For an SHM oscillator, the amplitude is 5 cm and its time period is 4 seconds. The minimum time taken by the particle to pass between points which are at distances 4 cm and 3 cm from the centre of oscillation on the same side of it will be

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

$A_2=5$

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

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

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

Given mg = kL
∴ Iα = (kLθ.L + k (L/2)²θ - mg (L/2)θ)
(mL²/3)α = kL² (3/4)θ (restoring torque)
α = (9k/4m)θ
∴ ω = (3/2)√ (k/m)

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 $=2n$