When the ball is in the tunnel at distance $x$ from the centre of the earth, then gravitational force acting on ball is

$F=\frac{Gm}{x^2}\times \left(\frac{4}{3}\pi x^3\mathrm{\rho }\right)=G\times \left(\frac{4}{3}\pi \mathrm{\rho }\right)mx$

Mass of the earth, $M=\frac{4}{3}\pi R^3\mathrm{\rho }$

or $\frac{4}{3}\pi \mathrm{\rho }=\frac{\mathrm{M}}{{\mathrm{R}}^3}$

$\therefore F=\frac{GMmx}{R^3} ie, F\propto x$

As this $F$ is directed towards the centre of earth $ie$, the mean position so the ball will execute periodic motion about the centre of earth

Here inertia factor=mass of ball$=m$

Spring factor$=\frac{GMm}{R^3}=\frac{gm}{R}$

$\therefore$ time period of oscillation of ball in the tunnel is

$T^{\text{'}}=2\pi \sqrt{\frac{\mathrm{i}\mathrm{n}\mathrm{e}\mathrm{r}\mathrm{t}\mathrm{i}\mathrm{a}\mathrm{ }\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{o}\mathrm{r}}{\mathrm{s}\mathrm{p}\mathrm{r}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{ }\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{o}\mathrm{r}}}$

$=2\mathrm{\pi }\sqrt{\frac{\mathrm{m}}{\mathrm{g}\mathrm{m}/\mathrm{R}}}$

$=2\pi \sqrt{\frac{R}{g}}$
Time spent by ball outside the tunnel on both the sides will be $=4\sqrt{2\text{h}/\text{g}}$

Therefore, total time period of oscillation of ball is

$=2\pi \sqrt{\frac{R}{g}}+4\sqrt{\frac{2h}{g}}$