Question

A particle of mass m is moving along the x-axis under the potential $V (x) = \frac{kx^2}{2} + \frac{\lambda}{x}$, where $k$ and $\lambda$ are positive constants of appropriate dimensions. The particle is slightly displaced from its equilibrium position. The particle oscillates with the angular frequency $\omega$ given by

• A. $3 \frac{k}{m}$
• B. $3 \frac{m}{k}$
• C. $\sqrt{\frac{k}{m}}$
• D. $\sqrt{ 3 \frac{m}{k}}$

Answer 1

Answer: (C)

$\sqrt{\frac{k}{m}}$

Answer 2

Given, $V(x)=\frac{k x^{2}}{2}+\frac{\lambda}{x}$
$\because$ In electrical analogy,
$\omega=\frac{1}{\sqrt{L C}}$
but in mechanical analogy $L$ and $C$
$\omega$ will be transformed into $m$ and $\frac{1}{k}$.
Here, $m$ is mass of particle.
Hence, angular frequency, $\omega=\frac{1}{\sqrt{m\left(\frac{1}{k}\right)}}$ or $\omega=\sqrt{\frac{k}{m}}$