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 :

CONCEPT:

The time period of the simple pendulum is written as;

T = 2 \(π \sqrt {\frac{l}{g}}\)

Here we have " l " as the length, and g is the acceleration due to gravity and T is the time period.

CALCULATION:

Given: Length, l = 2 m, and time period, t = 2 s

Now, using the time period of the simple pendulum we have;

T = 2 \(π \sqrt {\frac{l}{g}}\)    ---(1)

Now, on putting the given values in equation (1) we have;

2 = 2 \(\times π\sqrt{\frac{ 2}{g}}\)

1 =  \(π \sqrt{\frac{2}{g}}\)

Now, on squaring both sides we have;

1 =\(π^2 \times \frac{2}{g}\)

⇒ g = 2π² m/s²

Hence, option 4) is the correct answer.