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JEE PHYSICS MCQ: OSCILLATIONS

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If the time period of a 2m long simple pendulum is 2s, the acceleration due to gravity at the place, where pendulum is executing SHM is
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Correct option: C
2π2ms−22{\pi ^2}m{s^{ - 2}}
Option C (2π2ms−22{\pi ^2}m{s^{ - 2}}) is correct because the acceleration due to gravity is calculated using the formula T=2πlgT = 2\pi \sqrt{\frac{l}{g}}.

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