The given arrangement is in vertical plane and initially the springs are in natural length. The uniform rod is hinged at end O, about which it can freely rotate. The angular frequency of small oscillations of the rod is n/8 √(k/m). Find n. (Given: m = mass of rod, L = length of rod and k = mg/L)

$A_1=\sqrt[]{{\left(5\right)}^2+{\left(5\sqrt[]{3}\right)}^2}=10$

$A_2=5$

Then, $\frac{A_1}{A_2}=\frac{10}{5}=\frac{2}{1}$

$y=A\sin ?\frac{2\pi t}{T}$

$t_2-t_1=\frac{T}{2\pi }\left. [{\sin }^{-1}?\frac{x_1}{A}-{\sin }^{-1}?\frac{x_2}{A}\right]$

y = A sin (2πt/T)
t? - t? = (T/2π) [sin? ¹ (x? /A) - sin? ¹ (x? /A)]

Displacement equation of SHM of frequency ' $n$  '

$\mathrm{x}=\mathrm{A}\mathrm{s}\mathrm{i}\mathrm{n}?\left(\omega \mathrm{t}\right)=\mathrm{A}\mathrm{s}\mathrm{i}\mathrm{n}?\left(2\pi \mathrm{n}\mathrm{t}\right)$

Now,

Potential energy $U=\frac{1}{2}{\mathrm{k}\mathrm{x}}^2=\frac{1}{2}{\mathrm{K}\mathrm{A}}^2{\mathrm{s}\mathrm{i}\mathrm{n}}^2?2\pi \mathrm{n}\mathrm{t}$

$=\frac{1}{2}{\mathrm{k}\mathrm{A}}^2\left. [\frac{1-\mathrm{c}\mathrm{o}\mathrm{s}?2\pi 2\mathrm{n}\mathrm{t}}{2}\right]$

So frequency of potential energy $=2n$