A particle of mass m is fixed to one end of a light spring having force constant k and unstretched length. The other end is fixed. The system is given an angular speed ω about the fixed end of the spring such that it rotates in a circle in gravity free space. Then the stretch in the spring is: (a) (mω²)/(k - mω²) (b) (mω²)/(k - mω) (c) (mω²)/(k + mω²) (d) (mω²)/(k + mω)

Answer: (a). Spring force provides the centripetal force; radius of the circle is + x. kx = mω²( + x) x(k - mω²) = mω² ⇒ x = (mω²)/(k - mω²) So the answer is (a).

A particle of mass m is fixed to one end of a light spring having… | JEE Main 2020 (8 Jan, Shift 1)

Q42JEE Main 2020 · 8 JAN · SHIFT 1Difficult

Centripetal Acceleration & Force

A particle of mass m is fixed to one end of a light spring having force constant k and unstretched length $\ell$ . The other end is fixed. The system is given an angular speed $\omega$ about the fixed end of the spring such that it rotates in a circle in gravity free space. Then the stretch in the spring is:

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Answer: (a)

Spring force provides the centripetal force; radius of the circle is $\ell + x$ .
$kx = m\omega^2(\ell + x)$
$x(k - m\omega^2) = m\ell\omega^2 \Rightarrow x = \frac{m\ell\omega^2}{k - m\omega^2}$
So the answer is (a).

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