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 use the principles of simple harmonic motion (SHM). Here's a step-by-step solution:

### Step 1: Understand the Setup
The particle is released in a tunnel that runs through the Earth from one side to the other. As the particle moves towards the center of the Earth, it experiences a gravitational force that varies with distance from the center.

### Step 2: Gravitational Force Inside the Earth
According to the shell theorem, when a particle is inside a uniform spherical shell, it experiences no net gravitational force from the shell. Therefore, only the mass of the Earth that is at a radius less than the particle's distance from the center contributes to the gravitational force acting on it.

### Step 3: Expression for Gravitational Force
If \( r \) is the distance of the particle from the center of the Earth, the effective mass \( m_{\text{eff}} \) that contributes to the gravitational force is given by:
\[ m_{\text{eff}} = \frac{M}{R^3} r \]
where \( M \) is the total mass of the Earth and \( R \) is the radius of the Earth.

The gravitational force \( F \) acting on the particle is:
\[ F = -\frac{G M m}{r^2} \]
However, since we are considering the force inside the Earth, we can express it as:
\[ F = -\frac{G M m}{R^3} r \]
This shows that the force is directly proportional to the displacement \( r \) from the center, indicating simple harmonic motion.

### Step 4: Equation of Motion
The equation of motion for SHM can be written as:
\[ F = -k x \]
where \( k \) is a constant. From our expression for \( F \), we can identify:
\[ k = \frac{G M m}{R^3} \]
Thus, the motion of the particle can be described as SHM.

### Step 5: Time Period of Oscillation
The time period \( T \) of SHM is given by:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
Substituting \( k \) from the previous step:
\[ T = 2\pi \sqrt{\frac{m}{\frac{G M m}{R^3}}} \]
The mass \( m \) cancels out:
\[ T = 2\pi \sqrt{\frac{R^3}{G M}} \]

### Step 6: Relating \( g \) to \( G \) and \( M \)
We know that the acceleration due to gravity \( g \) at the surface of the Earth is given by:
\[ g = \frac{G M}{R^2} \]
From this, we can express \( G M \) as:
\[ G M = g R^2 \]

### Step 7: Final Expression for Time Period
Substituting \( G M \) back into the equation for \( T \):
\[ T = 2\pi \sqrt{\frac{R^3}{g R^2}} = 2\pi \sqrt{\frac{R}{g}} \]

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

---