Correct Answer - Option 4 : 0.1
Concept:
Free damped Vibration:
\(m\frac{{{d^2}x}}{{d{t^2}}} + c\frac{{dx}}{{dt}} + kx = 0\)
\(m\ddot x + c\dot x + kx = 0\)
Natural frequency:
\({\omega _n} = \sqrt {\frac{k}{m}}\)
Damping factor:
\(ξ = \frac{c}{{{c_c}}} = \frac{c}{{2\sqrt {km} }} \)
Where c = damping coefficient , m = mass, k = spring stiffness.
Calculation:
Given:
m = 175 kg, k = 700 N/cm = 70000 N/m , c = 7 Ncm-1sec = 700 Nm-1sec
\(ξ = \frac{c}{{{c_c}}} = \frac{c}{{2\sqrt {km} }} \)
\(ξ = \frac{c}{{2\sqrt {km} }} \)
\(ξ = \frac{700}{{2\sqrt {70000\times175} }} =0.1\)
ξ = 0.1