Answer
$13.9rad/s$
$0.452s$
Work Step by Step
We can determine the natural frequency and the period of vibration as follows:
$K=\frac{W}{\delta_{st}}$
$\implies K=\frac{20}{\frac{4in}{12in/ft}}=60lb/ft$
Now $\omega_n=\sqrt{\frac{K}{m}}$
We plug in the known values to obtain:
$\omega_n=\sqrt{\frac{60}{\frac{10}{32.2}}}=13.9rad/s$
and the period of vibration is given as
$T=\frac{2\pi}{\omega}$
$\implies T=\frac{2\pi}{13.9}$
$\implies T=0.452s$
