Q40 / 54 · JEE Advanced 2026 (Paper 2) · Physics
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 $\vec{F}(r) = -\dfrac{k}{r^2}\hat{r}$. 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:
- Correct answerA$\dfrac{2\pi\ell^3}{m k^2}$
- B$2\pi\sqrt{\dfrac{m}{k}}$
- C$\dfrac{2\pi\ell^3}{3 m k^2}$
- D$\dfrac{2\pi\ell^3}{5 m k^2}$
Solution
The equation of motion for the radial distance $r$ is given by:
$m \dfrac{d^2r}{dt^2} = F(r) + \dfrac{\ell^2}{mr^3} = -\dfrac{k}{r^2} + \dfrac{\ell^2}{mr^3}$
For a circular orbit of radius $r_0$, the radial acceleration is zero:
$-\dfrac{k}{r_0^2} + \dfrac{\ell^2}{mr_0^3} = 0 \Rightarrow r_0 = \dfrac{\ell^2}{mk}$
Let the particle be displaced by a small distance $x$ such that $r = r_0 + x$. The restoring force is:
$m \dfrac{d^2x}{dt^2} = -\dfrac{k}{(r_0+x)^2} + \dfrac{\ell^2}{m(r_0+x)^3}$
Using the binomial expansion for $x \ll r_0$:
$m \dfrac{d^2x}{dt^2} \approx -\dfrac{k}{r_0^2}\left(1 - \dfrac{2x}{r_0}\right) + \dfrac{\ell^2}{mr_0^3}\left(1 - \dfrac{3x}{r_0}\right)$
Since $\dfrac{k}{r_0^2} = \dfrac{\ell^2}{mr_0^3}$, the constant terms cancel out:
$m \dfrac{d^2x}{dt^2} \approx \left( \dfrac{2k}{r_0^3} - \dfrac{3\ell^2}{mr_0^4} \right) x$
Substituting $\dfrac{\ell^2}{mr_0^4} = \dfrac{k}{r_0^3}$ into the equation:
$m \dfrac{d^2x}{dt^2} = \left( \dfrac{2k}{r_0^3} - \dfrac{3k}{r_0^3} \right) x = -\dfrac{k}{r_0^3} x$
This represents simple harmonic motion with angular frequency $\omega = \sqrt{\dfrac{k}{mr_0^3}}$.
The time period is $T = \dfrac{2\pi}{\omega} = 2\pi \sqrt{\dfrac{mr_0^3}{k}}$.
Substituting $r_0 = \dfrac{\ell^2}{mk}$:
$T = 2\pi \sqrt{\dfrac{m}{k} \left( \dfrac{\ell^2}{mk} \right)^3} = 2\pi \sqrt{\dfrac{m \ell^6}{k m^3 k^3}} = \dfrac{2\pi \ell^3}{m k^2}$
Answer: $\dfrac{2\pi\ell^3}{m k^2}$
$m \dfrac{d^2r}{dt^2} = F(r) + \dfrac{\ell^2}{mr^3} = -\dfrac{k}{r^2} + \dfrac{\ell^2}{mr^3}$
For a circular orbit of radius $r_0$, the radial acceleration is zero:
$-\dfrac{k}{r_0^2} + \dfrac{\ell^2}{mr_0^3} = 0 \Rightarrow r_0 = \dfrac{\ell^2}{mk}$
Let the particle be displaced by a small distance $x$ such that $r = r_0 + x$. The restoring force is:
$m \dfrac{d^2x}{dt^2} = -\dfrac{k}{(r_0+x)^2} + \dfrac{\ell^2}{m(r_0+x)^3}$
Using the binomial expansion for $x \ll r_0$:
$m \dfrac{d^2x}{dt^2} \approx -\dfrac{k}{r_0^2}\left(1 - \dfrac{2x}{r_0}\right) + \dfrac{\ell^2}{mr_0^3}\left(1 - \dfrac{3x}{r_0}\right)$
Since $\dfrac{k}{r_0^2} = \dfrac{\ell^2}{mr_0^3}$, the constant terms cancel out:
$m \dfrac{d^2x}{dt^2} \approx \left( \dfrac{2k}{r_0^3} - \dfrac{3\ell^2}{mr_0^4} \right) x$
Substituting $\dfrac{\ell^2}{mr_0^4} = \dfrac{k}{r_0^3}$ into the equation:
$m \dfrac{d^2x}{dt^2} = \left( \dfrac{2k}{r_0^3} - \dfrac{3k}{r_0^3} \right) x = -\dfrac{k}{r_0^3} x$
This represents simple harmonic motion with angular frequency $\omega = \sqrt{\dfrac{k}{mr_0^3}}$.
The time period is $T = \dfrac{2\pi}{\omega} = 2\pi \sqrt{\dfrac{mr_0^3}{k}}$.
Substituting $r_0 = \dfrac{\ell^2}{mk}$:
$T = 2\pi \sqrt{\dfrac{m}{k} \left( \dfrac{\ell^2}{mk} \right)^3} = 2\pi \sqrt{\dfrac{m \ell^6}{k m^3 k^3}} = \dfrac{2\pi \ell^3}{m k^2}$
Answer: $\dfrac{2\pi\ell^3}{m k^2}$
Hence, the correct answer is option: 'A'.