Question

A block of mass 1 kg connected to a spring of stiffness 10 N m\(^{-1}\) is operating in a viscous medium such that the damping ratio (or damping factor) is equal to the ratio of the damped frequency to the natural frequency. The magnitude of the damping ratio for this system is _________ (rounded off to 2 decimal places).

Hint:

For systems with damping, the damping ratio \( \zeta \) helps determine the behavior of the oscillation. If \( \zeta = 1 \), the system is critically damped.

Answer 1

Step 1: Calculate Natural Frequency \[ \omega_n = \sqrt{\frac{k}{m}} = \sqrt{\frac{10}{1}} = \sqrt{10} \, {rad/s} \] Step 2: Relate Damping Ratio to Frequencies Given: \[ \zeta = \frac{\omega_d}{\omega_n} \] The damped frequency is: \[ \omega_d = \omega_n \sqrt{1 - \zeta^2} \] Substituting: \[ \zeta = \sqrt{1 - \zeta^2} \] Step 3: Solve for \(\zeta\) \[ \zeta^2 = 1 - \zeta^2 \implies 2\zeta^2 = 1 \implies \zeta = \frac{1}{\sqrt{2}} \approx 0.71 \] Final Answer The damping ratio is \(\boxed{0.71}\).