A rod of mass m and length L is pivoted at point O in a car whose acceleration towards left is a₀. The rod is free to rotate in vertical plane. In equilibrium state, the rod remains horizontal when other end is suspended by a spring of force constant k. The time period of small oscillations of rod is $\mathrm{T}=\frac{2\mathrm{\pi }}{\mathrm{D}\sqrt{3}}.$The value of D is ___________. (Given, $\left.\mathrm{k}=20{\mathrm{Nm}}^{-1},{\mathrm{a}}_0=10{\mathrm{ms}}^{-2},\mathrm{m}=1\mathrm{kg},\mathrm{L}=1\mathrm{m}\right)$

In equilibrium, $\mathrm{mg}\left(\frac{\mathrm{L}}{2}\right)=\mathrm{kyL}$

$\therefore \mathrm{y}=\frac{\mathrm{mg}}{2\mathrm{k}}(\text{ initial extension of spring) }$

During oscillations, consider a situation when the angular displacement of rod is $\mathrm{\theta }$ as shown in the figure.

${\mathrm{ma}}_0=$ Pseudo force
ln the reference frame of car, restoring torque is.

$\tau =-\left[m{a}_0\frac{L}{2}\sin \theta +my\left(\frac{L}{2}\cos \theta \right)-k(y-L\theta )L\cos \theta \right]$

$\text{ Using Eq. (i), }\mathrm{\tau }=-\left[\frac{{\mathrm{ma}}_0\mathrm{L}}{2}\mathrm{\theta }+{\mathrm{kL}}^2\mathrm{\theta }\right](\because \mathrm{\theta }\text{ is small })$

$\begin{array}{r}\mathrm{\tau }=-\left[\frac{{\mathrm{ma}}_0\mathrm{L}}{2}+{\mathrm{kL}}^2\right]\mathrm{\theta }\\ \therefore \mathrm{I\alpha }=-\left[\frac{{\mathrm{ma}}_0\mathrm{L}}{2}+\mathrm{kL}\right]\cdot \mathrm{L\theta }\\ \mathrm{\alpha }=-\frac{\left[\frac{{\mathrm{ma}}_0}{2}+\mathrm{kL}\right]}{{\mathrm{mL}}^2/3}\mathrm{\theta }=-\left[\frac{3{\mathrm{a}}_0}{2\mathrm{L}}+\frac{3\mathrm{k}}{\mathrm{m}}\right]\mathrm{\theta }\end{array}$

Using the given values, we get $\mathrm{\alpha }=-75\mathrm{\theta }$

$\begin{array}{l}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{\omega }=\sqrt{75}=5\sqrt{3}\Rightarrow \mathrm{T}=\frac{2\mathrm{\pi }}{\mathrm{\omega }}=\frac{2\mathrm{\pi }}{5\sqrt{3}}\\ \therefore \mathrm{D}=5\end{array}$