91Ó°ÊÓ

The differential equation of a coupled mass spring oscillator is

$\{\begin{array}{cc}{x}^{\text{'}\text{'}}+4x−4y=0,& x\left(0\right)=−1,x^{\text{'}}\left(0\right)=0\\ 3y^{\text{'}\text{'}}−6x+11y=0& y\left(0\right)=0,y^{\text{'}}\left(0\right)=0\end{array}$

The mechanism in Exercises 5.6 Problem 1 is

$\{\begin{array}{cc}{x}^{\text{'}\text{'}}+4x−4y=0,& x\left(0\right)=−1,x^{\text{'}}\left(0\right)=0\\ 3y^{\text{'}\text{'}}−6x+11y=0& y\left(0\right)=0,y^{\text{'}}\left(0\right)=0\end{array}$

Using the Laplace transform, we can obtain

$\{\begin{array}{l}L\{x"+4x-4y\}=0\\ L\{3y"-6x+11y\}=0\end{array}$

$⇒\{\begin{array}{l}{s}^{2}X\left(s\right)−sx\left(0\right)−x^{\text{'}}\left(0\right)+4X\left(s\right)−4Y\left(s\right)=0\\ 3\left[s^{2}Y\left(s\right)−sy\left(0\right)−y^{\text{'}}\left(0\right)\right]−6X\left(s\right)+11Y\left(s\right)=0\end{array}$

$⇒\{\begin{array}{l}{s}^{2}X\left(s\right)+s+4X\left(s\right)−4Y\left(s\right)=0\\ 3s^{2}Y\left(s\right)−6X\left(s\right)+11Y\left(s\right)=0\end{array}$

$⇒\{\begin{array}{l}(s^2+4)X\left(s\right)−4Y\left(s\right)=−s\\ (3s^2+11)Y\left(s\right)−6X\left(s\right)=0\end{array}$

$⇒\{\begin{array}{l}X\left(s\right)=\frac{4}{s^2+1}Y\left(s\right)−\frac{s}{s^2+1}\\ (3s^2+11)Y\left(s\right)−6X\left(s\right)=0\end{array}$

We now have

$\begin{array}{c}(3s^2+11)Y\left(s\right)−6\left(\frac{4}{s^2+1}Y\left(s\right)−\frac{s}{s^2+1}\right)=0\\ (3s^2+11)Y\left(s\right)−\frac{24}{s^2+1}Y\left(s\right)+\frac{6s}{s^2+1}=0\\ (3s^2+11−\frac{24}{s^2+1})Y\left(s\right)=−\frac{6s}{s^2+1}\\ \frac{(3s^2+11)(s^2+1)−24}{s^2+1}Y\left(s\right)=−\frac{6s}{s^2+1}\\ (3s^4+23s^2+20)Y\left(s\right)=6s\\ Y\left(s\right)=−\frac{6s}{3s^4+23s^2+20}\end{array}$

We can get partial fractions by using partial fractions.

$Y\left(s\right)=−\frac{6s}{17(s^2+1)}+\frac{6s}{17(s^2+\frac{20}{3})}$

As a result, if we use the inverse Laplace, we get

$y\left(t\right)=−\frac{6}{17}\cos t+\frac{6}{17}\cos \sqrt{\frac{20}{3}t}$

We now have X(s):

$\begin{array}{rcl}(3s^2+11)\frac{(−6s)}{3s^4+23s^2+20}−6X\left(s\right)& =& 0\\ X\left(s\right)& =& \frac{−s(3s^2+11)}{(s^2+1)(s^2+\frac{20}{3})}\end{array}$

We can get partial fractions by using partial fractions.

$X\left(s\right)=−\frac{8s}{17(s^2+1)}−\frac{9s}{17(s^2+\frac{20}{3})}$

As a result, if we use the inverse Laplace, we get

$x\left(t\right)=−\frac{8}{17}\cos t−\frac{9}{17}\cos \sqrt{\frac{20}{3}}t$

The solution is obtained as:

$x\left(t\right)=−\frac{8}{17}\cos t−\frac{9}{17}\cos \sqrt{\frac{20}{3}t}$

$y\left(t\right)=−\frac{6}{17}\cos t+\frac{6}{17}\cos \sqrt{\frac{20}{3}}t$