Question

A mass of 5 kg is suspended from a spring of stiffness constant 5 N/m. If the frequency of the natural oscillation is 72 times the frequency of the damped oscillation, find the damping constant.

Answer 1

The damping constant is $10 \sqrt{\frac{5183}{5184}}$ kg/s.

Answer 2

1. Calculate Natural Angular Frequency ($\omega_0$): $\omega_0 = \sqrt{\frac{k}{m}} = \sqrt{\frac{5 \text{ N/m}}{5 \text{ kg}}} = 1$ rad/s.
2. Determine Damped Angular Frequency ($\omega_d$): Since $f_0 = 72 f_d$, we have $\omega_0 = 72 \omega_d$, so $\omega_d = \frac{1}{72}$ rad/s.
3. Calculate Damping Factor ($\gamma$): $\omega_d^2 = \omega_0^2 - \gamma^2 \implies (\frac{1}{72})^2 = 1^2 - \gamma^2 \implies \gamma^2 = 1 - \frac{1}{5184} = \frac{5183}{5184} \implies \gamma = \sqrt{\frac{5183}{5184}}$ rad/s.
4. Find Damping Constant ($b$): $b = 2m\gamma = 2 \times 5 \text{ kg} \times \sqrt{\frac{5183}{5184}} = 10 \sqrt{\frac{5183}{5184}}$ kg/s.