A smooth-walled tunnel is made along the straight line connecting any two points on the earth's surface. A body is released at one end of the tunnel. Considering the earth to be a sphere of uniform density, calculate the time period of oscillation of the particle for its simple harmonic motion.

A smooth-walled tunnel is made along the straight line connecting any

1.	A smooth-walled tunnel is made along the straight line connecting any two points on the earth's surface. A body is released at one end of the tunnel. Considering the earth to be a sphere of uniform density, calculate the time period of oscillation of the particle for its simple harmonic motion.
Answer» Solution :Let a body of mass m be dropped into the tunnel AB and after some time, suppose the body reaches C. Force of attraction on the body at C. `F=(G4/3pir^(3)rhom)/R^(2)" "[rho="AVERAGE density of the earth"]` = `4/3piGrhomr=g/Rmr` `""[becauserho=(3g)/(4piRG)" "(R="radius of the earth")]` F acts along CO.. So component of F along CO, `Fcostheta=(mgr)/Rxxx/r=(mg)/Rx` `therefore` Acceleration of the body `a=(Fcostheta)/m=g/Rx""...(1)` Hence, `apropx" "[becauseg//R" is a CONSTANT"]` Therefore, (1) the acceleration of the body is DIRECTLY proportional to its distance from O and (2) this acceleration is always directed towards the centre O of the tunnel. Hence, it executes a simple harmonic motion and its time period `T=2pisqrt(x/a)""...(2)` From (1) and (2), `T=2pisqrt(R/g)`