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

$T=2\pi\sqrt{\frac{mK_{\circ}^2}{C}}$
We can determine the natural period of oscillation as follows: $\omega_n=\sqrt{\frac{K_{eq}}{m_{eq}}}$ $\implies \omega_n=\sqrt{\frac{C}{mK_{\circ}^2}}$ Now, $T=\frac{2\pi}{\omega_n}$ Thus: $\implies T=2\pi\sqrt{\frac{mK_{\circ}^2}{C}}$