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# 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 -----------------

509 Points
10 years ago

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