If the time period of a two meter long simple pendulum

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

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

  3. $2 \pi^{2} \mathrm{~ms}^{-2}$

  4. $16 \mathrm{~m} / \mathrm{s}^{2}$


Correct Option: , 3

Solution:

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

$2=2 \pi \sqrt{\frac{2}{g}}$

$\Rightarrow \mathrm{g}=2 \pi^{2}$

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