If the time period of a two-meter-long simple pendulum is $2 s$ , the acceleration due to gravity at the place where the pendulum is executing SHM is :

2 π^{2} {ms}^{- 2}

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16 {ms}^{- 2}

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9.8 {ms}^{- 2}

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4 π^{2} {ms}^{- 2}

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Solution

The correct option is A $2 π^{2} {ms}^{- 2}$
We know that,

$T =$ $2 π \sqrt{\frac{l}{g}}$

$\Rightarrow T^{2} = \frac{4 π^{2} l}{g}$

$\Rightarrow g = \frac{4 π^{2} l}{T^{2}}$

$\Rightarrow g = \frac{4 π^{2} \times 2}{(2)^{2}} = 2 π^{2} {ms}^{- 2}$

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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 the pendulum is executing SHM is :

The correct option is A 2π2 ms−2We know that, T = 2π√lg ⇒T2=4π2lg ⇒g=4π2lT2 ⇒g=4π2×2(2)2=2π2 ms−2

If the time period of a two-meter-long simple pendulum is $2 s$ , the acceleration due to gravity at the place where the pendulum is executing SHM is :

The correct option is A $2 π^{2} {ms}^{- 2}$
We know that,

$T =$ $2 π \sqrt{\frac{l}{g}}$

$\Rightarrow T^{2} = \frac{4 π^{2} l}{g}$

$\Rightarrow g = \frac{4 π^{2} l}{T^{2}}$

$\Rightarrow g = \frac{4 π^{2} \times 2}{(2)^{2}} = 2 π^{2} {ms}^{- 2}$