A vibrating system consists of a mass of 200 kg, a spring of stiffness 80 N/mm and a damper with damping coefficient of 800 N-s/m. What will be the natural frequency of damped vibration?

Concept:

The damped natural frequency of a vibrating system is given by

\( \omega_d = \omega_n \sqrt{1 - \zeta^2} \), where \( \omega_n = \sqrt{\frac{k}{m}} \) is the natural frequency and \( \zeta = \frac{c}{2\sqrt{km}} \) is the damping ratio.

Given:

Mass, m = 200 kg

Spring stiffness, k = 80 N/mm = 80000 N/m

Damping coefficient, c = 800 Ns/m

Calculation:

Undamped natural frequency, \( \omega_n = \sqrt{\frac{80000}{200}} = \sqrt{400} = 20~\text{rad/s} \)

Damping ratio, \( \zeta = \frac{800}{2\sqrt{80000 \cdot 200}} = \frac{800}{2 \cdot 4000} = 0.1 \)

Damped natural frequency, \( \omega_d = 20 \cdot \sqrt{1 - 0.1^2} = 20 \cdot \sqrt{0.99} = 19.8998~\text{rad/s} \)

Damped frequency in Hz, \( f_d = \frac{\omega_d}{2\pi} = \frac{19.8998}{6.283} \approx 3.17~\text{Hz} \)

Now, \( \frac{3\sqrt{11}}{\pi} = \frac{3 \times 3.3166}{3.14} \approx \frac{9.9498}{3.14} \approx 3.17~\text{Hz} \)