Imagine a narrow tunnel between the two diametrically opposite points of the earth. A particle of mass m is released in this tunnel . The time period of oscillation is

Options

A. `pisqrt((R)/(g))`
B. `(pi)/(2)sqrt((R)/(g))`
C. `2pisqrt((R)/(g))`
D. `(2)/(pi)sqrt((R)/(g))`

Explanation

To find the time period of oscillation for a particle of mass \( m \) released in a tunnel that goes through the Earth, we can follow these steps:

### Step 1: Understand the System
The particle will oscillate back and forth through the tunnel due to gravitational force. When it is at the center of the Earth, the gravitational force acting on it will be zero, and it will experience a restoring force when it moves away from the center.

### Step 2: Use the Concept of Simple Harmonic Motion (SHM)
The motion of the particle can be modeled as simple harmonic motion (SHM). In SHM, the time period \( T \) is given by the formula:
\[ T = 2\pi \sqrt{\frac{l}{g}} \]
where \( l \) is the effective length of the pendulum and \( g \) is the acceleration due to gravity.

### Step 3: Determine the Effective Length
In this case, the effective length \( l \) is the radius of the Earth \( R \). Therefore, we can substitute \( l \) with \( R \):
\[ T = 2\pi \sqrt{\frac{R}{g}} \]

### Step 4: Substitute Known Values
Here, \( R \) is the radius of the Earth (approximately \( 6.4 \times 10^6 \) meters) and \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)). However, we do not need to substitute the actual values since the question only asks for the expression for the time period.

### Final Expression
Thus, the time period of oscillation of the particle in the tunnel through the Earth is:
\[ T = 2\pi \sqrt{\frac{R}{g}} \]