When a particle of mass $m$ is attached to a vertical spring of spring constant $k$ and released, its motion is described by $y(t)=y_{0} \sin ^{2} \omega \mathrm{t}$, where ' $y$ ' is measured from the lower end of unstretched spring. Then $\omega$ is :
Correct Option: , 3
(3) $y=y_{0} \sin ^{2} \omega t$
$\Rightarrow y=\frac{y_{0}}{2}(1-\cos 2 \omega t)$ $\left(\because \sin ^{2} \omega t=\frac{1-\cos 2 \omega t}{2}\right)$
$\Rightarrow y-\frac{y_{0}}{2}=\frac{-y_{0}}{2} \cos 2 \omega t$
$\Rightarrow y=A \cos 2 \omega t$
$\therefore$ Amplitude $=\frac{y_{0}}{2}$
Angular velocity $=2 \omega$
For equilibrium of mass, $\frac{k y_{0}}{2}=m g \Rightarrow \frac{k}{m}=\frac{2 g}{y_{0}}$
Also, spring constant $k=m(2 \omega)^{2}$
$\Rightarrow 2 \omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{2 g}{y_{0}}} \Rightarrow \omega=\frac{1}{2} \sqrt{\frac{2 g}{y_{0}}}=\sqrt{\frac{g}{2 y_{0}}}$