If the earth were a homogeneous sphere of radius R and a straight hole bored in it through its centre, show that a particle dropped into the hole will execute shm and find out its period.

Explanation

Acceleration due to gravity at the surface of the earth
`g= (GM )/R^2`
M-Mass of the earth and R-radius
G-Gravitational constant
If `rho` is the density of earth then
`M = 4/3piR^2rho`
`:. g = 4/3 piRrhoG " "...(1)`
At a depth h below the surface of the earth
`g.=4/3 pi(R-h)rhoG" "...(2)`
Dividing eq. (2) by eq. (1)
`(g.)/g=(4/3pi(R-h)rhoG)/(4/3piRrhoG)`
`g.=g""((R-h))/R`
Let x be the distance of the particle from the centre of the earth. Then R-h = x
Then acceleration `=(d^2x)/(dt^2) =g.`
i.e.`(d^2x)/(dt^2) = -(gx)/R`
Negative sign is put to show that acceleration acts towards the centre of the earth which is opposite to the direction of increasing x.
`(d^2x)/(dt^2)+g/Rx=0`
`omega^2=g/R,omega=sqrt(g/R)`
Hence the particle dropped into the hole execute shm
Priod `T=(2pi)/omega=2pisqrt(R/g)`