Answer
$\omega_n=49.5 rad/s$;$T=0.127s$
Work Step by Step
We can determine the natural frequency and the period of vibration as follows:
$K=\frac{mg}{\delta _{st}}$
$\implies K=\frac{3\times 9.81}{0.06}=490.5N/m$
Now the natural frequency is given as
$\omega_n=\sqrt{\frac{K}{m}}$
We plug in the known values to obtain:
$\implies \omega_n=\sqrt{\frac{490.5}{0.2}}=49.5rad/s$
and $T=\frac{2\pi}{\omega_n}$
We plug in the known values to obtain:
$T=\frac{2\pi}{49.5}$
$\implies T=0.127s$
