A spring mass system (mass m, spring constant k and natural length l)… | JEE Main 2020 (9 Jan, Shift 2)

Q82JEE Main 2020 · 9 JAN · SHIFT 2Difficult

Spring Systems

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, rotates about its 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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Answer: (b)

Spring force provides the centripetal force: $kx = m\omega^2(l + x)$ .
$x = \frac{m\omega^2 l}{k - m\omega^2} \approx \frac{m\omega^2 l}{k}$ since $k \gg m\omega^2$ .
Relative change $\frac{x}{l} = \frac{m\omega^2}{k}$ , answer (b).

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Spring Systems question from JEE Main 2020 (9 Jan, Shift 2)

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