A tunnel is dug in the earth across one of its diameter. Two masses $‘m’\,\& \,‘2m’$ are dropped from the ends of the tunnel. The masses collide and stick to each other and perform $S.H.M.$ Then amplitude of $S.H.M.$ will be : [$R =$ radius of the earth]
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$2 m u-m u=3 m v$
$\Rightarrow v=u / 3 ; \frac{1}{2} m u^{2}=\frac{3 G M m}{2 R}-\frac{G M m}{R}$
$u=\sqrt{\frac{G m}{R}}$
$\Delta U(r)=+\frac{G M \times 3 m}{2 R^{3}} A^{2}=\frac{1}{2} \times 3 m \cdot v^{2}$
or $\frac{G M}{2 R^{3}} A^{2}=\frac{1}{2} \cdot \frac{1}{9} \cdot \frac{G M}{R}$
$\therefore \quad \frac{A^{2}}{R^{2}}=\frac{1}{9} \Rightarrow A=\frac{R}{3}$
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