Question:
A particle of mass \(m\), and angular momentum \(\ell\) is moving in a circular orbit of radius \(r_0\) under the influence of an attractive force \(\vec{F}(r) = -\frac{k}{r^2}\hat{r}\). Keeping its angular momentum unchanged, the particle is displaced radially by a small distance \(\delta r \ll r_0\), due to which its radial distance varies periodically. The corresponding time period is:
A particle of mass \(m\), and angular momentum \(\ell\) is moving in a circular orbit of radius \(r_0\) under the influence of an attractive force \(\vec{F}(r) = -\frac{k}{r^2}\hat{r}\). Keeping its angular momentum unchanged, the particle is displaced radially by a small distance \(\delta r \ll r_0\), due to which its radial distance varies periodically. The corresponding time period is:
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Small oscillations in a potential well are always harmonic. The effective potential method is the most efficient way to solve central force perturbation problems.
Updated On: May 20, 2026
- $\frac{2\pi \ell^3}{mk^2}$
- $2\pi \sqrt{\frac{m}{k}}$
- $\frac{2\pi \ell^3}{3mk^2}$
- $\frac{2\pi \ell^3}{5mk^2}$
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Question:
This problem deals with the stability of a circular orbit. When a particle in a central force field is perturbed, it undergoes radial oscillations. We need to find the period of these oscillations.
Step 2: Key Formula or Approach:
• Effective potential energy: $U_{eff}(r) = \frac{\ell^2}{2mr^2} + U(r)$, where $\vec{F} = -\frac{dU}{dr}\hat{r}$.
• Equilibrium radius $r_0$: Solve $\frac{dU_{eff}}{dr} = 0$.
• Oscillation frequency: $\omega = \sqrt{\frac{K_{eff}}{m}}$, where $K_{eff} = \frac{d^2 U_{eff}}{dr^2}\Big|_{r_0}$.
Step 3: Detailed Explanation:
• Potential energy $U(r)$: $\vec{F} = -\frac{k}{r^2} \implies U(r) = -\frac{k}{r}$.
• Effective potential: $U_{eff}(r) = \frac{\ell^2}{2mr^2} - \frac{k}{r}$.
• Find $r_0$: \[ \frac{dU_{eff}}{dr} = -\frac{\ell^2}{mr^3} + \frac{k}{r^2} = 0 \implies r_0 = \frac{\ell^2}{mk} \]
• Calculate the effective spring constant ($K_{eff}$): \[ \frac{d^2 U_{eff}}{dr^2} = \frac{3\ell^2}{mr^4} - \frac{2k}{r^3} \] Substitute $r_0 = \frac{\ell^2}{mk}$: \[ K_{eff} = \frac{3\ell^2}{m(\ell^2/mk)^4} - \frac{2k}{(\ell^2/mk)^3} = \frac{3m^3 k^4}{\ell^6} - \frac{2m^3 k^4}{\ell^6} = \frac{m^3 k^4}{\ell^6} \]
• Calculate frequency and time period: \[ \omega = \sqrt{\frac{K_{eff}}{m}} = \sqrt{\frac{m^2 k^4}{\ell^6}} = \frac{mk^2}{\ell^3} \] \[ T = \frac{2\pi}{\omega} = \frac{2\pi \ell^3}{mk^2} \]
Step 4: Final Answer:
The time period of radial oscillations is $\frac{2\pi \ell^3}{mk^2}$.
This problem deals with the stability of a circular orbit. When a particle in a central force field is perturbed, it undergoes radial oscillations. We need to find the period of these oscillations.
Step 2: Key Formula or Approach:
• Effective potential energy: $U_{eff}(r) = \frac{\ell^2}{2mr^2} + U(r)$, where $\vec{F} = -\frac{dU}{dr}\hat{r}$.
• Equilibrium radius $r_0$: Solve $\frac{dU_{eff}}{dr} = 0$.
• Oscillation frequency: $\omega = \sqrt{\frac{K_{eff}}{m}}$, where $K_{eff} = \frac{d^2 U_{eff}}{dr^2}\Big|_{r_0}$.
Step 3: Detailed Explanation:
• Potential energy $U(r)$: $\vec{F} = -\frac{k}{r^2} \implies U(r) = -\frac{k}{r}$.
• Effective potential: $U_{eff}(r) = \frac{\ell^2}{2mr^2} - \frac{k}{r}$.
• Find $r_0$: \[ \frac{dU_{eff}}{dr} = -\frac{\ell^2}{mr^3} + \frac{k}{r^2} = 0 \implies r_0 = \frac{\ell^2}{mk} \]
• Calculate the effective spring constant ($K_{eff}$): \[ \frac{d^2 U_{eff}}{dr^2} = \frac{3\ell^2}{mr^4} - \frac{2k}{r^3} \] Substitute $r_0 = \frac{\ell^2}{mk}$: \[ K_{eff} = \frac{3\ell^2}{m(\ell^2/mk)^4} - \frac{2k}{(\ell^2/mk)^3} = \frac{3m^3 k^4}{\ell^6} - \frac{2m^3 k^4}{\ell^6} = \frac{m^3 k^4}{\ell^6} \]
• Calculate frequency and time period: \[ \omega = \sqrt{\frac{K_{eff}}{m}} = \sqrt{\frac{m^2 k^4}{\ell^6}} = \frac{mk^2}{\ell^3} \] \[ T = \frac{2\pi}{\omega} = \frac{2\pi \ell^3}{mk^2} \]
Step 4: Final Answer:
The time period of radial oscillations is $\frac{2\pi \ell^3}{mk^2}$.
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