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 (GMe/Re)
gravitational potential at surface = -GMe/Re
accleration of each particle = -GMx/Re3 , 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(GMem/Re) = -GMem/Re 
here in the above eq mass cancels out so velocity is independent of mass of particle
 Vm = V2m = V = (GMe/Re)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 -GMe3mx/Re3
    F=3m(a)
   a (accleration)= -GMex/Re3
     -W2 x = -GMex/Re3                                  (a = -w2x)
      W = (GMe/Re3)1/2                .........3
now maximum velocity of particle is at mean position which is given by eq 2 ,
 Vmax = V/3
 Vmax = AW                             (from eq of shm)
    AW = V/3
    A = V/3W                           (A is amplitude of SHM)
from eq3 & eq1
  A = Re/3
thus ampitude will be 1/3rd times radius of earth