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 $\frac{1}{8}\sqrt{\frac{k}{m}}$. The value of $\delta$ is_________.

Answer 1

$T=2\pi \sqrt{\frac{m}{k}}\Rightarrow \delta =\frac{1}{2\pi }\sqrt{\frac{k}{m}}$

$\delta =\frac{1}{2\pi }\sqrt{\frac{(\frac{d^{2}v}{d{x}^2}{)}_{eqPosition}}{m}}$

N.I $\Rightarrow 0$

$U=\frac{1}{2}k{x}^2+\frac{kpq}{x^2}$

$\frac{du}{dx}=kx-\frac{2kpq}{x^3}=0$

$\frac{d^{2}u}{d{x}^2}=k+\frac{6kpq}{x^4}$

At equialibrium $\Rightarrow kl.=\frac{2kpq}{l^3}$

for $x=1eq-$

$\frac{d^{2}u}{d{x}^2}=k+\frac{6kpq}{x^4}=0$

$=k+3\frac{\left(2kpq\right)}{l^4}$

$=k\times 3k$

$=4k$

$\delta =\frac{1}{2\pi }\sqrt{\frac{4k}{m}}$

$=\frac{2}{2\pi }\sqrt{\frac{k}{m}}$

$=\frac{1}{\pi }\sqrt{\frac{k}{m}}$

$\delta =3.14$