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