Question

A spring mass system (mass $m$, spring constant $k$ and natural length of spring $l$) rests in equilibrium, on a horizontal disc. The free end of the spring is fixed at the centre of the disc. If the disc, together with spring mass system, rotates about it's axis with an angular velocity $\omega ,(k>>m{\omega }^2)$ the relative change in the length of the spring is best given by the option:

• A. $\sqrt{\frac{2}{3}\left(\frac{m\omega }{k}\right)}$
• B. $\frac{2m{\omega }^2}{k}$
• C. $\frac{m{\omega }^2}{k}$
• D. $\frac{m{\omega }^2}{3k}$

Answer 1

The correct option is C $\frac{m{\omega }^2}{k}$


The free body diagram in the frame of disc is as shown below.


$\therefore m{\omega }^2(l+x)=kx$

$\Rightarrow x=\frac{ml{\omega }^2}{k-m{\omega }^2}$

For $k>>m{\omega }^2$

$\Rightarrow \frac{x}{l}=\frac{m{\omega }^2}{k}$

Hence, $\left(C\right)$ is the correct answer.