 # When a particle of mass m is attached to a vertical spring of spring constant k Question:

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

1. (1) $\frac{1}{2} \sqrt{\frac{g}{y_{0}}}$

2. (2) $\sqrt{\frac{g}{y_{0}}}$

3. (3) $\sqrt{\frac{g}{2 y_{0}}}$

4. (4) $\sqrt{\frac{2 g}{y_{0}}}$

Correct Option: , 3

Solution:

(3) $y=y_{0} \sin ^{2} \omega t$

$\Rightarrow y=\frac{y_{0}}{2}(1-\cos 2 \omega t)$                  $\left(\because \sin ^{2} \omega t=\frac{1-\cos 2 \omega t}{2}\right)$

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

$\Rightarrow y=A \cos 2 \omega t$

$\therefore$ Amplitude $=\frac{y_{0}}{2}$

Angular velocity $=2 \omega$

For equilibrium of mass, $\frac{k y_{0}}{2}=m g \Rightarrow \frac{k}{m}=\frac{2 g}{y_{0}}$

Also, spring constant $k=m(2 \omega)^{2}$

$\Rightarrow 2 \omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{2 g}{y_{0}}} \Rightarrow \omega=\frac{1}{2} \sqrt{\frac{2 g}{y_{0}}}=\sqrt{\frac{g}{2 y_{0}}}$  