Question

Tunnel is dug through the earth from one side to the other side along with a diameter. The motion of a particle into the tunnel is simple harmonic motion. Find the time period, neglect all the frictional forces and assume that the earth has a uniform density. Assume that g = 6.67 x 10-11 nm2 kg-2 ; density of the earth = 5.51 x 103 kg m-3.

Answer 1

Let us assume that a tunnel is made along the earth diameter.  

Let a particle of mass m is placed at a distance y from the center of earth in that tunnel.  

There will be gravitational attraction force experienced by the particle due to mass of earth of radius y.  

F = GMmy/R³

Here G = 6.67x10^(-11) Nm²/Kg²

M = Mass of earth = density * volume = d*V = d*4πR³/3

F = 4πGdR³ym/3*R³ = 4πGdym/3

Linear Force F = ma.  

Hence a acceleration at a distance y is given by  

a = 4πGdy/3  

a = Frequency² * distance

Frequency² = w² = 4πGd/3

W = √(4πGd/3)

Time Period = 2π/w

T = 2π√(3/4πGd)

  = √(3π/Gd)

  = √(3*3.14/5.51*1000*6.67*10^-11)

  = 5062 seconds

  = 84.4 minutes.

Answer 2

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