QJEEA-2024-03: JEE Advanced 2024 JEEA-2024-03 Solution: Q.3 Two beads, each with charge 𝑞 and mass 𝑚, are on a…

Understanding the Problem

• We have two charged beads on a circular hoop.
• One bead is fixed, and the other oscillates.
• We need to find the angular frequency (ω) of these oscillations.

Solution

1. Forces:

• The force between the two beads is Coulomb's force: F = k * q^2 / r^2, where k = 1 / (4πε₀).
• Let the equilibrium position of the oscillating bead be directly opposite the fixed bead.
• When the oscillating bead is displaced by a small distance x along the hoop, the distance between the beads is approximately 2R.
• The coulomb force is therefore F = q^2 / (4πε₀(2R)^2) = q^2 / (16πε₀R^2)
• The component of this force that acts along the hoop is responsible for the oscillation.
• Let the angle between the equilibrium position and the displaced position be θ. Therefore Sin(θ) = x / (2R).
• The force along the hoop is therefore Fr = -Fsin(θ) = -(q^2 / (16πε₀R^2)) * (x / (2R)) = -q^2x / (32πε₀R^3)

1. Acceleration:

• Using Newton's second law (F = ma), we find the acceleration (a) of the oscillating bead:
• a = Fr / m = (-q^2x / (32πε₀R^3)) / m = -q^2x / (32πε₀mR^3)

1. Angular Frequency:

• For simple harmonic motion, acceleration is related to angular frequency (ω) and displacement (x) by: a = -ω^2x.
• Comparing this with our expression for acceleration, we get:
• ω^2 = q^2 / (32πε₀mR^3)

Therefore, the correct answer is (B) q^2/(32πε₀R^3m).