A particle of mass m is released through a narrow tunnel dug between two diametrically opposite ends of Earth. Show that it will exceute SHM and calculate its time period?

A particle of mass m is released through a narrow tunnel dug between t

1.	A particle of mass m is released through a narrow tunnel dug between two diametrically opposite ends of Earth. Show that it will exceute SHM and calculate its time period?
Answer» Solution :When the particle is at a distance X from the centre of earth, the force acting on it `F= (Gmm)/(R^3) x ` (or) ma `=-(GMm)/(R^3) x` (or) `a=- ((GM)/(R^3))x` Comparing with SHM equation `, omega^2 = (GM)/(R^3) ` (or) `T= 2pi SQRT((R^3)/(GM)) ` (or) `2pi sqrt((R)/(g))`