JEE Advanced 2020 — Paper 1 — Question 14

Potential energy of a system after a small displacement.

$\mathrm{U}=\frac{1}{2} \mathrm{kx}^{2}+\frac{\mathrm{k}^{\prime} \mathrm{pq}}{(\ell+\mathrm{x})^{2}}$

$\frac{\mathrm{dU}}{\mathrm{dx}}=\mathrm{kx}-\frac{2 \mathrm{k}^{\prime} \mathrm{pq}}{(\ell+x)^{3}}$

$\frac{\mathrm{d}^{2} \mathrm{U}}{\mathrm{dx}^{2}}=\mathrm{k}+\frac{6 \mathrm{k}^{\prime} \mathrm{pq}}{(\ell+\mathrm{x})^{4}}=\mathrm{k}+3 \mathrm{k}=4 \mathrm{k}$

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

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

So, $\delta=3.14$

and for angular frequency

$\omega=\frac{1}{\delta} \sqrt{\frac{k}{m}}$

$\delta=0.5$