A particle of mass m is dropped from a great height h above the hole in the earth dug along its diameter.

Options

A. The motion of the particle is simple harmonic
B. The motion of the particle is periodic
C. The speed of the particle at the centre of earth equals `sqrt((2GM)/((R+h)))` , where R and M are the radius and mass of the earth respectively
D. The speed of the particle at the centre of earth equals `sqrt((GM(R+3h))/(R(R+h)))`, where R and M are the radius and mass of the earth respectively

Explanation

To solve the problem of a particle of mass \( m \) dropped from a height \( h \) above a hole dug through the Earth along its diameter, we can analyze the motion of the particle as it falls and then moves through the Earth.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- A particle is dropped from a height \( h \) above the Earth's surface into a hole that goes through the Earth's center. We need to analyze its motion as it falls and passes through the Earth.

2. **Potential Energy at Height \( h \)**:
- The potential energy \( U \) of the particle at height \( h \) above the Earth's surface is given by:
\[ U_h = mgh \]
- Here, \( g \) is the acceleration due to gravity at the Earth's surface.

3. **Potential Energy at the Center of the Earth**:
- When the particle reaches the center of the Earth, we need to consider the gravitational potential energy at that point. The potential energy at the center can be derived from the gravitational potential energy formula:
\[ U_c = -\frac{GMm}{R} \]
- Where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.

4. **Using Conservation of Energy**:
- The total mechanical energy is conserved. Therefore, the potential energy at height \( h \) will equal the potential energy at the center plus the kinetic energy at the center:
\[ mgh = -\frac{GMm}{R} + \frac{1}{2}mv^2 \]
- Rearranging this equation gives us:
\[ \frac{1}{2}mv^2 = mgh + \frac{GMm}{R} \]

5. **Finding the Velocity \( v \)**:
- We can simplify the equation to find the velocity \( v \) when the particle reaches the center:
\[ v^2 = 2gh + \frac{2GM}{R} \]
- Thus, the velocity \( v \) is:
\[ v = \sqrt{2gh + \frac{2GM}{R}} \]

6. **Final Expression**:
- The final expression for the velocity of the particle when it reaches the center of the Earth is:
\[ v = \sqrt{2g(h + \frac{R}{2})} \]

### Conclusion:
The particle, when dropped from height \( h \), will have a velocity given by the derived formula when it reaches the center of the Earth.