Time Period of Small Radial Oscillations in Central Force Motion

To find the time period of small radial oscillations of the particle, we analyze the effective radial force and potential acting on the particle in central force motion.

The equation of motion for the radial coordinate $r$ of a particle of mass $m$ with angular momentum $\ell$ under a central force $\vec{F}(r) = -\frac{k}{r^2} \hat{r}$ is given by: $m \ddot{r} = F_{\text{eff}}(r) = -\frac{k}{r^2} + \frac{\ell^2}{m r^3}$

1. Equilibrium Condition (Circular Orbit)

For a stable circular orbit of radius $r_0$, the radial acceleration is zero ($\ddot{r} = 0$). Thus, the net effective force must be zero: $F_{\text{eff}}(r_0) = -\frac{k}{r_0^2} + \frac{\ell^2}{m r_0^3} = 0$

Solving for $r_0$: $\frac{k}{r_0^2} = \frac{\ell^2}{m r_0^3} \implies r_0 = \frac{\ell^2}{mk}$

2. Small Oscillations about Equilibrium

Let the radial distance be displaced slightly by $r = r_0 + \delta r$, where $\delta r \ll r_0$. Using a first-order Taylor series expansion of $F_{\text{eff}}(r)$ around $r = r_0$: $F_{\text{eff}}(r_0 + \delta r) \approx F_{\text{eff}}(r_0) + \left.\frac{d F_{\text{eff}}}{dr}\right|_{r=r_0} \delta r$

Since $F_{\text{eff}}(r_0) = 0$, the equation of motion for small displacement $\delta r$ simplifies to: $m \frac{d^2(\delta r)}{dt^2} = \left.\frac{d F_{\text{eff}}}{dr}\right|_{r=r_0} \delta r$

Calculating the derivative of $F_{\text{eff}}(r)$: $\frac{d F_{\text{eff}}}{dr} = \frac{d}{dr}\left(-\frac{k}{r^2} + \frac{\ell^2}{m r^3}\right) = \frac{2k}{r^3} - \frac{3\ell^2}{m r^4}$

Evaluating this derivative at $r = r_0$: $\left.\frac{d F_{\text{eff}}}{dr}\right|_{r=r_0} = \frac{2k}{r_0^3} - \frac{3\ell^2}{m r_0^4}$

Substituting $\frac{\ell^2}{m} = k r_0$: $\left.\frac{d F_{\text{eff}}}{dr}\right|_{r=r_0} = \frac{2k}{r_0^3} - \frac{3(k r_0)}{r_0^4} = \frac{2k}{r_0^3} - \frac{3k}{r_0^3} = -\frac{k}{r_0^3}$

3. Frequency and Time Period of Oscillations

The equation for the small radial displacement becomes: $m \frac{d^2(\delta r)}{dt^2} = -\frac{k}{r_0^3} \delta r \implies \frac{d^2(\delta r)}{dt^2} + \frac{k}{m r_0^3} \delta r = 0$

This represents a simple harmonic motion (SHM) with angular frequency $\omega$: $\omega = \sqrt{\frac{k}{m r_0^3}}$

The corresponding time period $T$ of the radial oscillations is: $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{m r_0^3}{k}}$

Substituting $r_0 = \frac{\ell^2}{mk}$ into the expression for $T$: $T = 2\pi \sqrt{\frac{m}{k} \left(\frac{\ell^2}{mk}\right)^3} = 2\pi \sqrt{\frac{m}{k} \cdot \frac{\ell^6}{m^3 k^3}} = 2\pi \sqrt{\frac{\ell^6}{m^2 k^4}} = \frac{2\pi \ell^3}{mk^2}$

Thus, the correct option is A.