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Question

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:


A

2π2ms-2

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B

16ms-2

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C

9.8ms-2

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D

π2ms-2

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Solution

The correct option is A

2π2ms-2


Step 1: Given

Length of the pendulum: l=2m

Time period of the pendulum: T=2s

Step 2: Formula Used

T=2πlg

Where g is the acceleration due to gravity,

Step 3: Calculate the acceleration due to gravity where SHM is executed using the formula

2=2π2g4=2π22gg=8π24g=2π2

Hence, the acceleration due to gravity at the place where the pendulum is executing S.H.M. is 2π2ms-2.


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