Question

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.]

(A) \(q^{2} /\left(4 \pi \varepsilon_{0} R^{3} m\right)\)

(B) \(q^{2} /\left(32 \pi \varepsilon_{0} R^{3} m\right)\)

(C) \(q^{2} /\left(8 \pi \varepsilon_{0} R^{3} m\right)\)

(D) \(q^{2} /\left(16 \pi \varepsilon_{0} R^{3} m\right)\)

Answer 1

Correct option is (B) \(q^{2} /\left(32 \pi \varepsilon_{0} R^{3} m\right)\)

As the hoop mass is not given so it must not move or else its inertia must have some effect.

Here \(r=2 R \cos \phi\)

Also \(\theta=2 \phi\)

\(\Rightarrow \theta=\frac{\phi}{2}\)

And \(\theta=\frac{x}{R}\)

If \(\theta\) is the small angular displacement of free charge, then \(F(\phi)=\frac{K q^{2}}{r^{2}}\)

So, restoring force towards mean position is \(F_{(R)}=\frac{K q^{2}}{r^{2}} \sin \phi\)

\(a_{R}=\frac{F_{(R)}}{m}=\frac{-K q^{2}}{m r^{2}} \cdot \sin \phi=\frac{-K q^{2} \cdot \sin \phi}{m 4 R^{2} \cos ^{2} \phi}\)

\(\Rightarrow a_{R}=-\frac{K q^{2}}{4 m R^{2} \cos ^{2}\left(\frac{\phi}{2}\right)} \cdot \sin \left(\frac{\theta}{2}\right) \approx \frac{-K q^{2}}{4 m R^{2}} \frac{1}{2} \cdot \frac{x}{R}=\omega^{2} \cdot x\)

So, \(\omega^{2}=\frac{q^{2}}{32 \pi \varepsilon_{0} m R^{3}}\)