## Engineering Mechanics: Statics & Dynamics (14th Edition)

$T=1.45s$
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 $T=\frac{2\pi}{\omega_n}$ $\implies T=\frac{2\pi}{4.3359}$ $\implies T=1.45s$