Answer
$T=1.45s$
Work Step by Step
The natural period of vibration can be determined as follows:
$\omega_n=\sqrt{\frac{K}{m}}$
$\omega_n=\sqrt{\frac{mg}{m\delta_{st}}}$
$\implies \omega_n=\sqrt{\frac{g}{\delta_{st}}}$
We plug in the known values to obtain:
$\omega_n=\sqrt{\frac{28.32}{\frac{18in}{12in/ft}}}$
$\implies \omega_n=4.3359rad/s$
Now $T=\frac{2\pi}{\omega_n}$
$\implies T=\frac{2\pi}{4.3359}$
$\implies T=1.45s$
