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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) $2\pi^2$ ms$^{-2}$
(2) 16 m/s$^2$
(3) 9.8 ms$^{-2}$
(4) $\pi^2$ ms$^{-2}$
$T = 2\pi\sqrt{\frac{l}{g}}$. $g = \frac{4\pi^2 \times 2}{4} = 2\pi^2$ ms$^{-2}$.
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