Oscillations · Question 4

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 [ 0 is the permittivity of free space.]

1. q2/(4 0 R3m)
2. q2/(32 0 R3m)
3. q2/(8 0 R3m)
4. q2/(16 0 R3m)

Answer

(2) q2/(32 0 R3m)

Solution

Equilibrium: free bead diametrically opposite the glued bead.

Let the free bead be displaced by angle (at the centre) from equilibrium. Angle between beads at centre = + , separation =2R + 2=2R 2.

U= q24 0 2R ( /2) q28 0 R (1+ 28 ).

So U=const+ 12 ( q232 0 R ) 2, while K= 12mR2 2.

2= q2/(32 0 R)mR2= q232 0 R3m.

Answer: (B)