When a particle of mass m is attached to a vertical spring of...

$y = {y}_{0}\sin^{2}\omega t$

$y = \dfrac{{y}_{0}}{2} ( 1 - \cot 2\omega t )$

$y - \dfrac{{y}_{0}}{2} = - \dfrac{{y}_{0}}{2}\cos 2\omega t$

$Y = Acos2\omega t$

$2\omega = \sqrt{\dfrac{k}{m}}$

maximum displacemnet $= {y}_{0} = \dfrac{mg}{k}$

${y}_{0} \times ( 2\omega{)}^{2} = 2g$

$\omega = \sqrt{\dfrac{g}{2{y}_{0}}}$