Question

Assume that the earth is a solid sphere of uniform density and a tunnel is dug along its diameter throughout the earth. It is found that when a particle is released in this tunnel, it executes a simple harmonic motion. The mass of the particle is $100 \,g$. The time period of the motion of the particle will be (approximately) (Take $g=10 \, m s ^{-2}$, radius of earth $=6400 \,km$ )

• A. 1 hour 24 minutes
• B. 1 hour 40 minutes
• C. 12 hours
• D. 24 hours

Answer 1

Answer: (A)

Let at some time particle is at a distance $x$ from centre of Earth, then at that position field
$ E =\frac{ GM }{ R ^3 }x $
$ \therefore $ Acceleration of particle
$ \vec{ a }=-\frac{ GM }{ R ^3} \vec{ x }$
$ \Rightarrow \omega=\sqrt{\frac{ GM }{ R ^3}}=\sqrt{\frac{ g }{ R }}$
Now $T =\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{ R }{ g }}$
$ \Rightarrow T =2 \times 3.14 \times \sqrt{\frac{6400 \times 10^3}{10}} $
$ =2 \times 3.14 \times 800 \,\sec \approx 1 $ hour $ 24 $ minutes