A simple pendulum is taken at a place where its separation from the earth's surface is equal to the radius of the earth. Calculate the time period of small oscillation if the length of the string is `1.0m`. Take `g = pi^(2) m//s^(2)` at the surface of the earth.

Explanation

To solve the problem of calculating the time period of a simple pendulum at a height equal to the radius of the Earth, we will follow these steps:

### Step 1: Understand the given information
- Length of the pendulum (L) = 1.0 m
- Gravitational acceleration at the surface of the Earth (g) = π² m/s²
- Height (h) = Radius of the Earth (R)

### Step 2: Determine the gravitational acceleration at height h
The formula for gravitational acceleration at a height h above the Earth's surface is given by:
\[ g' = \frac{g}{(1 + \frac{h}{R})^2} \]
Since h = R, we can substitute this into the equation:
\[ g' = \frac{g}{(1 + \frac{R}{R})^2} = \frac{g}{(1 + 1)^2} = \frac{g}{4} \]

### Step 3: Substitute the value of g
Substituting the value of g = π² m/s² into the equation for g':
\[ g' = \frac{\pi^2}{4} \text{ m/s}^2 \]

### Step 4: Calculate the time period of the pendulum
The formula for the time period (T) of a simple pendulum is given by:
\[ T = 2\pi \sqrt{\frac{L}{g'}} \]
Substituting L = 1.0 m and g' = \frac{\pi^2}{4} into the formula:
\[ T = 2\pi \sqrt{\frac{1.0}{\frac{\pi^2}{4}}} \]

### Step 5: Simplify the expression
\[ T = 2\pi \sqrt{\frac{4}{\pi^2}} = 2\pi \cdot \frac{2}{\pi} = 4 \text{ seconds} \]

### Final Answer
Thus, the time period of small oscillations of the pendulum is:
\[ \boxed{4 \text{ seconds}} \]

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