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 $\frac{2}{\sqrt{3}}$ times the frequency of the damped oscillation, find the damping constant.

Answer 1

5 kg/s

Answer 2

1. Calculate the natural angular frequency: $\omega_0 = \sqrt{\frac{k}{m}} = \sqrt{\frac{5 \text{ N/m}}{5 \text{ kg}}} = 1$ rad/s.
2. Determine the damped angular frequency: Given $f_0 = \frac{2}{\sqrt{3}} f_d$, which implies $\omega_0 = \frac{2}{\sqrt{3}} \omega_d$. Thus, $\omega_d = \frac{\sqrt{3}}{2} \omega_0 = \frac{\sqrt{3}}{2}$ rad/s.
3. Calculate the damping factor: Using the relation $\omega_d^2 = \omega_0^2 - \gamma^2$, we find $\gamma^2 = \omega_0^2 - \omega_d^2 = 1^2 - (\frac{\sqrt{3}}{2})^2 = 1 - \frac{3}{4} = \frac{1}{4}$. So, $\gamma = \frac{1}{2}$ rad/s.
4. Find the damping constant: The damping constant $b$ is related by $b = 2m\gamma$. Substituting the values, $b = 2 \times 5 \text{ kg} \times \frac{1}{2} \text{ s}^{-1} = 5$ kg/s.