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

`T=2pisqrt((R)/(g))`
Oscillations of a particle dropped in a tunnel along the diameter of the earth.
Consider earth to be a sphere of radius R and centre O. A straight tunnel is dug along the diameter of the earth. Let .g. be the value of acceleration due to gravity at the surface of the earth.
Suppose a body of mass .m. is dropped into the tunnel and it is at point P. i.e., at a depth d below the surface of the earth at any instant.
If g. is acceleration due to gravity at P.
then `g.=g(1-(d)/(R))=g((R-d)/(R))`

If y is distance of the body from the centre of the earth, then
`R-d=y`
`therefore g.=g((y)/(R))`
Force acting on the body a point P is
`F=-mg.=-(mg)/(R)y" i.e., "Fpropy`
Negative sign indicates that the force acts in the opposite direction of displacement.
Thus the body will execute SHM with force constant, `k=(mg)/(R)`
The period of oscillation of the body will be `T=2pisqrt((m)/(k))=2pisqrt((m)/(mg//R))`
`T=2pisqrt((R)/(g))`