Let's break down the problem: **Given:** - Mass, \( m = 20 \, \text{kg} \) - Stiffness, \( k = 2 \, \text{kN/m} = 2000 \, \text{N/m} \) - Damping factor, \( \xi = 0.32 \) - We need the value of \( k \) (frequency ratio) at which the magnification factor is maximum. **Step 1: Natural Frequency** \[ \omega_n = \sqrt{\frac{k}{m}} = \sqrt{\frac{2000}{20}} = \sqrt{100} = 10 \, \text{rad/s} \] **Step 2: Magnification Factor** Magnification factor for a damped system is: \[ M = \frac{1}{\sqrt{(1 - r^2)^2 + (2\xi r)^2}} \] where \( r = \frac{\omega}{\omega_n} \) is the frequency ratio. **Step 3: Maximum Magnification Factor** Maximum magnification (resonance) for a damped system occurs at: \[ r = \sqrt{1 - 2\xi^2} \] Plug in the given value of \(\xi = 0.32\): \[ r = \sqrt{1 - 2(0.32)^2} = \sqrt{1 - 2 \times 0.1024} = \sqrt{1 - 0.2048} = \sqrt{0.7952} \approx 0.89 \] **Step 4: Answer** So, the value of \( k \) (frequency ratio for maximum magnification) up to 2 decimal places is: \[ \boxed{0.89} \] **Option:** 0.89 Let me know if you need the explanation or formula derivation!