When a particle of mass m is attached to a vertical spring of spring constant k and released, its motion is described by $y (t) = y_{0} \sin^{2} ω t,$ where $' y'$ is measured from the lower end of unstretched spring. Then $ω$ is :

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$\frac{1}{2} \sqrt{\frac{g}{y_{0}}}$

$\sqrt{\frac{g}{y_{0}}}$

$\sqrt{\frac{g}{2 y_{0}}}$

$\sqrt{\frac{2 g}{y_{0}}}$

answer is C.

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Detailed Solution

$\begin{array}{l} y (t) = y_{o} \sin^{2} (ω t) = \frac{y_{o}}{2} (1 - \cos 2 ω t) \\ \Rightarrow y - \frac{y_{o}}{2} = - \frac{y_{o}}{2} \cos 2 ω t \end{array}$

So elongation in the spring in equilibrium is $\frac{y_{o}}{2}$ .
Therefore $m g = \frac{k y_{o}}{2}$

$\begin{array}{l} \Rightarrow \sqrt{\frac{k}{m}} = \sqrt{\frac{2 g}{y_{o}}} = 2 ω \\ \Rightarrow ω = \sqrt{\frac{g}{2 y_{o}}} \end{array}$

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