A simple pendulum is taken to a planet where the acceleration due to gravity is 6 times that on earth. What will be the time period of the pendulum on the planet?

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Solution

At earth, $T = 2 π \sqrt{\frac{l}{g}}$ ..(1)
At planet, $T^{'} = 2 π \sqrt{\frac{l}{6 g}}$ ..(2)
$(1) \div (2)$
$\frac{T}{T^{'}} = \sqrt{6}$
$T^{'} = \frac{T}{\sqrt{6}}$

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A simple pendulum is taken to a planet where the acceleration due to gravity is 6 times that on earth. What will be the time period of the pendulum on the planet?

At earth, T=2π√lg ..(1)At planet, T′=2π√l6g..(2)(1)÷(2)TT′=√6T′=T√6

At earth, $T = 2 π \sqrt{\frac{l}{g}}$ ..(1)
At planet, $T^{'} = 2 π \sqrt{\frac{l}{6 g}}$ ..(2)
$(1) \div (2)$
$\frac{T}{T^{'}} = \sqrt{6}$
$T^{'} = \frac{T}{\sqrt{6}}$