JEE Advanced · 2024 · 26 May 2024 · Paper 1 · Elasticity & SHM

Two beads, each with charge

q

and mass

m

, are on a horizontal, frictionless, non-conducting, circular hoop of radius

R

. One of the beads is glued to the hoop at some point, while the other one performs small oscillations about its equilibrium position along the hoop. The square of the angular frequency of the small oscillations is given by [

ε_{0}

is the permittivity of free space]

(A) $q^{2} / (4 π ε_{0} R^{3} m)$
(B) $q^{2} / (32 π ε_{0} R^{3} m)$ ✓ Correct
(C) $q^{2} / (8 π ε_{0} R^{3} m)$
(D) $q^{2} / (16 π ε_{0} R^{3} m)$

Answer: (B)

Solution

At equilibrium, the beads are diametrically opposite. For small angular displacement

θ

, the restoring force is

q E sin (θ /2)

where

E = \frac{1}{4 π ε _{0}} \frac{q}{4 R ^{2} c o s ^{2} ( θ /2 )}

. The restoring torque is

τ = q E R sin (θ /2) = I α

. For small

θ

sin (θ /2) \approx θ /2

and

cos (θ /2) \approx 1

, giving

\frac{q ^{2} R}{32 π ε _{0} R ^{2}} θ = m R^{2} α

. This yields

ω^{2} = \frac{q ^{2}}{32 π ε _{0} m R ^{3}}

JEE Advanced · 2024 · 26 May 2024 · Paper 1 · Single correct option

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JEE Advanced · 2024 · 26 May 2024 · Paper 1 · Elasticity & SHM

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