Let's break down the problem: **Given:** - Mass, \( m = 150\, \text{kg} \) - Spring stiffness, \( k = 45\, \text{kN/m} = 45000\, \text{N/m} \) - Damping is 30% of the critical value (\( c = 0.3c_c \)) We need to **calculate the logarithmic decrement (\( \delta \))**. --- ## Step 1: Find the undamped natural frequency (\( \omega_n \)): \[ \omega_n = \sqrt{\frac{k}{m}} \] \[ \omega_n = \sqrt{\frac{45000}{150}} = \sqrt{300} \approx 17.32 \ \text{rad/s} \] --- ## Step 2: Find the damping ratio (\( \xi \)): Damping is 30% of critical value: \[ \xi = \frac{c}{c_c} = 0.3 \] --- ## Step 3: Logarithmic Decrement Formula (\( \delta \)): \[ \delta = \frac{2\pi\xi}{\sqrt{1-\xi^2}} \] Plug in \( \xi = 0.3 \): \[ \delta = \frac{2\pi \times 0.3}{\sqrt{1 - (0.3)^2}} \] \[ = \frac{2\pi \times 0.3}{\sqrt{1 - 0.09}} \] \[ = \frac{2\pi \times 0.3}{\sqrt{0.91}} \] \[ = \frac{1.884}{0.954} \] \[ \approx 1.97 \] --- **Final Answer:** The value of logarithmic decrement is **1.97**. So, the correct option is: > **3. 1.97**