A mass of 0.5 kg is attached to one end of a massless spring whose… | JEE Main 2022 (30 Jul, Shift 2)

Q29JEE Main 2022 · 30 JUL · SHIFT 2Medium

Centripetal Acceleration & Force

A mass of $0.5\text{ kg}$ is attached to one end of a massless spring whose natural length is $1\text{ m}$ and spring constant is $800\text{ N/m}$ The other end of the spring is fixed while the mass moves in a circular path in a horizontal plane with an angular speed of $20\text{ rad/s}$ . The extension in the length of the spring will be:

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Answer: (c)

Spring force gives the centripetal force: $kx = m\omega^2(l + x)$ .
$m\omega^2 = 0.5 \times 400 = 200$ , so $800x = 200(1 + x)$ .
$600x = 200 \Rightarrow x = \frac{1}{3}\text{ m}$
So the answer is (c) $\frac{1}{3}\text{ m}$ .

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