A particle of mass $$ $m$ is fixed to one end of a light spring having force constant $$ $k$ and unstretched length $\mathcal{l}$ $$ . The other end is fixed. The system is given an angular speed $\omega$ $$ 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:

Correct Option - 4


Detailed Solution:

$\mathrm{}m{?}^2(\mathcal{l}+x)=kx$

$\begin{array}{rr}& \left(\frac{\mathcal{l}}{x}+1\right)=\frac{k}{m{?}^2}\\ & x=\frac{\mathcal{l}m{?}^2}{k-m{?}^2}\end{array}$

 

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