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 holewill execute a simple harmonic motion such that its time period is T= 2pi sqrt((R)/(g))

Consider the Earth as a homogenous sphere of radius R and a straight h

1.	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 holewill execute a simple harmonic motion such that its time period is T= 2pi sqrt((R)/(g))
Answer» SOLUTION :`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))`