Consider the Earth as a homogenous sphere of radius R and a straight hole is bored in it through its centre. Show that a particle dropped into the hole will execute a simple harmonic motion such that its time period is `T= 2pi sqrt((R)/(g))`

Explanation

Gravitational field
intensity (E)
`(Gmx)/(R^(3))`
Force `=((Gmx)/(R^(3)))M`
Note : Gravitational
field intensity
`=(F)/(M)`
Acceleration `a=(F)/(M)`
`=((Gm)/(R^(3)))x " "....(1)`
Since we know that
`a=- omega^(2) x " "...(2)`
(here , negative has no mathematical meaning. It can be neglected)
Comparing equation (1) & (2) we get,
`omega^(2)=(Gm)/(R^(3))`
`:.T=(2pi)/(omega)= 2pi sqrt((R^(3))/(Gm))`
`rArr 2 pi sqrt((R)/(Gm//R^(2)))`
since `(gm)/(R^(2))=g`
Time period, `T=2pi sqrt((R)/(g))`
Hence proved.