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$\overrightarrow{p}$pointing towards the charge $q$ is fixed at the origin, the spring gets stretched to a length $l$ 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

Let us extend the spring $x$ distance from its mean point .
$\Delta l\to x$
At $l$ : $F_e$ =$F_{sp}$

$kl$ = $\frac{2kpq}{l^3}$

The net force at displaced position :

$F_{net}=F_{sp}-F_e=k(l+x)-\frac{q\left(2kp\right)}{(l+x{)}^3}$
$F_{net}=k(x+l)$ – $\frac{q\left(2kp\right)}{l^3(1+\frac{x}{l}{)}^3}$

$F_{net}=kx$ + $kl-q\left(\frac{2kp}{l^3}\right)(1-\frac{3x}{l})$

$F_{net}=kx+kl-q\left(\frac{2kp}{l^3}\right)+\frac{2kpq}{l^3}.\frac{3x}{l}$
$F_{net}=kx+kl\frac{3x}{l}=4kx$
$k_{eq}=4k$
$T=2\pi \sqrt{\frac{m}{4k}}=\pi \sqrt{\frac{m}{k}}$
$f=\frac{1}{\pi }\sqrt{\frac{k}{m}}$
So,$\delta =\pi =3.14$