A particle of mass m is fixed to one end of a light spring having force constant k and unstretched length. The other end is fixed. The system is given an angular speed 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) \frac{\mathrm{m} \ell \omega^{2}}{\mathrm{k}+\mathrm{m} \omega^{2}}

(C) \frac{\mathrm{m} \ell \omega^{2}}{\mathrm{k}-\mathrm{m} \omega}

(D) \frac{\mathrm{m} \ell \omega^{2}}{\mathrm{k}+\mathrm{m} \omega}

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JEE Mains · Physics · STD 11 - Newtons laws of motion

JEE Mains 2020
Medium
MCQ

A particle of mass \(\mathrm{m}\) is fixed to one end of a light spring having force constant \(\mathrm{k}\) and unstretched length \(\ell .\) 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

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Correct Answer

(B) \(\frac{\mathrm{m} \ell \omega^{2}}{\mathrm{k}-\mathrm{m} \omega^{2}}\)

Step-by-step Solution

Detailed explanation

\(\mathrm{kx}=\mathrm{m}\ell \omega^{2}+\mathrm{m} \mathrm{x} \omega^{2}\)

\(\mathrm{x}=\frac{\mathrm{m}\ell \omega^{2}}{\mathrm{k}-\mathrm{m} \omega^{2}}\)

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