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(C) Let the displacement from equilibrium be $x = \Delta \ell$.At equilibrium length $\ell$,the spring force $F_{sp} = k\ell$ balances the electric fo

Vedclass · Accepted Answer

$3.14$

Vedclass · Answer

(C) Let the displacement from equilibrium be $x = \Delta \ell$.At equilibrium length $\ell$,the spring force $F_{sp} = k\ell$ balances the electric force $F_e = \frac{2kpq}{\ell^3}$. Thus,$k\ell = \frac{2kpq}{\ell^3}$.When displaced by $x$,the net restoring force is $F_{net} = F_{sp} - F_e = k(\ell + x) - \frac{2kpq}{(\ell + x)^3}$.Using the binomial approximation $(1 + \frac{x}{\ell})^{-3} \approx 1 - \frac{3x}{\ell}$ for $x \ll \ell$:$F_{net} = k\ell + kx - \frac{2kpq}{\ell^3}(1 + \frac{x}{\ell})^{-3} \approx k\ell + kx - \frac{2kpq}{\ell^3}(1 - \frac{3x}{\ell})$.Substituting $k\ell = \frac{2kpq}{\ell^3}$:$F_{net} = k\ell + kx - k\ell(1 - \frac{3x}{\ell}) = k\ell + kx - k\ell + 3kx = 4kx$.The effective spring constant is $k_{eff} = 4k$.The frequency of oscillation is $f = \frac{1}{2\pi} \sqrt{\frac{k_{eff}}{m}} = \frac{1}{2\pi} \sqrt{\frac{4k}{m}} = \frac{2}{2\pi} \sqrt{\frac{k}{m}} = \frac{1}{\pi} \sqrt{\frac{k}{m}}$.Comparing this with $\frac{1}{\delta} \sqrt{\frac{k}{m}}$,we get $\delta = \pi \approx 3.14$.

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