Multi-Degree of Freedom Systems - Examples 1-5

Example 1 - Derive the differential equation of motion for the system shown in the following figure.

$$\begin{align} & {{m}_{1}}{{{\ddot{x}}}_{1}}=-k{{x}_{1}}-5k\left( {{x}_{1}}-{{x}_{3}} \right)-k\left( {{x}_{1}}-{{x}_{2}} \right)+{{F}_{1}}\left( t \right) \\ & {{m}_{2}}{{{\ddot{x}}}_{2}}=-k\left( {{x}_{2}}-{{x}_{1}} \right)-k\left( {{x}_{2}}-{{x}_{3}} \right)+{{F}_{2}}\left( t \right) \\ & \underline{{{m}_{3}}{{{\ddot{x}}}_{3}}=-5k\left( {{x}_{3}}-{{x}_{1}} \right)-k\left( {{x}_{3}}-{{x}_{2}} \right)+{{F}_{3}}\left( t \right)} \\ & {{m}_{1}}{{{\ddot{x}}}_{1}}+\left( 7{{x}_{1}}-{{x}_{2}}-5{{x}_{3}} \right)k={{F}_{1}}\left( t \right) \\ & {{m}_{2}}{{{\ddot{x}}}_{2}}+\left( -{{x}_{1}}+2{{x}_{2}}-{{x}_{3}} \right)k={{F}_{2}}\left( t \right) \\ & {{m}_{3}}{{{\ddot{x}}}_{3}}+\left( -5{{x}_{1}}-1{{x}_{2}}+7{{x}_{3}} \right)k={{F}_{3}}\left( t \right) \\ \end{align}$$

Or in matrix form

$$\left[ \begin{matrix} {{m}_{1}} & 0 & 0 \\ 0 & {{m}_{2}} & 0 \\ 0 & 0 & {{m}_{3}} \\ \end{matrix} \right]\left( \begin{matrix} {{{\ddot{x}}}_{1}} \\ {{{\ddot{x}}}_{2}} \\ {{{\ddot{x}}}_{3}} \\ \end{matrix} \right)+k\left[ \begin{matrix} 7 & -1 & -5 \\ -1 & 2 & -1 \\ -5 & -1 & 7 \\ \end{matrix} \right]\left( \begin{matrix} {{x}_{1}} \\ {{x}_{2}} \\ {{x}_{3}} \\ \end{matrix} \right)=\left( \begin{matrix} {{F}_{1}}\left( t \right) \\ {{F}_{2}}\left( t \right) \\ {{F}_{3}}\left( t \right) \\ \end{matrix} \right)$$