Let's break down the problem step by step: --- ## **Step 1: Understand the System** - **Given:** - Mass, \( m = 2 \) kg - Spring constant, \( K = 400 \) N/m - Lengths: \( l = 200 \) mm = .2 m, \( a = 75 \) mm = .075 m This is a rotational vibrating system with a damper and spring attached *not at the pivot* but at a distance \( a \) from it. --- ## **Step 2: Equation of Motion** Let \(\theta\) be the angular displacement of the rod (measured from equilibrium). - **Moment of Inertia about O:** For a point mass at the end of a rod: \[ I = ml^2 \] - **Spring Force:** When rod rotates by \(\theta\), point at distance \(a\) moves vertically by \(a\theta\). So, force in spring = \(K(a\theta)\) (Hooke's law). Moment about O due to spring: \[ M_{spring} = K(a\theta) \cdot a = Ka^2 \theta \] - **Damping Force:** Damper at \(a\): vertical velocity = \(a\dot{\theta}\). Damping force = \(c(a\dot{\theta})\). Moment about O due to damper: \[ M_{damper} = c(a\dot{\theta}) \cdot a = ca^2\dot{\theta} \] --- ## **Step 3: Write the Differential Equation** Sum of moments about O (using D'Alembert's Principle): \[ I\ddot{\theta} + ca^2\dot{\theta} + Ka^2\theta = \] Substitute \( I = ml^2 \): \[ ml^2\ddot{\theta} + ca^2\dot{\theta} + Ka^2\theta = \] Or, \[ \boxed{ \ddot{\theta} + \frac{ca^2}{ml^2}\dot{\theta} + \frac{Ka^2}{ml^2}\theta = } \] --- ## **Step 4: Expressions for Critical Damping Coefficient and Frequency** This is a standard second-order ODE: \[ \ddot{\theta} + 2\zeta\omega_n\dot{\theta} + \omega_n^2\theta = \] - **Natural Frequency (\(\omega_n\)):** \[ \omega_n = \sqrt{\frac{Ka^2}{ml^2}} \] - **Critical Damping Coefficient (\(c_{cr}\)):** \[ 2\zeta\omega_n = \frac{ca^2}{ml^2} \] For critical damping, \(\zeta = 1\): \[ 2\omega_n = \frac{c_{cr}a^2}{ml^2} \implies c_{cr} = \frac{2ml^2}{a^2}\omega_n \] Substitute \(\omega_n\): \[ c_{cr} = \frac{2ml^2}{a^2} \sqrt{\frac{Ka^2}{ml^2}} \] Simplifying: \[ c_{cr} = 2\sqrt{Km}\frac{l}{a} \] --- ## **Step 5: Frequency of Vibration** - **Natural Frequency (\(f_n\)):** \[ f_n = \frac{\omega_n}{2\pi} = \frac{1}{2\pi}\sqrt{\frac{Ka^2}{ml^2}} \] --- ## **Step 6: Substitute Values** Given: - \(m = 2\) kg - \(K = 400\) N/m - \(l = .2\) m - \(a = .075\) m ### **Calculate \(\omega_n\):** \[ \omega_n = \sqrt{\frac{400 \times (.075)^2}{2 \times (.2)^2}} \] \[ = \sqrt{\frac{400 \times .005625}{2 \times .04}} \] \[ = \sqrt{\frac{2.25}{.08}} \] \[ = \sqrt{28.125} \] \[ \approx 5.306~\text{rad/s} \] ### **Calculate \(c_{cr}\):** \[ c_{cr} = 2 \sqrt{400 \times 2} \cdot \frac{.2}{.075} \] \[ = 2 \cdot \sqrt{800} \cdot \frac{.2}{.075} \] \[ \sqrt{800} \approx 28.284 \] \[ c_{cr} = 2 \cdot 28.284 \cdot 2.6667 \] \[ = 56.568 \cdot 2.6667 \approx 151.05~\text{N·s/m} \] ### **Calculate \(f_n\):** \[ f_n = \frac{5.306}{2\pi} \approx .844~\text{Hz} \] --- ## **Final Answers** ### **1. Differential Equation:** \[ \boxed{ml^2\ddot{\theta} + ca^2\dot{\theta} + Ka^2\theta = } \] ### **2. Critical Damping Coefficient:** \[ \boxed{c_{cr} = 2\sqrt{Km}\frac{l}{a} = 151.05~\text{N·s/m}} \] ### **3. Frequency of Vibration:** \[ \boxed{f_n = \frac{1}{2\pi}\sqrt{\frac{Ka^2}{ml^2}} = .844~\text{Hz}} \] If you need further clarifications or a breakdown of any step, just ask!