Let's solve **System A** step by step using the **Newton (force-balance) method**. --- # **System A Solution** ### **Step 1: Define the System and Coordinates** - Vertical rigid rod of length \( L \), mass \( m \) (or just consider mass at the top \( M \)). - Two springs (stiffness \( K \)) attached at distances \( a \) and \( b \) from the pivot. - Small angular displacement \( \theta \) (assume small-angle approximations). ### **Step 2: Free-Body Diagram and Forces** For small \( \theta \), the horizontal displacement of each spring at points \( a \) and \( b \) is: - At \( a \): \( x_a = a\theta \) - At \( b \): \( x_b = b\theta \) Spring forces (restoring, oppose motion): - At \( a \): \( F_a = -K (a\theta) \) - At \( b \): \( F_b = -K (b\theta) \) ### **Step 3: Write the Equation of Motion (Torque Balance about Pivot)** - **Moment of inertia** for a point mass \( M \) at distance \( L \) from pivot: \( I = ML^2 \) - For distributed rod of mass \( m \): \( I_\text{rod} = \frac{1}{3}mL^2 \) - Here, let's use \( M \) at the top for simplicity. Sum torques due to the two springs: \[ \sum \tau = I \ddot{\theta} \] \[ (-K a\theta) \cdot a + (-K b\theta) \cdot b = M L^2 \ddot{\theta} \] \[ -K a^2 \theta - K b^2 \theta = M L^2 \ddot{\theta} \] \[ -K(a^2 + b^2)\theta = M L^2 \ddot{\theta} \] ### **Step 4: Rearranging the Equation** \[ M L^2 \ddot{\theta} + K(a^2 + b^2)\theta = 0 \] This is the governing differential equation. ### **Step 5: Find the Natural Frequency** Standard form: \[ \ddot{\theta} + \left[\frac{K(a^2 + b^2)}{M L^2}\right] \theta = 0 \] So, \[ \omega_n^2 = \frac{K(a^2 + b^2)}{M L^2} \] \[ \boxed{\omega_n = \sqrt{\frac{K(a^2 + b^2)}{M L^2}}} \] ### **Step 6: Condition for Stability** - System is stable if \(\omega_n^2 > 0\) - \(K > 0\), and \(a^2 + b^2 > 0\) always if \(a, b \neq 0\) - **So the system is stable for positive spring stiffness and nonzero attachment points.** --- ## **Summary for System A** ### **Governing Differential Equation:** \[ M L^2 \ddot{\theta} + K(a^2 + b^2)\theta = 0 \] ### **Natural Frequency:** \[ \boxed{\omega_n = \sqrt{\frac{K(a^2 + b^2)}{M L^2}}} \] ### **Stability Condition:** \[ K > 0, \quad a^2 + b^2 > 0 \] --- ## **System B** ### **Step 1: Define the System and Coordinates** - Two masses (\( M_1, M_2 \)), connected by a rigid, massless rod: - \( M_1 \) at bottom, \( M_2 \) at top. - Pivot at distance \( L_2 \) from \( M_2 \), \( L_1 \) from \( M_1 \). - Small angular displacement \( \theta \). ### **Step 2: Forces and Torques** The system is essentially an inverted pendulum with mass at both ends. - **Moment of inertia about the pivot:** \[ I = M_1 L_1^2 + M_2 L_2^2 \] - **Restoring torque by gravity:** - \( M_1 \) at a distance \( L_1 \) below pivot: \( -M_1 g L_1 \sin\theta \) - \( M_2 \) at a distance \( L_2 \) above pivot: \( +M_2 g L_2 \sin\theta \) For **small \( \theta \)**, \(\sin\theta \approx \theta\): \[ \tau = -M_1 g L_1 \theta + M_2 g L_2 \theta \] ### **Step 3: Equation of Motion** \[ I \ddot{\theta} = \tau \] \[ (M_1 L_1^2 + M_2 L_2^2) \ddot{\theta} = -M_1 g L_1 \theta + M_2 g L_2 \theta \] \[ (M_1 L_1^2 + M_2 L_2^2) \ddot{\theta} + [M_1 g L_1 - M_2 g L_2]\theta = 0 \] ### **Step 4: Natural Frequency** \[ \ddot{\theta} + \frac{M_1 g L_1 - M_2 g L_2}{M_1 L_1^2 + M_2 L_2^2}\theta = 0 \] So, \[ \omega_n^2 = \frac{M_1 g L_1 - M_2 g L_2}{M_1 L_1^2 + M_2 L_2^2} \] \[ \boxed{\omega_n = \sqrt{\frac{M_1 g L_1 - M_2 g L_2}{M_1 L_1^2 + M_2 L_2^2}}} \] ### **Step 5: Stability Condition** - System is stable if \( \omega_n^2 > 0 \): \[ M_1 g L_1 > M_2 g L_2 \] or \[ \boxed{M_1 L_1 > M_2 L_2} \] --- ## **Summary for System B** ### **Governing Differential Equation:** \[ (M_1 L_1^2 + M_2 L_2^2) \ddot{\theta} + [M_1 g L_1 - M_2 g L_2]\theta = 0 \] ### **Natural Frequency:** \[ \boxed{\omega_n = \sqrt{\frac{M_1 g L_1 - M_2 g L_2}{M_1 L_1^2 + M_2 L_2^2}}} \] ### **Stability Condition:** \[ \boxed{M_1 L_1 > M_2 L_2} \] --- ### **Let me know if you want a full derivation for distributed mass in System A, or if you have follow-up questions!**