JEE Mains · Physics · STD 11 - Newtons laws of motion
- JEE Mains 2020
- Medium
- MCQ
A spring mass system (mass \(m\), spring constant \(\mathrm{k}\) and natural length \(l\) ) rest 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 \(\omega,\left(\mathrm{k} > > \mathrm{m} \omega^{2}\right)\) the relative change in the length of the spring is best given by the option
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(\frac{m \omega^{2}}{k}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{k} \Delta \ell=\mathrm{m} \omega^{2}\left(\ell_{0}+\Delta \ell\right)\)
\(\Delta \ell=\frac{m \omega^{2} \ell_{0}}{k-m \omega^{2}} \approx \frac{m \omega \ell_{0}}{k}\)
\(\frac{\Delta \ell}{\ell_{0}}=\) Relative change \(=\frac{\mathrm{m} \omega^{2}}{\mathrm{k}}\)
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