Show that the period of oscillation of simple pendulum at depth `h` below earth's surface is inversely proportional to `sqrt(R - h)` , where `R` is the radius of earth. Find out the time period of a second pendulum at a depth `R//2` from the earth's surface ?

Explanation

To solve the problem, we need to show that the period of oscillation of a simple pendulum at a depth \( h \) below the Earth's surface is inversely proportional to \( \sqrt{R - h} \), where \( R \) is the radius of the Earth. We will also find the time period of a second pendulum at a depth of \( \frac{R}{2} \).

### Step-by-Step Solution:

1. **Understanding the Time Period of a Simple Pendulum**:
The time period \( T \) of a simple pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity.

2. **Acceleration Due to Gravity at Depth \( h \)**:
At a depth \( h \) below the Earth's surface, the acceleration due to gravity \( g' \) is given by:
\[ g' = g \left(1 - \frac{h}{R}\right) \]
where \( g \) is the acceleration due to gravity at the surface of the Earth.

3. **Finding the New Time Period \( T' \)**:
The new time period \( T' \) at depth \( h \) can be expressed as:
\[ T' = 2\pi \sqrt{\frac{L}{g'}} \]
Substituting \( g' \) into the equation, we have:
\[ T' = 2\pi \sqrt{\frac{L}{g \left(1 - \frac{h}{R}\right)}} \]

4. **Relating \( T' \) to \( T \)**:
Since the time period at the surface \( T \) is:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
We can express \( T' \) in terms of \( T \):
\[ T' = T \sqrt{\frac{g}{g \left(1 - \frac{h}{R}\right)}} = T \sqrt{\frac{1}{1 - \frac{h}{R}}} \]

5. **Inversely Proportional Relationship**:
From the above expression, we can see that:
\[ T' \propto \frac{1}{\sqrt{1 - \frac{h}{R}}} \]
This means:
\[ T' \propto \frac{1}{\sqrt{R - h}} \]
Thus, we have shown that the period of oscillation of a simple pendulum at depth \( h \) is inversely proportional to \( \sqrt{R - h} \).

6. **Finding the Time Period at Depth \( \frac{R}{2} \)**:
Now, we need to find the time period \( T' \) at a depth of \( \frac{R}{2} \):
\[ h = \frac{R}{2} \]
Substituting \( h \) into the relationship we derived:
\[ T' = T \sqrt{\frac{R}{R - \frac{R}{2}}} = T \sqrt{\frac{R}{\frac{R}{2}}} = T \sqrt{2} \]

7. **Calculating the Time Period of a Second Pendulum**:
A second pendulum has a time period \( T = 2 \) seconds at the surface. Therefore:
\[ T' = 2 \sqrt{2} \approx 2 \times 1.414 = 2.828 \text{ seconds} \]

### Final Answer:
The time period of a second pendulum at a depth of \( \frac{R}{2} \) from the Earth's surface is approximately \( 2.828 \) seconds.