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 -----------------

gravitational potential at center  = -3/2 (GM_e/R_e)

gravitational potential at surface = -GMe_/R_e

accleration of each particle = -GMx/R_e³ , this is independent of mass so both of these particle will meet at center of earth...

applying energy conservation for mass m

 mV²/2 - 3/2(GM_em/R_e) = -GM_em/R_e 

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

 V_m = V_2m = V = (GM_e/R_e)^1/2                    _{................1}

now applying momentam conservation at center of earth

-mV + 2mV = (m+2m)V¹                           (due to opposite direction one of the velocity is -ve)

  3mV¹ = mV

      V¹ = V/3        ...............2   

now this combined mass will perform SHM because force acting on this mass is -GM_e3mx/R_e³

    F=3m(a)

   a (accleration)= -GM_ex/R_e³

     -W² x = -GM_ex/R_e³                                  (a = -w²x)

      W = (GM_e/R_e³)^1/2                .........3

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

 V_max = V/3

 V_max = AW                             (from eq of shm)

    AW = V/3

    A = V/3W                           (A is amplitude of SHM)

from eq3 & eq1

  A = R_e/3

thus ampitude will be 1/3rd times radius of earth