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Physics Oscillations Physics Class 11th Oscillations

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

Option 1 -

9.8 ms^-2

Option 2 -

16 m/s²

Option 3 -

2π² ms^-2

Option 4 -

π² ms^-2

2 Views | Posted a month ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

a month ago

Correct Option - 3

Detailed Solution:

$T = 2 π \sqrt[]{\frac{l}{g}}$

$\Rightarrow 2 = 2 π \sqrt[]{\frac{2}{g}}$

$\Rightarrow g = 2 π^{2} m / s^{2}$

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

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