Question

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]

• A. $R$
• B. $\frac{R}{2}$
• C. $\frac{R}{3}$
• D. $\frac{2R}{3}$

Answer 1

The correct option is C $\frac{R}{3}$

gravitational potential at center $=\frac{-3}{2}\left(\frac{G{M}_e}{R_e}\right)$

gravitational potential at surface $=\frac{-G{M}_e}{R_e}$

acceleration of each particle $=\frac{-G{M}_x}{R_e^3}$, this is independent of mass so both of these particles will meet at the center of earth...

applying energy conservation for mass $m$

$\frac{m{V}^2}{2}-\frac{3}{2}\left(\frac{G{M}_{e}m}{R_e}\right)=\frac{-G{M}_{e}m}{R_e}$

here in the above eq mass cancels out so velocity is independent of mass of a particle

$Vm=V^{2}m=V=(\frac{G{M}_e}{R_e}{)}^{\frac{1}{2}}................1$

now applying momentam conservation at center of the earth

$-mV+2mV=(m+2m)V_1$ (due to opposite direction one of the velocity is -ve)

$3m{V}_1=mV$

$V_1=\frac{V}{3}...............2$

now this combined mass will perform SHM because force acting on this mass is $\frac{-G{M}_e^{3}mx}{R^3}$

$F=3m\left(a\right)$

a (accleration)$=\frac{-G{M}_{e}x}{R^3}$

$-W2x=\frac{-GMex}{R_e^3}(a=-w^{2}x)$

$W={\left(\frac{G{M}_e}{R^3}\right)}^{\frac{1}{2}}.........3$

now the maximum velocity of a particle is at a mean position which is given by eq 2,

$V_{max}=\frac{V}{3}$

$V_{max}=AW$ (from eq of shm)

$AW=\frac{V}{3}$

$A=\frac{V}{3}W$ (A is amplitude of SHM)

from $eq3$ & $eq1$

$A=\frac{R}{3}$

thus amplitude will be ${\frac{1}{3}}^{rd}$ times radius of earth