Question

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):
A) $R$.
B) $\dfrac{R}{2}$.
C) $\dfrac{R}{3}$.
D) $\dfrac{{2R}}{3}$.

Answer 1

A simple harmonic motion is a periodic motion in which there is a restoring force which is acting towards the equilibrium position of the body. Also the restoring force is proportional to the displacement of the body executing simple harmonic motion.

Formula used: The formula of gravitational energy at centre is given by,
${P_c} = \dfrac{{ - 3}}{2}\left( {\dfrac{{G{M_e}m}}{{{R_e}}}} \right)$.
Where G is the universal gravitational constant m is the mass of the body ${M_e}$ is the mass of the earth and ${R_e}$ is the radius of the earth.
The formula of the gravitational energy of body on the earth surface is given by,
${P_s} = \left( {\dfrac{{G{M_e}m}}{{{R_e}}}} \right)$
Where G is the universal gravitational constant m is the mass of the body ${M_e}$ is the mass of the earth and ${R_e}$ is the radius of the earth.

Complete step-by-step answer:
It is given in the problem that a tunnel is dig into the earth along the diameter and two particles are dropped from two opposite sides of mass m and 2m the two masses collide and stick to each other and then start oscillating and we need to find the value of the amplitude of the simple harmonic motion.
The acceleration of each particle is independent of the mass of the particles therefore the two particles will meet at the centre.
The energy of the particle of mass m at initially,
${E_i} = - \left( {\dfrac{{G{M_e}m}}{{{R_e}}}} \right) + 0$
The energy of the particle of mass m at finally,
${E_f} = \dfrac{{ - 3}}{2}\left( {\dfrac{{G{M_e}m}}{{{R_e}}}} \right) + \dfrac{1}{2}m{V_1}^2$
Conserve the energy of the particle of mass m,
$\Rightarrow - \left( {\dfrac{{G{M_e}m}}{{{R_e}}}} \right) = \dfrac{{ - 3}}{2}\left( {\dfrac{{G{M_e}m}}{{{R_e}}}} \right) + \dfrac{1}{2}m{V_1}^2$
$\Rightarrow - \left( {\dfrac{{G{M_e}}}{{{R_e}}}} \right) = \dfrac{{ - 3}}{2}\left( {\dfrac{{G{M_e}}}{{{R_e}}}} \right) + \dfrac{1}{2}{V_1}^2$
$\Rightarrow \dfrac{1}{2}{V_1}^2 = \dfrac{3}{2}\left( {\dfrac{{G{M_e}}}{{{R_e}}}} \right) - \left( {\dfrac{{G{M_e}}}{{{R_e}}}} \right)$
$\Rightarrow {V_1}^2 = \left( {\dfrac{{G{M_e}}}{{{R_e}}}} \right)$
$\Rightarrow {V_1} = \sqrt {\dfrac{{G{M_e}}}{{{R_e}}}}$
The energy of particle of mass 2m initially is,
${E_i} = - \left( {\dfrac{{G{M_e}2m}}{{{R_e}}}} \right) + 0$
The energy of the particle of mass 2m at finally,
${E_f} = \dfrac{{ - 3}}{2}\left( {\dfrac{{G{M_e}2m}}{{{R_e}}}} \right) + \dfrac{1}{2}\left( {2m} \right)\left( {{V_2}^2} \right)$
The energy conservation for the mass 2m.
$\Rightarrow - \left( {\dfrac{{G{M_e}2m}}{{{R_e}}}} \right) + 0 = \dfrac{{ - 3}}{2}\left( {\dfrac{{G{M_e}2m}}{{{R_e}}}} \right) + \dfrac{1}{2}\left( {2m} \right)\left( {{V_2}^2} \right)$
$\Rightarrow - \left( {\dfrac{{G{M_e}2}}{{{R_e}}}} \right) + 0 = \dfrac{{ - 3}}{1}\left( {\dfrac{{G{M_e}}}{{{R_e}}}} \right) + \left( {{V_2}^2} \right)$
$\Rightarrow - 2\left( {\dfrac{{G{M_e}}}{{{R_e}}}} \right) + 0 = - 3\left( {\dfrac{{G{M_e}}}{{{R_e}}}} \right) + \left( {{V_2}^2} \right)$
$\Rightarrow \left( {{V_2}^2} \right) = 3\left( {\dfrac{{G{M_e}}}{{{R_e}}}} \right) + - 2\left( {\dfrac{{G{M_e}}}{{{R_e}}}} \right)$
$\Rightarrow {V_2} = \sqrt {\dfrac{{G{M_e}}}{{{R_e}}}}$
Momentum conservation for the final velocity of both of the particles V’ will be,
$\Rightarrow mV - 2mV = \left( {m + 2m} \right)V'$
$\Rightarrow - mV = 3mV'$
$\Rightarrow - V = 3V'$
$\Rightarrow V' = \dfrac{{ - V}}{3}$
Energy conservation after collision is equal to,
$\Rightarrow {E_i} = P.{E_i} + K.{E_i}$
$\Rightarrow {E_i} = - \left( {\dfrac{3}{2}} \right) \times \left( {\dfrac{{3mG{M_e}}}{{{R_e}}}} \right) + \left( {\dfrac{1}{2} \times 3m} \right) \times \left( {\dfrac{1}{3}\sqrt {\dfrac{{G{M_e}}}{{{R_e}}}} } \right)$
Energy conservation after collision for final energy,
$\Rightarrow {E_f} = P.{E_f} + K.{E_f}$
$\Rightarrow {E_f} = - \dfrac{{3G{M_e}}}{{2{R_e}}}\left( {3{\operatorname{R} ^2}_e - {r^2}} \right) \times 3m + 0$
Applying energy conservation,
$\Rightarrow {E_i} = {E_f}$
$\Rightarrow \left( {\dfrac{{ - 3}}{2}} \right) \times \left( {\dfrac{{G{M_e}}}{{{R_e}}}} \right) \times \left( {3m} \right) + \left( {\dfrac{1}{2}} \right) \times \left( {3m} \right) \times \left( {\dfrac{1}{9}} \right) \times \left( {\dfrac{{G{M_e}}}{{{R_e}}}} \right) = - \left( {\dfrac{{3G{M_e}}}{{2{R_e}}}} \right) \times \left( {3{R^2} - {r^2}} \right) \times 3m$
$\Rightarrow \left( { - \dfrac{{9G{M_e}m}}{{2{R_e}}}} \right) + \left( {\dfrac{{G{M_e}m}}{{6{R_e}}}} \right) = - \left( {\dfrac{{3G{M_e}}}{{2{R_e}}}} \right) \times \left( {3{R_e}^2 - {r^2}} \right) \times 3m$
After solving we get,
$\Rightarrow r = \dfrac{R}{3}$
So the amplitude of oscillation will be$A = \dfrac{R}{3}$.

The correct answer for this problem is option C.

Note: It is advisable to students to remember the formula of the gravitational energy at centre of centre and the formula of gravitational energy at the surface of the earth as it can help to solve these problems.