Compare the time periods of two pendulums of lengths 1 m and 9 m.

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Solution

$T_{1} = 2 π \sqrt{\frac{1}{g}}$
$T_{2} = 2 π \sqrt{\frac{9}{g}} = 3 \times 2 π \sqrt{\frac{1}{g}} = 3 T_{1}$
therefore time period of second pendulum is thrice of first pendulum.

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