Question

JEE ADVANCED PHYSICS QUESTION SEPTEMBER 2020

Answer 1

At equilibrium, the restoring force in the spring system gets balanced by the electrostatic force.

$\sf{\longrightarrow kl=F_q}$

$\sf{\longrightarrow kl=qE}$

Along axial line, $\sf{E=\dfrac{2Kp}{l^3}}$ where $\sf{K=\dfrac{1}{4\pi\epsilon_0}.}$ Then,

$\sf{\longrightarrow kl=\dfrac{2Kpq}{l^3}}$

$\sf{\longrightarrow k=\dfrac{2Kpq}{l^4}}$

As the point mass is slightly displaced by $\sf{\Delta l,}$ the net force in the system,

$\sf{\longrightarrow F_{net}=F_q\,\!'-k(l+\Delta l)}$

$\sf{\longrightarrow F_{net}=\dfrac{2Kpq}{(l+\Delta l)^3}-k(l+\Delta l)}$

$\sf{\longrightarrow F_{net}=\dfrac{2Kpq}{l^3}\left[1+\dfrac{\Delta l}{l}\right]^{-3}-k(l+\Delta l)}$

As $\sf{\Delta l<<l,}$

$\sf{\longrightarrow F_{net}=kl\left[1-\dfrac{3\Delta l}{l}\right]-k(l+\Delta l)}$

$\sf{\longrightarrow F_{net}=-4k\Delta l}$

That is,

$\sf{\longrightarrow -k'\Delta l=-4k\Delta l}$

$\sf{\longrightarrow k'=4k}$

Then, frequency,

$\sf{\longrightarrow \nu=\dfrac{1}{2\pi}\sqrt{\dfrac{k'}{m}}}$

$\sf{\longrightarrow \dfrac{1}{\delta}\sqrt{\dfrac{k}{m}}=\dfrac{1}{\pi}\sqrt{\dfrac{k}{m}}}$

$\sf{\longrightarrow\underline{\underline{\delta=\pi}}}$