How is the time period of a simple pendulum affected if the acceleration due to gravity is reduced to one-fourth?

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Solution

$T_{1} = 2 π \sqrt{\frac{l}{g}}$
When acceleration due to gravity is reduced to one -fourth, then
$T_{2} = 2 π \sqrt{\frac{l}{\frac{g}{4}}} = 2 π \sqrt{\frac{4 l}{g}} = 2 \times 2 π \sqrt{\frac{l}{g}} = 2 T_{1}$
Hence the time period becomes double when acceleration due to gravity reduced to one-fourth.

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How is the time period of a simple pendulum affected if the acceleration due to gravity is reduced to one-fourth?

T1=2π√lgWhen acceleration due to gravity is reduced to one -fourth, thenT2=2π ⎷lg4=2π√4lg=2×2π√lg=2T1Hence the time period becomes double when acceleration due to gravity reduced to one-fourth.