Q 11-13-126JEE MainJEE Main 2020 (6 Sep, Shift 2)Medium
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 the unstretched spring. Then $\omega$ is:
Answer: (C) $\sqrt{\dfrac{g}{2y_0}}$
$y = y_0\sin^2\omega t = \dfrac{y_0}{2} - \dfrac{y_0}{2}\cos2\omega t$
This is SHM of angular frequency $2\omega$ about the mean position $y = \dfrac{y_0}2$, which must be the equilibrium extension: $\dfrac{y_0}{2} = \dfrac{mg}{k}$.
$$2\omega = \sqrt{\frac km} = \sqrt{\frac{g}{y_0/2}} = \sqrt{\frac{2g}{y_0}} \Rightarrow \omega = \sqrt{\frac{g}{2y_0}}$$
Solution by Sreeraj P, M.Sc Physics