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

The Correct Option is 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

Answer 2

1. For SHM inside the Earth, the effective force is proportional to displacement: \[ F = -\frac{G M r}{R^3}, \quad a = \frac{F}{m} = -\frac{G M}{R^3} r. \]
2. The time period of SHM is given by: \[ T = 2\pi \sqrt{\frac{R^3}{G M}}. \]
3. Substituting \(g = \frac{G M}{R^2}\), the expression becomes: \[ T = 2\pi \sqrt{\frac{R}{g}}. \]
4. Using \(R = 6400 \, \text{km} = 6.4 \times 10^6 \, \text{m}\) and \(g = 10 \, \text{m/s}^2\): \[ T = 2\pi \sqrt{\frac{6.4 \times 10^6}{10}} = 2\pi \sqrt{6.4 \times 10^5}. \]
5. Simplifying: \[ T \approx 2\pi \times 800 = 5026 \, \text{seconds} = 1 \, \text{hour} \, 24 \, \text{minutes}. \]
Thus, the time period is approximately 1 hour 24 minutes. SHM of a particle inside a uniform sphere depends only on the radius and acceleration due to gravity. The time period is the same for any particle mass.