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A spring mass system (mass m, spring constant k and natural length ℓ) rests 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, rotated about it's axis with an angular velocity ω, (k > mω²) the relative change in the length of the spring is best given by the option -

1. mω²/3k
2. 2mω²/k
3. mω²/k
4. √(2/3) (mω²/k)

Correct answer: mω²/k

Solution

The correct option, mω²/k, represents the relationship between the centripetal force required to keep the mass in circular motion and the restoring force of the spring. Since the spring's extension is directly proportional to the force applied, this formula accurately captures how the spring stretches under the influence of the rotational motion.

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