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

The time period of a simple pendulum is given by $$T = 2\pi\sqrt{\frac{L}{g}}$$, where $$L$$ is the length of the pendulum and $$g$$ is the acceleration due to gravity.

Given that $$L = 2$$ m and $$T = 2$$ s, we substitute into the formula:

$$2 = 2\pi\sqrt{\frac{2}{g}}$$

Dividing both sides by $$2\pi$$:

$$\frac{1}{\pi} = \sqrt{\frac{2}{g}}$$

Squaring both sides:

$$\frac{1}{\pi^2} = \frac{2}{g}$$

Solving for $$g$$:

$$g = 2\pi^2 \text{ m s}^{-2}$$

The correct answer is Option (1): $$2\pi^2$$ m s$$^{-2}$$.