A tunnel is dug in the earth across one of its diameter. Two masses `m` and `2m` are dropped from the two ends of the tunel. The masses collide and stick each other. They perform `SHM`, the ampulitude of which is `(R =` radius of earth)

Options

A. `R`
B. `R//2`
C. `R//3`
D. `2R//3`

Explanation

To solve the problem, we will analyze the situation step by step.

### Step 1: Understanding the System
We have a tunnel dug through the Earth along its diameter. Two masses, `m` and `2m`, are dropped from either end of the tunnel. When they meet, they collide and stick together, forming a single mass of `3m`.

**Hint:** Visualize the scenario with a diagram showing the tunnel and the two masses.

### Step 2: Forces Acting on the Masses
As the masses fall towards the center of the Earth, they experience gravitational force. Inside the Earth, the gravitational force acting on a mass at a distance `r` from the center is given by:

\[ F = -\frac{G M(r) m}{r^2} \]

where \( M(r) \) is the mass of the Earth within radius `r`. By the shell theorem, this force can be simplified to:

\[ F = -\frac{G M_e m}{R^3} r \]

where \( M_e \) is the total mass of the Earth, and \( R \) is the radius of the Earth.

**Hint:** Recall the shell theorem and how gravity behaves inside a spherical shell.

### Step 3: Equation of Motion
The acceleration `a` of the masses can be expressed as:

\[ a = -\frac{G M_e}{R^3} r \]

This is the equation of motion for simple harmonic motion (SHM), where:

\[ \omega^2 = \frac{G M_e}{R^3} \]

**Hint:** Identify the form of the SHM equation and relate it to the standard form \( a = -\omega^2 x \).

### Step 4: Amplitude of SHM
When the two masses collide and stick together, they will oscillate about the center of the tunnel. The amplitude of their oscillation will be equal to the maximum distance they can reach from the center, which is equal to the radius of the Earth \( R \).

**Hint:** Think about the maximum displacement in SHM, which is the amplitude.

### Step 5: Conclusion
Thus, the amplitude of the resulting SHM after the collision of the two masses is:

\[ \text{Amplitude} = R \]

**Final Answer:** The amplitude of the SHM performed by the combined mass after the collision is \( R \) (the radius of the Earth).