When a particle of mass $m$ is attached to a vertical spring of spring constant $k$ and released, its motion is described by $y ( t )= y _{0} \sin ^{2} \omega t ,$ where $'y'$ is measured from the lower end of unstretched spring. Then $\omega$ is
JEE MAIN 2020, Diffcult
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$y = y _{0} \sin ^{2} \omega t$
$y =\frac{ y _{0}}{2}(1-\cos 2 \omega t )$
$y -\frac{ y _{0}}{2}=-\frac{ y _{0}}{2} cos 2 \omega t$
Amplitude : $\frac{y_{0}}{2}$
$\frac{y_{0}}{2}=\frac{m g}{K}$
$2 \omega=\sqrt{\frac{K}{m}}=\sqrt{\frac{2 g}{y_{0}}}$
$\omega=\sqrt{\frac{g}{2 y_{0}}}$
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