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:

(D) $\frac{\mathrm{m} \ell \omega^{2}}{\mathrm{k}-\mathrm{m} \omega^{2}}$

(A) $\frac{\mathrm{m} \ell \omega^{2}}{\mathrm{k}+\mathrm{m} \omega}$

(B) $\frac{\mathrm{m} \ell \omega^{2}}{\mathrm{k}+\mathrm{m} \omega^{2}}$

(C) $\frac{\mathrm{m} \ell \omega^{2}}{\mathrm{k}-\omega \mathrm{m}}$

JEE Main · 2020 · 08 Jan 2020 · Shift 1 · Elasticity & SHM

A particle of mass

m

is fixed to one end of a light spring having force constant

k

and unstretched length

ℓ

. 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:

Answer: (D)

Let the spring stretch by

x

, so the mass travels on a circle of radius

ℓ + x

. The spring force provides the centripetal force,

k x = m ω^{2} (ℓ + x)

. Expanding,

k x - m ω^{2} x = m ω^{2} ℓ

, so

x (k - m ω^{2}) = m ω^{2} ℓ

and the stretch is

x = \frac{m ℓ ω ^{2}}{k - m ω ^{2}}

JEE Main · 2020 · 08 Jan 2020 · Shift 1 · Single correct option