Question

Do Problem 31 if the spring constants are 6k,2k and 3k .

Short answer

The characteristics vibration frequencies are${\omega }_1=\sqrt{\frac{4k}{m}}$and ${\omega }_2=\sqrt{\frac{9k}{m}}$.

Step-by-step solution

Definition of eigenvalue

For a matrix A , all possible values of $\mathit{\lambda }$which satisfies the equation $\left|A-\lambda I\right|\mathbf{=}\mathbf{0}$, where I is the identity matrix, are considered as the eigenvalues of a matrix.

Find characteristics vibration frequencies

The figure with system of masses and springs is shown below:

Now, x-y is the compression or extension of the middle spring. So, the potential energy of the middle spring is given by .

For the other two springs, the potential energy is given by $\frac{1}{2}\left(6k\right){\left(x\right)}^2$and$\frac{1}{2}\left(3k\right){\left(y\right)}^2$ .

So, the total potential energy is $V=\frac{1}{2}\left(6k\right){\left(x\right)}^2+\frac{1}{2}\left(2k\right){\left(x-y\right)}^2+\frac{1}{2}\left(3k\right){\left(y\right)}^2$.

Here, $V$is a bi-variable function, so, the forces applied on two masses are $-\frac{\partial V}{\partial x}$and $-\frac{\partial V}{\partial y}$.

Substitute $x\text{'}\text{'}=\frac{d^{2}x}{d{t}^2}$and $x\text{'}=\frac{dx}{dt}$. Then the motions represented by $\frac{-\partial V}{\partial x}=-k\left(8x-2y\right)$and $\frac{-\partial V}{\partial y}=-k\left(-2x+5y\right)$.

Let $\omega$be the frequency for $x$and y. Then$x\text{'}\text{'}=-{\omega }^{2}x,y\text{'}\text{'}=--{\omega }^{2}y$. So, from above equations, we have $-m{\omega }^{2}x=-k\left(8x-2y\right)$and $-m{\omega }^{2}y=-k\left(-2x+5y\right)$.

Now, substitute $\lambda =\frac{m{\omega }^2}{k}$. We get $\lambda x=8x-2y$and $\lambda y=\left(-2x+5y\right)$.

Now, write the above equation in the matrix form as $\left(\begin{array}{cc}8& -2\\ -2& 5\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\lambda \left(\begin{array}{c}x\\ y\end{array}\right)$. So, the characteristic equation becomes$\left|\begin{array}{cc}8-\lambda & -2\\ -2& 5-\lambda \end{array}\right|=0$. Solve this equation for $\lambda$:

localid="1664374045167" $\begin{array}{rcl}\left(8-\lambda \right)\left(5-\lambda \right)-\left(-2\right)\left(-2\right)& =& 0\\ 40-13\lambda +{\lambda }^2-4& =& 0\\ {\lambda }^2-13\lambda +36& =& 0\\ \left(\lambda -4\right)\left(\lambda -9\right)& =& 0\end{array}$

So, the eigenvalues are 4 and 9.

Now, if localid="1664374052831" $\begin{array}{rcl}\lambda & =& 4\end{array}$, then:

localid="1664374061150" $\begin{array}{rcl}{\omega }_1& =& \sqrt{\frac{k\lambda }{m}}\\ & =& \sqrt{\frac{4k}{m}}\end{array}$

Now, if localid="1664374068591" $\begin{array}{rcl}\lambda & =& 9\end{array}$, then:

$\begin{array}{rcl}{\omega }_2& =& \sqrt{\frac{k\lambda }{m}}\\ & =& \sqrt{\frac{9k}{m}}\end{array}$

Thus, the characteristic frequencies are

$\begin{array}{rcl}{\omega }_1& =& \sqrt{\frac{4\lambda }{m}}\\ {\omega }_2& =& \sqrt{\frac{9k}{m}}\end{array}$

Now, on substituting$\begin{array}{rcl}\lambda & =& 1\end{array}$in$\begin{array}{rcl}\lambda x& =& 8x-2y\end{array}$, we get$\begin{array}{rcl}2x& =& y\end{array}$. Substituting $\begin{array}{rcl}\lambda & =& 9\end{array}$in$\begin{array}{rcl}\lambda y& =& \left(-2x+5y\right)\end{array}$, we get$\begin{array}{rcl}x& =& 2y\end{array}$.

So, at frequency$\begin{array}{rcl}& & {\omega }_1\end{array}$, the two masses oscillate back and forth together like$\begin{array}{rcl}& \to & \to \end{array}$and then $\begin{array}{rcl}& \leftarrow & \leftarrow \end{array}$.

At frequency$\begin{array}{rcl}& & {\omega }_2\end{array}$, the two masses oscillate in opposite directions like$\begin{array}{rcl}& \leftarrow & \to \end{array}$ and then like$\begin{array}{rcl}& \to & \leftarrow \end{array}$.