Q: 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_{o} sin^{2} ω t , Where ‘y’ is measured from the lower end of unstretched spring . The ω is

(a) $\displaystyle \frac{1}{2} \sqrt{\frac{g}{y_o}} $

(b) $\displaystyle \sqrt{\frac{g}{y_o}} $

(c) $\displaystyle \sqrt{\frac{g}{2 y_o}} $

(d) $\displaystyle \sqrt{\frac{2 g}{y_o}} $

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Sol: $\displaystyle y(t) = y_o sin^2 \omega t $

$\displaystyle y(t) = \frac{y_o}{2} (1-cos2 \omega t) $

Here angular frequency of oscillation ω’ = 2 ω

$\displaystyle \omega’ = \sqrt{\frac{k}{m}} $

$\displaystyle 2 \omega = \sqrt{\frac{k}{m}} $

$\displaystyle \omega = \frac{1}{2} \sqrt{\frac{k}{m}} $

Also $\large mg = k (Amplitude ) = k \frac{y_o}{2}$

$\displaystyle \frac{k}{m} = \frac{2 g}{y_o}$

$\displaystyle \omega = \sqrt{\frac{g}{2 y_o}} $