Question

One end of a spring of negligible unstretched length and spring constant $k$ is fixed at the origin $(0,0)$. a point particle of mass $m$ carrying a positive charge $q$ is attached at its other end. The entire system is kept on a smooth horizontal surface. When a point dipole $\vec{p}$ pointing towards the charge $q$ is fixed at the origin, the spring gets stretched to a length I and attains a new equilibrium position (see figure below). If the point mass is now displaced slightly by $\Delta l < < l$ from its equilibrium position and released, it is found to oscillate at frequency $\frac{1}{\delta} \sqrt{\frac{k}{m}}$. The value of $\delta$ is ______.

Answer 1

Potential energy of a system after a small displacement.
$U=\frac{1}{2} k x^{2}+\frac{k^{\prime} p q}{(\ell+x)^{2}}$
$\frac{d U}{d x}=k x-\frac{2 k^{\prime} p q}{(\ell+x)^{3}} $
$\frac{d^{2} U}{d x^{2}}=k+\frac{6 k^{\prime} p q}{(\ell+x)^{4}}=k+3 k=4 k$
$f=\frac{1}{2 \pi} \sqrt{\frac{4 k}{m}}$
$f=\frac{1}{\pi} \sqrt{\frac{k}{m}}$
So, $ \delta=3.14$
and for angular frequency
$\omega =\frac{1}{\delta} \sqrt{\frac{ k }{ m }} $
$\delta =0.5$