Physics · Simple Harmonic Motion

JEE Advanced 2024 — Paper 1 — Question 20

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 [ $\varepsilon_{0}$ is the permittivity of free space.]

Option A:
$q^{2} /\left(4 \pi \varepsilon_{0} R^{3} m\right)$
Option B:
$q^{2} /\left(32 \pi \varepsilon_{0} R^{3} m\right)$
Correct
Option C:
$q^{2} /\left(8 \pi \varepsilon_{0} R^{3} m\right)$
Option D:
$q^{2} /\left(16 \pi \varepsilon_{0} R^{3} m\right)$

Answer: B

Step-by-step solution

So, $a_{t}=$ tangential acceleration of bead $=\frac{k q q^{2} \sin \left(\frac{\theta}{2}\right)}{4 m R^{2} \cos ^{2}\left(\frac{\theta}{2}\right)}$ So, $\mathrm{a}_{\mathrm{t}} \approx \frac{\mathrm{kq}^{2}}{4 \mathrm{mR}^{2}}\left(\frac{\theta}{2}\right)=\frac{\mathrm{q}^{2} \mathrm{x}}{32 \pi \varepsilon_{0} \mathrm{R}^{3} \mathrm{~m}}$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2024
Paper: Paper 1
Subject: Physics
Chapter: Simple Harmonic Motion
Topic: Forced and Damped Oscillations

A block of mass 5 kg moves along the x -direction subject to the force F=(-20 x+10) N , with the value of x in metre. At time t=0 s , it is…

Answer: B

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