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Q. The time period of a simple pendulum, when it is made to oscillate on the surface of moon

J & K CETJ & K CET 2004

Solution:

Time period of a simple pendulum of length $l$ is given by
$T=2 n \sqrt{\frac{l}{g}}$
where $g$ is acceleration due to gravity.
On moon $g_{m}=\frac{g}{6}$
$\therefore T'=2 \pi \sqrt{\frac{l}{g / 6}}=2 \pi \sqrt{\frac{6 l}{g}}$
$=\sqrt{6} .2 \pi \sqrt{\frac{l}{g}}$
$\Rightarrow T'=\sqrt{6} T$
Hence, time period increases on the surface of moon.