Question

Assume that a narrow tunnel is dug between two diametrically opposite points on the earth's surface. If a particle is released in this tunnel, it will execute simple harmonic motion. What will be the time period of SHM of this particle?

• A. \( \frac{1}{2\pi} \sqrt{\frac{R^3}{GM}} \)
• B. \( \frac{1}{2\pi} \sqrt{\frac{GM}{R^3}} \)
• C. \( 2\pi \sqrt{\frac{R^3}{GM}} \)
• D. \( 2\pi R \sqrt{\frac{1}{GM}} \)

Hint:

The time period for SHM inside a spherical shell (such as the Earth) depends on the Earth's radius \( R \) and gravitational constant \( G \). The mass of the particle cancels out in the calculation, so the time period is independent of the particle's mass.

Answer 1

The Correct Option is C

Step 1: Understand the scenario.
In this case, the particle is released inside a tunnel that runs through the Earth and is assumed to execute simple harmonic motion (SHM) under the influence of gravity. This SHM occurs because the force acting on the particle is proportional to its displacement from the center of the Earth, as the gravitational force inside the Earth follows the form of SHM.
Step 2: Force inside the Earth.
The gravitational force at a distance \( r \) from the center of the Earth (where \( r \leq R \), \( R \) being the Earth's radius) is given by: \[ F = - \frac{GMm}{R^3} r \] Where:
\( F \) is the force,
\( G \) is the gravitational constant,
\( M \) is the mass of the Earth,
\( m \) is the mass of the particle,
\( r \) is the distance from the center of the Earth, and
\( R \) is the Earth's radius.
This equation indicates that the force is proportional to \( r \), which is characteristic of simple harmonic motion.
Step 3: Equation of motion.
For SHM, the general equation is: \[ F = -kr \] where \( k \) is the force constant. From the above equation for force, we can identify \( k = \frac{GMm}{R^3} \).
Step 4: Time period of SHM.
The time period \( T \) for SHM is given by the formula: \[ T = 2\pi \sqrt{\frac{m}{k}} \] Substituting \( k = \frac{GM}{R^3} \) into this equation: \[ T = 2\pi \sqrt{\frac{m}{\frac{GM}{R^3}}} = 2\pi \sqrt{\frac{R^3}{GM}} \] Thus, the time period of SHM is \( 2\pi \sqrt{\frac{R^3}{GM}} \), which corresponds to option (3).