Q22E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Recompute the coupled mass–spring oscillator motion in Problem 1, Exercises 5.6 (page 287), using Laplace transforms.

 

Step-by-Step Solution

Answer

$$\begin{gathered} x\left(t\right)=-\frac{8}{17}\cos t-\frac{9}{17}\cos \sqrt{\frac{20}{3}t} \\ y\left(t\right)=-\frac{6}{17}\cos t+\frac{6}{17}\cos \sqrt{\frac{20}{3}}t \end{gathered}$$

Step 1:Given Information

The differential equation of a coupled mass spring oscillator is 

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

Step 2: Determining the coupled mass–spring oscillator motion

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

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

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

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

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

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

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

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

Step 3: Determining the Result

The solution is obtained as:

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

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