Question

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 :

• A. 16 m/s$^2$
• B. π$^2$ ms$^{-2}$
• C. 9.8 ms$^{-2}$
• D. 2π$^2$ ms$^{-2}$

Hint:

When $T=2\pi$, $g$ is simply equal to the length $l$. In this case, since $T=2$, the factor of $\pi$ remains in the result for $g$.

Answer 1

The Correct Option is D

Step 1: The time period of a simple pendulum is $T = 2\pi\sqrt{\frac{l}{g}}$.
Step 2: Given $T = 2$ s and $l = 2$ m.
Step 3: Substitute values: $2 = 2\pi\sqrt{\frac{2}{g}} \Rightarrow 1 = \pi\sqrt{\frac{2}{g}}$.
Step 4: Squaring both sides: $1 = \pi^2 \frac{2}{g} \Rightarrow g = 2\pi^2$ ms$^{-2}$.