Question

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. $\sqrt{\frac{2g }{ y _{0}}}$

Answer 1

Answer: (B)

$y=y_{0} \sin ^{2} \omega t$
$y =\frac{ y _{0}}{2}(1-\cos 2 \omega t )$
$y-\frac{y_{0}}{2}=-\frac{y_{0}}{2} \cos 2 \omega t$
Amplitude : $\frac{y_{0}}{2}$

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