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 $\Delta |\ll |$ from its equilibrium position and released, it is found to oscillate at frequency $\frac{\delta \pi }{1}\sqrt{\frac{k}{m}}.$ The value of $\delta$ is

Answer 1

Answer: 1

as we know, for spring block system,
$f=\frac{1}{2\pi }\sqrt{\frac{K}{m}}$
so we can also observe $k^{{}^{\prime }}$ as $\frac{d^{2}U}{d{x}^2}|$ from eq.
$U=\frac{1}{2}k{x}^2$
$\frac{dU}{dx}=Kx$
$\frac{d^{2}U}{d{x}^2}=K$
So we can say,
$f=\frac{1}{2\pi }\sqrt{\frac{d^{2}U/d{x}^2}{m}}$
so by above concept let us find $P.E.,(u)$ of system from eqbm. As given, negligible unstretched length, so spring length is elongated length only.
$U=\frac{1}{2}k{x}^2+\frac{p\times q}{4\pi {ϵ}_0{x}^2}\dots ..$ (i)
$\frac{dU}{dx}=kx-\frac{2pq}{4\pi {\in }_0{x}^3}$
(for $eq{b}^m,\frac{dU}{dx}=0$ ) given $eq{b}^m$ at $x=\ell$
$\therefore K\ell =\frac{2pq}{4\pi {\in }_0{\ell }^3}\dots$ (2)
diff. eq (1) again, $\frac{d^{2}U}{d{x}^2}=K+\frac{6pq}{4\pi {\in }_0{x}^4}$
as $eq{b}^m$ at $x=\ell ,{\therefore \frac{d^{2}U}{d{x}^2}|}_{\text{at }n=\ell }=K+\frac{6pq}{4\pi {\in }_0{\ell }^4}$
(from (2) put value of $\frac{2pq}{4\pi {ϵ}_0{\ell }^3}$ )
${\frac{d^{2}U}{d{x}^2}|}_{\text{at }x=\ell }=k+3K=4K$
$\therefore f=\frac{1}{2\pi }\sqrt{\frac{d^{2}U/d{x}^2}{m}}=\frac{1}{2\pi }\sqrt{\frac{4K}{m}}=\frac{1}{\pi }\sqrt{\frac{k}{m}}$
$\therefore \delta =1$ Ans.