Question

Time period of simple harmonic motion of a body dropped in a tunnel dug along diameter of earth

Answer 1

Answer:t AB be the imaginary tunnel dug across a diameter of the Earth and O be its centre as shown in. Let P be the position at any instant of the body dropped from one end of the tunnel, where OP=x.  The Earth can be considered to be made of two parts. Its outer shell-I does not exert any force on the particle. The gravitational force on the particle is only due to the sphere-II of radius x. If ρ is the density of the Earth, then mass of the sphere-II =4π3x3ρ.                   According to Newton?s law of gravitation, the force of attraction acting on the body of mass m and situated at P is given                 F=G[(4π/3)x3ρ]mx2=(4π/3)Gρxm                 This force acts towards O.                 Acceleration produced in the body i.e.,                 a=Fm=(4π/3)Gρxmm                 or a=(4π/3)Gρx                 or a∝x              [as (4π/3)Gρ is a constant]                 Thus, the acceleration is directly proportional to x, the displacement of the body from the fixed point O. More so, this acceleration is directed towards the fixed point. Hence, the body executes SHM along the tunnel AB.                 Note: The time period of the body is given by                 T=2πdisplacement (x) acceleration (a)−−−−−−−−−−−−−−√                 or T=2πx(4π/3)Gρx−−−−−−−−−√=2π34πGρ−−−−−√                 or T=3πGρ−−−√                          ? (i)                 As G=6.67×10−11Nm2/kg2                               and ρ=5.52×103kg/m3,                 T=3×3.14(6.67×10−11)(5.52×103)s−−−−−−−−−−−−−−−−−−−−−−√                 =5060s=84min20s

Explanation: