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 $100g$. The time period of the motion of the particle will be (approximately) (Take $g=10m{s}^{-2}$, radius of earth $=6400km$ )

• 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