Let's solve both systems step-by-step. --- # (a) Linear Mechanical System 1 ### **Step 1: Free-Body Diagram** - **For mass \( m \):** - To the right: Applied force \( F_a(t) \) - To the left: Spring force \( kx_1 \) (Hooke's law) - To the left: Damper force \( b\dot{x}_1 \) (damping force proportional to velocity) ### **Step 2: Newton’s Second Law** \[ \sum F = m\ddot{x}_1 \] \[ F_a(t) - kx_1 - b\dot{x}_1 = m\ddot{x}_1 \] ### **Step 3: Equation of Motion** \[ m\ddot{x}_1 + b\dot{x}_1 + kx_1 = F_a(t) \] --- ## **(b) Linear Mechanical System 2** ### **Step 1: Free-Body Diagrams** - **For mass \( m_1 \) (bottom block):** - To the right: Applied force \( F_a(t) \) - To the left: Damper force \( b(\dot{x}_1 - \dot{x}_2) \) (relative velocity between \( m_1 \) and \( m_2 \)) - **For mass \( m_2 \) (top block):** - To the right: Damper force \( b(\dot{x}_1 - \dot{x}_2) \) - To the left: Spring force \( kx_2 \) (assuming \( x_2 \) is measured from the equilibrium position of the spring) ### **Step 2: Newton’s Second Law** #### **For \( m_1 \):** \[ m_1 \ddot{x}_1 = F_a(t) - b(\dot{x}_1 - \dot{x}_2) \] #### **For \( m_2 \):** \[ m_2 \ddot{x}_2 = b(\dot{x}_1 - \dot{x}_2) - kx_2 \] ### **Step 3: Equations of Motion** \[ \boxed{ \begin{align*} m_1 \ddot{x}_1 + b(\dot{x}_1 - \dot{x}_2) &= F_a(t) \\ m_2 \ddot{x}_2 + b(\dot{x}_2 - \dot{x}_1) + kx_2 &= \end{align*} } \] --- ## **Final Answers** ### (a) **Equation of Motion:** \[ \boxed{ m\ddot{x}_1 + b\dot{x}_1 + kx_1 = F_a(t) } \] ### (b) **Equations of Motion:** \[ \boxed{ \begin{align*} m_1 \ddot{x}_1 + b(\dot{x}_1 - \dot{x}_2) &= F_a(t) \\ m_2 \ddot{x}_2 + b(\dot{x}_2 - \dot{x}_1) + kx_2 &= \end{align*} } \] Let me know if you need the free-body diagrams sketched or further explanation!