ExamBro
ExamBro
JEE Mains · Physics · STD 11 - 4.1 newtons laws of motion

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

  1. A \(\frac{2 m \omega^{2}}{k}\)
  2. B \(\frac{m \omega^{2}}{3 k}\)
  3. C \(\sqrt{\frac{2}{3}}\left(\frac{\mathrm{m} \omega^{2}}{\mathrm{k}}\right)\)
  4. D \(\frac{m \omega^{2}}{k}\)
Verified Solution

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 \(=\frac{\mathrm{m} \omega^{2}}{\mathrm{k}}\)
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app