JEE Mains · Physics · STD 11 - 4.1 newtons laws of motion
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
- A \(\frac{\mathrm{m} \ell \omega^{2}}{\mathrm{k}+\mathrm{m} \omega^{2}}\)
- B \(\frac{\mathrm{m} \ell \omega^{2}}{\mathrm{k}-\mathrm{m} \omega^{2}}\)
- C \(\frac{\mathrm{m} \ell \omega^{2}}{\mathrm{k}-\mathrm{m} \omega}\)
- D \(\frac{\mathrm{m} \ell \omega^{2}}{\mathrm{k}+\mathrm{m} \omega}\)
Answer & Solution
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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