DISTANCE x.Taking moment about '0'Kx × 2L = mg × L\(x = \FRAC{{mg}}{{2k}}\)From similar triangle property \(\frac{x}{{2L}} = \frac{\delta }{L}\)\(\delta = \frac{x}{2} = \frac{{mg}}{{4k}}\)Using static DEFLECTION of the mass 'm'\({\omega _n} = \sqrt {\frac{g}{\delta }} = \sqrt {\frac{{4K}}{m}} \)