If the time period of a two meter long simple pendulum is 2 s,
Question:

If the time period of a two meter long simple pendulum is $2 \mathrm{~s}$, the acceleration due to gravity at the place where pendulum is executing S.H.M. is :

  1. (1) $2 \pi^{2} \mathrm{~ms}^{-2}$

  2. (2) $16 \mathrm{~m} / \mathrm{s}^{2}$

  3. (3) $9.8 \mathrm{~ms}^{-2}$

  4. (4) $\pi^{2} \mathrm{~ms}^{-2}$


Correct Option: 1

Solution:

(1)

$\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{I}}{\mathrm{g}}}$

$\mathrm{T}^{2}=\frac{4 \pi^{2} \mathrm{I}}{\mathrm{g}}$

$\mathrm{g}=\frac{4 \pi^{2} \mathrm{I}}{\mathrm{T}^{2}}$

$=\frac{4 \pi^{2} \times 2}{(2)^{2}}=2 \pi^{2} \mathrm{~ms}^{-2}$

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