Question:
A spring mass system (mass \(m\), spring constant \(k\) 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 \(\omega\), (\(k \gg m\omega^2\)) the relative change in the length of the spring is best given by the option :
A spring mass system (mass \(m\), spring constant \(k\) 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 \(\omega\), (\(k \gg m\omega^2\)) the relative change in the length of the spring is best given by the option :
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The spring force supplies the centripetal force on the mass moving in a circle of radius \(l + x\). Use \(k \gg m\omega^2\) to simplify.
Updated On: Nov 5, 2026
- \(\frac{m\omega^2}{3k}\)
- \(\frac{m\omega^2}{k}\)
- \(\frac{2m\omega^2}{k}\)
- \(\sqrt{\frac{2}{3}}\left(\frac{m\omega^2}{k}\right)\)
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The Correct Option is B
Solution and Explanation
Let the extension be \(x\). The mass moves in a circle of radius \(l + x\), and the spring force provides the centripetal force: \(kx = m\omega^2(l + x)\).
\(x(k - m\omega^2) = m\omega^2 l\), so \(\frac{x}{l} = \frac{m\omega^2}{k - m\omega^2}\).
Since \(k \gg m\omega^2\), \(k - m\omega^2 \approx k\), hence \(\frac{x}{l} \approx \frac{m\omega^2}{k}\).
Answer: (2) \(\frac{m\omega^2}{k}\)
\(x(k - m\omega^2) = m\omega^2 l\), so \(\frac{x}{l} = \frac{m\omega^2}{k - m\omega^2}\).
Since \(k \gg m\omega^2\), \(k - m\omega^2 \approx k\), hence \(\frac{x}{l} \approx \frac{m\omega^2}{k}\).
Answer: (2) \(\frac{m\omega^2}{k}\)
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