Let's solve each part step by step. --- ## (a) **Diagram and Forces** **Diagram Description:** - The bar is pivoted at \( O \). - At distances \( l_1 \) and \( l_2 \) from the pivot, springs with stiffness \( k_1 \) and \( k_2 \) are attached perpendicular to the bar. - At the end of the bar (distance \( l_3 \)), a mass \( m \) is attached. **Forces:** - **Static (\(\Theta\)):** Springs exert restoring forces perpendicular to the bar, proportional to their extensions (which depend on the static angular displacement \(\Theta\)). - **Dynamic (\(\theta\)):** For small angular displacement \(\theta\) from equilibrium, restoring torques are generated by both springs. **Restoring forces for small angular displacement \(\theta\):** - Extension at \( k_1 \): \( l_1 \theta \) - Extension at \( k_2 \): \( l_2 \theta \) - Forces: \( F_1 = -k_1 l_1 \theta \), \( F_2 = -k_2 l_2 \theta \) - Torques: \( \tau_1 = -k_1 l_1^2 \theta \), \( \tau_2 = -k_2 l_2^2 \theta \) --- ## (b) **Equation of Motion and Natural Frequency** ### **Step 1: Moment of Inertia** The mass is at a distance \( l_3 \): \[ I = m l_3^2 \] ### **Step 2: Equation of Motion (Summing Torques)** \[ I \ddot{\theta} + (k_1 l_1^2 + k_2 l_2^2) \theta = \] ### **Step 3: Natural Frequency** This is a standard SHM equation: \[ \ddot{\theta} + \frac{k_1 l_1^2 + k_2 l_2^2}{m l_3^2} \theta = \] So, \[ \omega_n = \sqrt{\frac{k_1 l_1^2 + k_2 l_2^2}{m l_3^2}} \] --- ## (c) **Damping Ratio and Torsional Damping Coefficient** ### **Step 1: Logarithmic Decrement** Given: \[ \frac{\theta_n}{\theta_{n+1}} = 1.6 \implies \delta = \ln(1.6) = .470 \] ### **Step 2: Damping Ratio (\(\zeta\))** \[ \delta = 2\pi\zeta / \sqrt{1 - \zeta^2} \] Let \( x = \zeta \): \[ .470 = \frac{2\pi x}{\sqrt{1-x^2}} \] \[ .470 \sqrt{1-x^2} = 2\pi x \] \[ .2209 (1-x^2) = 4\pi^2 x^2 \] \[ .2209 - .2209 x^2 = 39.4784 x^2 \] \[ .2209 = (39.4784 + .2209) x^2 \] \[ x^2 = \frac{.2209}{39.6993} \] \[ x = \sqrt{.00557} \approx .0747 \] So, **damping ratio** \( \zeta \approx .075 \) --- ### **Step 3: Find Torsional Damping Coefficient \( c \)** \[ \zeta = \frac{c}{2 I \omega_n} \] - \( m = 2 \) kg - \( l_1 = l_2 = l_3 = .2 \) m - \( k_1 = k_2 = 10 \) N/m #### Calculate \( I \): \[ I = m l_3^2 = 2 \times (.2)^2 = 2 \times .04 = .08 \text{ kg m}^2 \] #### Calculate \( \omega_n \): \[ k_1 l_1^2 + k_2 l_2^2 = 10 \times (.2)^2 + 10 \times (.2)^2 = 2 \times 10 \times .04 = .8 \] \[ \omega_n = \sqrt{ \frac{.8}{.08} } = \sqrt{10} = 3.162 \text{ rad/s} \] #### Solve for \( c \): \[ c = 2 \zeta I \omega_n = 2 \times .075 \times .08 \times 3.162 \] \[ = .15 \times .08 \times 3.162 = .012 \times 3.162 = .0379 \text{ Nms/rad} \] --- ## **Final Answers** ### (a) **Diagram provided above; forces explained.** ### (b) **Equation of motion:** \[ I \ddot{\theta} + (k_1 l_1^2 + k_2 l_2^2)\theta = \] **Natural frequency:** \[ \omega_n = \sqrt{ \frac{k_1 l_1^2 + k_2 l_2^2}{m l_3^2} } \] ### (c) **Damping ratio:** \( \zeta \approx .075 \) **Torsional damping coefficient:** \[ \boxed{c = .038 \ \text{Nms/rad}} \] --- Let me know if you need a sketch or further clarification!