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

NTA AbhyasNTA Abhyas 2022

Solution:

Given, $V \left(\right. x \left.\right) = \frac{k x^{2}}{2} + \frac{\lambda }{x^{'}}$
$F=\frac{- \partial V \left(\right. x \left.\right)}{\partial x}= \, -kx$
$a=\frac{- k}{m}x$
$\omega ^{2}=\frac{k}{m}$
$\omega =\sqrt{\frac{k}{m}}$