Question 120 of 210

A particle of mass $m$ is fixed to one end of a light spring having force constant $k$ and unstretched length $l.$ The other end is fixed. The system is given an angular speed $\omega$ about the fixed end of the spring such that it rotates in a circle in gravity free space. Then the stretch in the spring is:

A$\dfrac{ml{\omega }^{2}}{k-\omega m}$B$\dfrac{ml{\omega }^{2}}{k-m{\omega }^{2}}$C$\dfrac{ml{\omega }^{2}}{k+m{\omega }^{2}}$D$\dfrac{ml{\omega }^{2}}{k+m\omega }$

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Answer: (B) $\dfrac{ml{\omega }^{2}}{k-m{\omega }^{2}}$

As we know, the spring force will give the necessary centripetal force for rotation.
So,
$m{\omega }^{2}\left(l+x\right)=kx$
$\left(\dfrac{l}{x}+1\right)=\dfrac{k}{m{\omega }^{2}}$
Thus, the stretch in the spring
$x=\dfrac{lm{\omega }^{2}}{k-m{\omega }^{2}}$