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 ${}^{\prime }{y}^{\prime }$ is measured from the lower end of unstretched spring. Then $\omega$ is :

• A. $\sqrt{\frac{g}{y_0}}$
• B. $\sqrt{\frac{g}{2y_0}}$
• 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 : $\frac{y_0}{2}$

$\frac{y_0}{2}=\frac{mg}{K}$
$2\omega =\sqrt{\frac{K}{m}}=\sqrt{\frac{2g}{y_0}}$
$\omega =\sqrt{\frac{g}{2y_0}}$