If the time period of a two meter long simple pendulum is 2s, the acceleration due to gravity at the place where pendulum is executing S.H.M. is:

Options

A. `pi^(2)ms^(-2)`
B. `9.8ms^(-2)`
C. `2pi^(2) ms^(-2)`
D. `16 m//s^(-2)`

Answer

To find the acceleration due to gravity \( g \) at the place where a simple pendulum of length \( L = 2 \) meters has a time period \( T = 2 \) seconds, we can use the formula for the time period of a simple pendulum:

\[ T = 2\pi \sqrt{\frac{L}{g}} \]

### Step-by-step Solution:

1. **Write down the formula for the time period of a simple pendulum**:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]

2. **Substitute the known values into the formula**:
Here, \( T = 2 \) seconds and \( L = 2 \) meters. Substitute these values into the equation:
\[ 2 = 2\pi \sqrt{\frac{2}{g}} \]

3. **Square both sides to eliminate the square root**:
\[ 2^2 = (2\pi)^2 \left(\frac{2}{g}\right) \]
This simplifies to:
\[ 4 = 4\pi^2 \left(\frac{2}{g}\right) \]

4. **Rearrange the equation to solve for \( g \)**:
Multiply both sides by \( g \) and divide by \( 4 \):
\[ g = 4\pi^2 \left(\frac{2}{4}\right) \]
Simplifying gives:
\[ g = 2\pi^2 \]

5. **Final expression for \( g \)**:
Thus, the acceleration due to gravity is:
\[ g = 2\pi^2 \, \text{m/s}^2 \]

### Conclusion:
The acceleration due to gravity at the place where the pendulum is executing simple harmonic motion is \( g = 2\pi^2 \, \text{m/s}^2 \).