7.3 Aplicação: duplo massa mola

Considere o duplo sistema massa-mola, onde as molas possuem constantes $k_1$ e $k_2$ e as massas envolvidas são $m_1$ e $m_2$. Desconsiderando o amortecimento, temos o seguinte sistema: $$\begin{array}{rcll}{m}_1{ẍ}_1\left(t\right)& =& -k_1{x}_1\left(t\right)+k_2\left[x_2\left(t\right)-x_1\left(t\right)\right]+f_1\left(t\right)& \\ m_2{ẍ}_2\left(t\right)& =& -k_2\left[x_2\left(t\right)-x_1\left(t\right)\right]+f_2\left(t\right),& \end{array}$$

onde $x_1$ e $x_2$ representam o deslocamento de cada uma das massas e $f_1$ e $f_2$ são as forças externas aplicadas. Tomando transformada de Laplace, obtemos: $$\begin{array}{rcll}{m}_1\left(s^2{X}_1\left(s\right)-{ẋ}_1\left(0\right)-s{x}_1\left(0\right)\right)& =& -\left(k_1+k_2\right)X_1\left(s\right)+k_2{X}_2\left(s\right)+F_1\left(s\right)& \\ m_2\left(s^2{X}_2\left(s\right)-{ẋ}_2\left(0\right)-s{x}_2\left(0\right)\right)& =& -k_2{X}_2\left(s\right)+k_2{X}_1\left(s\right)+F_2\left(s\right)& \end{array}$$

isto é: $$\begin{array}{rcll}\left(m_1{s}^2+k_1+k_2\right)X_1\left(s\right)-k_2{X}_2\left(s\right)& =& F_1\left(s\right)+m_1{ẋ}_1\left(0\right)+s{m}_1{x}_1\left(0\right)& \\ -k_2{X}_1\left(s\right)+\left(m_2{s}^2+k_2\right)X_2\left(s\right)& =& F_2\left(s\right)+m_2{ẋ}_2\left(0\right)+s{m}_2{x}_2\left(0\right)& \end{array}$$

A representação matricial do sistema é: $$\begin{array}{rcll}\left[\begin{array}{cc}\hfill m_1{s}^2+k_1+k_2\hfill & \hfill -k_2\hfill \\ \hfill -k_2\hfill & \hfill m_2{s}^2+k_2\hfill \end{array}\right]\left[\begin{array}{c}\hfill X_1\left(s\right)\hfill \\ \hfill X_2\left(s\right)\hfill \end{array}\right]=\left[\begin{array}{c}\hfill F_1\left(s\right)+m_1{ẋ}_1\left(0\right)+s{m}_1{x}_1\left(0\right)\hfill \\ \hfill F_2\left(s\right)+m_2{ẋ}_2\left(0\right)+s{m}_2{x}_2\left(0\right)\hfill \end{array}\right]& & & \end{array}$$

e sua solução pode ser escrita como: $$\begin{array}{rcll}\left[\begin{array}{c}\hfill X_1\left(s\right)\hfill \\ \hfill X_2\left(s\right)\hfill \end{array}\right]& =& \frac{1}{P\left(s\right)}\left[\begin{array}{cc}\hfill m_2{s}^2+k_2\hfill & \hfill k_2\hfill \\ \hfill k_2\hfill & \hfill m_1{s}^2+k_1+k_2\hfill \end{array}\right]\left[\begin{array}{c}\hfill F_1\left(s\right)+m_1{ẋ}_1\left(0\right)+s{m}_1{x}_1\left(0\right)\hfill \\ \hfill F_2\left(s\right)+m_2{ẋ}_2\left(0\right)+s{m}_2{x}_2\left(0\right)\hfill \end{array}\right],& \text{(7.17)}\end{array}$$

onde $P\left(s\right)=m_1{m}_2{s}^4+\left(m_1{k}_2+m_2{k}_1+m_2{k}_2\right)s^2+k_1{k}_2$. Vamos resolver um caso particular onde $m_1=m_2=1$, $f_1=f_2=0$, $k_1=6$ e $k_2=4$, temos o seguinte sistema massa-mola: $$\begin{array}{rcll}{ẍ}_1\left(t\right)& =& -6x_1\left(t\right)+4\left[x_2\left(t\right)-x_1\left(t\right)\right]& \\ {ẍ}_2\left(t\right)& =& -4\left[x_2\left(t\right)-x_1\left(t\right)\right],& \end{array}$$

