The period of a seconds pendulum on a planet, | vedclass

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(C) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$.For a seconds pendulum on Earth,$T = 2 	ext{ s}$.The acceleration

Vedclass · Accepted Answer

$2 \sqrt{3}$ seconds

Vedclass · Answer

(C) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$.For a seconds pendulum on Earth,$T = 2 	ext{ s}$.The acceleration due to gravity on a planet is $g' = \frac{GM'}{R'^2}$.Given $M' = 3M$ and $R' = 3R$,we have:$\frac{g'}{g} = \frac{M'}{M} \cdot \frac{R^2}{R'^2} = 3 \cdot \frac{1}{3^2} = \frac{3}{9} = \frac{1}{3}$.The time period on the planet is $T' = 2\pi \sqrt{\frac{L}{g'}}$.Thus,$\frac{T'}{T} = \sqrt{\frac{g}{g'}} = \sqrt{3}$.$T' = T \cdot \sqrt{3} = 2 \sqrt{3} 	ext{ seconds}$.

The period of a seconds pendulum on a planet,whose mass and radius are three times that of Earth,is

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