{y}\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right) \\ \left(\begin{array}{cc}-1/3& -2/3\\ -1/2& 1\end{array}\right)\left(\begin{array}{c}\mathrm{x}\\ \mathrm{y}\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right) \\ \left(\begin{array}{cc}-\frac{1}{3}\mathrm{x}-& \frac{2}{3}\mathrm{y}\\ -\frac{1}{2}\mathrm{x}& -\mathrm{y}\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right) \end{gathered}$$

Then,

$$\begin{gathered} \Rightarrow -\frac{1}{3}x-\frac{2}{3}y=0 \\ \Rightarrow x=-2y,-\frac{1}{2}x-y=0 \\ \Rightarrow x=-2y \end{gathered}$$

The eigenvector corresponding to eigenvalue $\lambda =2/3$is given by $\left(A-\lambda I\right)X=0$

$$\begin{gathered} \left(A-\frac{2}{3}I\right)X=O\left(\mathrm{null}\mathrm{matrix}\right) \\ \left(\begin{array}{cc}5/3& -2/3\\ -1/2& 1\end{array}\right)-\frac{2}{3}\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right) \\ \left(\begin{array}{cc}1& -2/3\\ -1/2& 1/3\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right) \\ \left(\begin{array}{c}x-\frac{2}{3}y\\ -\frac{1}{2}x+\frac{1}{3}y\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right) \\ \mathrm{Then},\begin{array}{}\end{array} \\ \Rightarrow \mathrm{x}-\frac{2}{3}\mathrm{y}=0 \\ \Rightarrow 3\mathrm{x}=2\mathrm{y},-\frac{1}{2}\mathrm{x}+\frac{1}{3}\mathrm{y}=0 \\ \Rightarrow 3\mathrm{x}=2\mathrm{y}. \end{gathered}$$

Interpret the above results

Interpret the above results as follows:

When

$$\begin{gathered} \lambda =2 \\ \Rightarrow \frac{{m_{{\omega }_1}}^2}{k}=2 \\ \Rightarrow {\omega }_1=\sqrt{\frac{2k}{m}} \end{gathered}$$

and eigenvector x=-2y that means both the masses will oscillate with characteristic frequency ${\omega }_1=\sqrt{\frac{2k}{m}}$, back and forth together in same direction (as

$x=-2y)\mathrm{like}\leftarrow \to \mathrm{and}\to \leftarrow$

When

$$\begin{gathered} \lambda =\frac{2}{3} \\ \Rightarrow \frac{{m_{{\omega }_2}}^2}{k}=\frac{2}{3} \\ \Rightarrow {\omega }_2=\sqrt{\frac{2k}{3m}} \end{gathered}$$

and eigenvector 3x=2y that means both the masses will oscillate with characteristic frequency ${\omega }_2=\sqrt{\frac{2k}{3m}}$, back and forth together in opposite directions (as 3x=2y) Like $\to \to$and $\leftarrow \leftarrow$.