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
- A \(\pi^{2}\, ms ^{-2}\)
- B \(9.8\, ms ^{-2}\)
- C \(2 \pi^{2}\, ms ^{-2}\)
- D \(16\, m / s ^{2}\)
Answer & Solution
Correct Answer
(C) \(2 \pi^{2}\, ms ^{-2}\)
Step-by-step Solution
Detailed explanation
\(T =2 \pi \sqrt{\frac{l}{ g }}\) \(2=2 \pi \sqrt{\frac{2}{ g }}\) \(\Rightarrow g =2 \pi^{2}\)
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