Question

Compare the time periods of a simple pendulum two different places, where the acceleration due to gravity, $g$ are $9.8{ms}^{-2}$ and $4.9{ms}^{-2}$ respectively.

Answer 1

time period $T=2\pi \sqrt{(}\frac{l}{g})$

for compairng time period we take the ratio for both value of g , i.e, $\frac{T_1}{T_2}=\sqrt{(}\frac{g_2}{g_1})=\sqrt{(}\frac{4.9}{9.8})=\frac{1}{\sqrt{2}}$

this shows $T_2=\sqrt{2}{T}_1$