Question

Find the characteristic frequencies and the characteristic modes of vibration as in Example 7 for the following arrays of masses and springs, reading from top to bottom in a diagram like Figure 12.3.

3k,m,2k,m

Short answer

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

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

Characteristic frequency ${\omega }_2=\sqrt{\frac{6k}{m}}$and mode of vibrations x=-2y or ,$\leftarrow \to$ and $\to \leftarrow$.

Step-by-step solution

Given information

The masses $\left(m_1=2m,m_2=m\right)$and spring system $k_1=3k,k_2=2k$, is as shown in figure here.

Kinetic energy of the spring at compression or extension

The kinetic energy of the spring at compression or extension x(say) is given by,$\mathit{T}\mathbf{=}{\mathit{m}}_{\mathbf{1}}\stackrel{\mathbf{,}}{{\mathbf{x}}^{\mathbf{2}}}$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$\mathit{m}\stackrel{\mathbf{,}\mathbf{,}}{\mathbf{x}}\mathbf{=}\mathbf{-}\frac{\mathbf{\partial }\mathbf{V}}{\mathbf{\partial }\mathbf{x}}$.

Total kinetic energy of the masses- springs

A mass- springs system of masses $m_1=m,m_2=m$, and $k_1=k,k_2=2k$. Suppose at any time x and y are 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

$$\begin{gathered} T=\frac{1}{2}{m}_1\stackrel{,}{x^2}+\frac{1}{2}{m}_1\stackrel{,}{y^2} \\ =\frac{1}{2}m\stackrel{,}{x^2}+\frac{1}{2}m\stackrel{,}{y^2} \\ =\frac{1}{2}m\left(\stackrel{,}{x^2+y^2}\right).......\left(1\right) \end{gathered}$$

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

$T=\frac{1}{2}mr{\text{'}}^{T}r\text{'}$

Here

$\mathrm{T}=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right),\mathrm{r}\text{'}=\left(\begin{array}{c}\mathrm{x}\\ \mathrm{y}\end{array}\right),\mathrm{and}\mathrm{r}{\text{'}}^{\mathrm{T}}=\left(\mathrm{x}\mathrm{y}\right)$

The matrix of the system is $\mathrm{T}=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$.......(2)

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

Equation of motion for masses

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

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

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

$$\begin{gathered} m_2\stackrel{,,}{y}=-\frac{\partial V}{\partial y} \\ =-k\left\{-2x+2y\right\} \\ \stackrel{,,}{y}=-{\omega }^{2}y\left(\mathrm{For}\mathrm{SHM}\mathrm{of}\mathrm{mass}m\right) \\ -m{\omega }^{2}y=-k\left\{-2x+2y\right\}.....\left(5\right) \end{gathered}$$

Therefore,

In matrix form of combined equation from equations, (4) and (5), we have

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

$$\begin{gathered} \left(\frac{m{\omega }^2}{k}\right)\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\left(\begin{array}{cc}5& -2\\ -2& 2\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\mathrm{Or} \\ \mathrm{\lambda Tr}=\mathrm{Vr},\mathrm{where}\frac{{\mathrm{m\omega }}^2}{\mathrm{k}}=\mathrm{\lambda } \\ \mathrm{and}\left(\begin{array}{cc}5& -2\\ -2& 2\end{array}\right)=\mathrm{V}......\left(6\right) \end{gathered}$$

, where

and

The equation (6) may be written as

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

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

Thus, we have $\mathrm{M}={\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)}^{-1}\left(\begin{array}{cc}5& -2\\ -2& 2\end{array}\right)=\left(\begin{array}{cc}5& -2\\ -2& 2\end{array}\right)$.

Determine eigenvalues and eigenvectors of coefficient matrix

Now, determine eigenvalues and eigenvectors of coefficient matrix $M=\left(\begin{array}{cc}5& -2\\ -2& 2\end{array}\right)$.

