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 $=6400 \,km$ )
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.
