A tunnel is dug in the earth across one of its diameter. Two masses m and 2m are dropped from the two ends of the tunnel. The masses collide and stick to each other. They perform SHM, The amplitude of which is (R-Radius of earth)

Velocity of each mass can be calculated by using energy conservation.
$\begin{array}{l}-\frac{3}{2}\text{G}\frac{Mm}{R}+\frac{1}{2}{\text{mv}}^2=\frac{-GMm}{R}\\ \frac{1}{2}{\text{mv}}^2=\frac{GMm}{2R}\\ \text{V = }\sqrt{\frac{GM}{R}}\end{array}$
$\begin{array}{l}\text{After collision}\\ 2\text{m}\sqrt{\frac{Gm}{R}}-m\sqrt{\frac{Gm}{R}}=3m{V}^1\\ {\text{⇒V}}^1=\frac{1}{3}\sqrt{\frac{GM}{R}}\end{array}$

$\begin{array}{l}\mathrm{Total} \mathrm{energy} \mathrm{at} \mathrm{mean} \mathrm{position}=\mathrm{Total} \mathrm{energy} \mathrm{at} \mathrm{extreme} \mathrm{position}\\ \frac{1}{2}{\text{3mV}}^{1^2}+\frac{-3}{2}\frac{3\mathrm{GMm}}{\mathrm{R}}=-\frac{3\mathrm{GMm}}{{\mathrm{R}}^3}(\frac{3}{2}{\mathrm{R}}^2-\frac{{\text{A}}^2}{2})\end{array}$