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

Answer: 3.14

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