ExamBro
ExamBro
JEE Mains · Physics · STD 11 - 13. oscillations

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

  1. A \(\pi^{2}\, ms ^{-2}\)
  2. B \(9.8\, ms ^{-2}\)
  3. C \(2 \pi^{2}\, ms ^{-2}\)
  4. D \(16\, m / s ^{2}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(2 \pi^{2}\, ms ^{-2}\)

Step-by-step Solution

Detailed explanation

\(T =2 \pi \sqrt{\frac{l}{ g }}\) \(2=2 \pi \sqrt{\frac{2}{ g }}\) \(\Rightarrow g =2 \pi^{2}\)