Correct Answer - Option 1 : 0.471 and 1.19 Hz
Concept:
The damped natural frequency is given as, \({{\rm{\omega }}_{\rm{d}}} = {\rm{\;}}{{\rm{\omega }}_{\rm{n}}}\sqrt {1 - {{\rm{ξ }}^2}} \)
where, ξ = Damping factor and it is written as, \({\rm{ξ = }}\frac{{\rm{C}}}{{2\sqrt {{\rm{km}}} }}\)
Calculation:
Given:
k = 3600 N/m, C = 400 Ns/m, m = 50 kg
Damping factor:
\({\rm{ξ }} = \frac{{400}}{{2\sqrt {3600{\rm{\;}} \times {\rm{\;}}50} }}\)
\({\bf{ξ }}\) = 0.471
Since damped natural frequency, \({{\rm{\omega }}_{\rm{d}}} = {\rm{\;}}{{\rm{\omega }}_{\rm{n}}}\sqrt {1 - {{\rm{ξ }}^2}} \)
\({{\rm{\omega }}_{\rm{n}}} = \sqrt {\frac{{\rm{k}}}{{\rm{m}}}} \)
\({{\rm{\omega }}_{\rm{n}}} = \sqrt {\frac{{3600}}{{50}}} \) = 8.485 rad/s
Now, \({{\rm{\omega }}_{\rm{d}}} = {\rm{\;}}8.458{\rm{\;}}\sqrt {1 - {{0.471}^2}} \)
\({{\rm{\omega }}_{\rm{d}}}\) = 7.461 rad/s
\({{\rm{f}}_{\rm{d}}} = {\rm{\;}}\frac{{{{\rm{\omega }}_{\rm{d}}}}}{{2{\rm{\pi }}}} = \frac{{7.461}}{{2{\rm{\pi }}}}\)
\({{\rm{f}}_{\rm{d}}}\) = 1.187
\({{\rm{f}}_{\rm{d}}}\) ≈ 1.19 Hz
