Assume that a tunnel is dug along a cord of earth at a perpendicular distance `R/2` from earth's centre where `R` is the radius of earth. The wall of tunnel is frictionless. Find the time period of particle excuting SHM in tunnel

Answer

To find the time period of a particle executing simple harmonic motion (SHM) in a tunnel dug along a chord of the Earth at a perpendicular distance \( \frac{R}{2} \) from the Earth's center, we can follow these steps:

### Step 1: Understand the gravitational force inside the Earth
When a particle is inside a spherical shell of uniform density, the gravitational force acting on it is due only to the mass of the Earth that is at a radius smaller than the distance of the particle from the center.

Using the shell theorem, the gravitational force \( F \) acting on a particle of mass \( m \) at a distance \( r \) from the center of the Earth is given by:
\[ F = -\frac{GM(r)}{r^2} m \]
where \( M(r) \) is the mass of the Earth enclosed within radius \( r \).

### Step 2: Calculate the mass \( M(r) \)
The mass \( M(r) \) of the Earth within radius \( r \) can be expressed as:
\[ M(r) = \rho \cdot \frac{4}{3} \pi r^3 \]
where \( \rho \) is the density of the Earth. The average density of the Earth can be calculated using:
\[ \rho = \frac{M}{\frac{4}{3} \pi R^3} \]
where \( M \) is the total mass of the Earth and \( R \) is its radius.

### Step 3: Substitute \( M(r) \) into the force equation
At a distance \( r = \frac{R}{2} \):
\[ M\left(\frac{R}{2}\right) = \rho \cdot \frac{4}{3} \pi \left(\frac{R}{2}\right)^3 = \rho \cdot \frac{4}{3} \pi \cdot \frac{R^3}{8} = \frac{1}{6} M \]
Thus, the force acting on the particle at \( r = \frac{R}{2} \) becomes:
\[ F = -\frac{G \cdot \frac{1}{6} M}{\left(\frac{R}{2}\right)^2} m = -\frac{G \cdot \frac{1}{6} M \cdot 4}{R^2} m = -\frac{2GM}{3R^2} m \]

### Step 4: Relate the force to SHM
The force can be expressed in terms of the displacement \( x \) from the equilibrium position:
\[ F = -kx \]
where \( k \) is the spring constant. From the previous step, we have:
\[ k = \frac{2GM}{3R^2} \]

### Step 5: Find the time period of SHM
The time period \( T \) of SHM is given by:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
Substituting for \( k \):
\[ T = 2\pi \sqrt{\frac{m}{\frac{2GM}{3R^2}}} = 2\pi \sqrt{\frac{3mR^2}{2GM}} \]

### Step 6: Final expression for the time period
Thus, the final expression for the time period \( T \) is:
\[ T = 2\pi \sqrt{\frac{3R^2}{2g}} \]
where \( g = \frac{GM}{R^2} \) is the acceleration due to gravity at the surface of the Earth.

### Conclusion
The time period of the particle executing SHM in the tunnel is:
\[ T = 2\pi \sqrt{\frac{3R}{2g}} \]