13.Oscillations
hard

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

A

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

B

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

C

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

D

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

(JEE MAIN-2020)

Solution

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

$y =\frac{ y _{0}}{2}(1-\cos 2 \omega t )$

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

Amplitude : $\frac{y_{0}}{2}$

$\frac{y_{0}}{2}=\frac{m g}{K}$

$2 \omega=\sqrt{\frac{K}{m}}=\sqrt{\frac{2 g}{y_{0}}}$

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

Standard 11
Physics

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