Usando (7.17), temos: $$\begin{array}{rcll}\left[\begin{array}{c}\hfill X_1\left(s\right)\hfill \\ \hfill X_2\left(s\right)\hfill \end{array}\right]& =& \frac{1}{s^4+14s^2+24}\left[\begin{array}{cc}\hfill s^2+4\hfill & \hfill 4\hfill \\ \hfill 4\hfill & \hfill s^2+10\hfill \end{array}\right]\left[\begin{array}{c}\hfill {ẋ}_1\left(0\right)+s{x}_1\left(0\right)\hfill \\ \hfill {ẋ}_2\left(0\right)+s{x}_2\left(0\right)\hfill \end{array}\right].& \end{array}$$

Para completar o sistema, impondo as seguintes condições iniciais: $x_1\left(0\right)=x_2\left(0\right)=0$, ${ẋ}_1\left(0\right)=1$ e ${ẋ}_2\left(0\right)=-1$: $$\begin{array}{rcll}\left[\begin{array}{c}\hfill X_1\left(s\right)\hfill \\ \hfill X_2\left(s\right)\hfill \end{array}\right]& =& \frac{1}{s^4+14s^2+24}\left[\begin{array}{cc}\hfill s^2+4\hfill & \hfill 4\hfill \\ \hfill 4\hfill & \hfill s^2+10\hfill \end{array}\right]\left[\begin{array}{c}\hfill 1\hfill \\ \hfill -1\hfill \end{array}\right].& \end{array}$$

Logo,

e

Usamos frações parciais para escrever $$\begin{array}{rcll}\frac{s^2}{s^4+14s^2+24}=\frac{s^2}{\left(s^2+2\right)\left(s^2+12\right)}& =& \frac{A}{s^2+2}+\frac{B}{s^2+12}& \\ & =& \frac{A\left(s^2+12\right)+B\left(s^2+2\right)}{\left(s^2+2\right)\left(s^2+12\right)}& \\ & =& \frac{\left(A+B\right)s^2+12A+2B}{\left(s^2+2\right)\left(s^2+12\right)},& \end{array}$$

ou seja, $A+B=1$ e $12A+2B=0$. Logo, $B=-6A$ e $A-6A=1$, ou seja, $A=-\frac{1}{5}$ e $B=\frac{6}{5}$. Portanto,

e, calculando a transformada inversa, temos: $$\begin{array}{rcll}x\left(t\right)& =& \frac{6}{5\sqrt{12}}\mathrm{sen}<!--FUNCTION\; APPLICATION-->\left(\sqrt{12}t\right)-\frac{1}{5\sqrt{2}}\mathrm{sen}<!--FUNCTION\; APPLICATION-->\left(\sqrt{2}t\right)& \\ & =& \frac{\sqrt{3}}{5}\mathrm{sen}<!--FUNCTION\; APPLICATION-->\left(2\sqrt{3}t\right)-\frac{\sqrt{2}}{10}\mathrm{sen}<!--FUNCTION\; APPLICATION-->\left(\sqrt{2}t\right).& \end{array}$$

Da mesma forma,

e $$\begin{array}{rcll}y\left(t\right)& =& -\frac{3}{5\sqrt{12}}\mathrm{sen}<!--FUNCTION\; APPLICATION-->\left(\sqrt{12}t\right)-\frac{2}{5\sqrt{2}}\mathrm{sen}<!--FUNCTION\; APPLICATION-->\left(\sqrt{2}t\right)& \\ & =& -\frac{\sqrt{3}}{10}\mathrm{sen}<!--FUNCTION\; APPLICATION-->\left(2\sqrt{3}t\right)-\frac{\sqrt{2}}{5}\mathrm{sen}<!--FUNCTION\; APPLICATION-->\left(\sqrt{2}t\right).& \end{array}$$

A figura 7.2 apresenta os gráficos de $x\left(t\right)$ e $y\left(t\right)$: