Question

What is the frequency of the body at the centre of the earth?

Answer 1

`f = 1.975xx10^-4 Hz`

or the period is:
`T =1/f = 5064s = 84.4` minutes

Explanation:

If an object with mass m is dropped into a tunnel passes from one end of the earth to the other end through the center of the earth, the equation of motion of the object, assuming resistance free, is:

`F_"net"=ma`

The only source of force is the gravitational force, which always point toward the center of the earth regardless which hemisphere side it is in, and the motion is essential dimensional; therefore

`cancelma= -G(cancelmM)/r^2`

`because a= (dv)/(dt) = d/dt (dr//dt)=(d^2r)/dt^2`

`therefore (d^2r)/dt^2=-GM/r^2`

When m is located at r from the center of the earth, the mass portion of the earth contributes to the gravitational attraction is `M=rhoV= rho(4/3pir^3)`

`(d^2r)/dt^2=-G((4pi)/3rhor^cancel3)/cancel(r^2)=-((4pi)/3Grho)r`

This equation looks just like a simple harmonic motion. Hence a solution to differential equation above is

`r=Rcos(omegat)`

where R is radius of the earth and `omega` is the angular frequency.

Taking derivative twice on r, the differential equation becomes
`-omega^2cancelr=-((4pi)/3Grho)cancelr`

`omega^2=((4pi)/3Grho)`
`omega=sqrt( (4pi)/3Grho)`
`f = omega/(2pi) =1/(2pi)sqrt( (4pi)/3Grho)=sqrt( (Grho)/(3pi))`

`f=sqrt(6.67xx10^-11 (Nm^2)/(kg^2)*5.51xx10^3(kg)/m^3//3pi`

`f =1.975 xx10^-4 1/s= 1.975xx10^-4 Hz`

or the period is

`T =1/f = 5064s = 84.4` minutes

I hope this is what you are looking for.