The characteristic matrix of matrix M is

$$\begin{gathered} \mathrm{M}-\mathrm{\lambda I}=\left(\begin{array}{cc}5& -2\\ -2& 2\end{array}\right)-\mathrm{\lambda }\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right) \\ =\left(\begin{array}{cc}5-\mathrm{\lambda }& -2\\ -2& 2-\mathrm{\lambda }\end{array}\right) \end{gathered}$$

Hence, the characteristic equation of matrix Ais

$$\begin{gathered} \left|\mathrm{M}-\mathrm{\lambda I}\right|=0 \\ \left(\begin{array}{cc}5-\lambda & -2\\ -2& 2-\lambda \end{array}\right)={\lambda }^2-7\lambda +6 \\ \left(\lambda -1\right)\left(\lambda -6\right)=0 \\ \lambda -1,\lambda =6. \end{gathered}$$

Thus, the eigenvalues of matrix M are localid="1659334192643" $\lambda -1$and $\lambda =6.$.

The eigenvector X corresponding to eigenvalue $\lambda =1$is given by (M- $\lambda$I) X=0

localid="1659333722020" $$\begin{gathered} \left(\mathrm{M}-1.\mathrm{I}\right)X=\mathrm{O} \\ \left(\begin{array}{cc}5& -2\\ -2& 2\end{array}\right)-1\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\left(\begin{array}{c}\mathrm{x}\\ \mathrm{y}\end{array}\right)=\left(\begin{array}{cc}4& -2\\ -2& 1\end{array}\right)\left(\begin{array}{c}\mathrm{x}\\ \mathrm{y}\end{array}\right) \\ \left(\begin{array}{cc}4\mathrm{x}& -2\mathrm{y}\\ -2\mathrm{x}& +\mathrm{y}\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right) \\ 4\mathrm{x}-2\mathrm{y}=0 \\ \mathrm{Solve}\mathrm{further} \\ \Rightarrow 2\mathrm{x}=\mathrm{y},-2\mathrm{x}+\mathrm{y}=0 \\ \Rightarrow 2\mathrm{x}=\mathrm{y} \end{gathered}$$

Solve further

The eigenvector corresponding to eigenvalue$\lambda =7$is given by (M- $\lambda \mathrm{I})\mathrm{X}=0$

$$\begin{gathered} \left(\mathrm{M}-7\mathrm{I}\right)\mathrm{X}=\mathrm{O} \\ \left(\begin{array}{cc}5& -2\\ -2& 2\end{array}\right)-6\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\left(\begin{array}{c}\mathrm{x}\\ \mathrm{y}\end{array}\right)=\left(\begin{array}{cc}-1& -2\\ -2& -4\end{array}\right)\left(\begin{array}{c}\mathrm{x}\\ \mathrm{y}\end{array}\right) \\ \left(\begin{array}{cc}-\mathrm{x}& -2\mathrm{y}\\ -2\mathrm{x}& -4\mathrm{y}\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right) \\ -\mathrm{x}-2\mathrm{y}=0 \\ \mathrm{Solve}\mathrm{further} \\ \Rightarrow \mathrm{x}=-2\mathrm{y},-2\mathrm{x}+4\mathrm{y}=0 \\ \Rightarrow \mathrm{x}=-2\mathrm{y} \end{gathered}$$

Interpret the above results

Interpret the above results as follows:

When

$$\begin{gathered} \lambda =1 \\ \Rightarrow \frac{{m_{{\omega }_1}}^2}{k}=1 \\ \Rightarrow {\omega }_1=\sqrt{\frac{k}{m}}=\sqrt{\frac{k}{m}} \end{gathered}$$

and eigenvector 2x=y that means both the masses will oscillate with characteristic frequency role="math" localid="1659334062216" ${\omega }_1=\sqrt{\frac{k}{m}}$, back and forth together in the same directions (as 2x=y ) like$\to \to$and $\leftarrow \leftarrow$.

When

$$\begin{gathered} \lambda =6 \\ \Rightarrow \frac{{m_{{\omega }_2}}^2}{k}=6 \\ \Rightarrow {\omega }_2=\sqrt{\frac{6k}{2m}} \end{gathered}$$

and eigenvector x=-2y that means both the masses will oscillate with characteristic frequency ${\omega }_2=\sqrt{\frac{6k}{m}}$, back and forth together in the opposite directions (as x=-2y ) like,$\leftarrow \to$and$\to \leftarrow$