The acceleration due to gravity on the surface of earth is `9.8 ms^(-2)`.Time period of a simple pendulum on earth and moon are 3.5 second and 8.4 second respectively. Find the acceleration due to gravity on the moon .
Hint `: T_(e) = 2pi sqrt((L)/(g_(e))) T_(m)= 2pi sqrt((L)/(g_(m)))`
`(T_(e)^(2))/(T_(m)^(2))= (g_(m))/(g_(e))`
`g_(m) = (T_(e)^(2))/(T_(m)^(2))g_(e)`

Explanation

To find the acceleration due to gravity on the moon, we can use the relationship between the time periods of a simple pendulum on Earth and the moon. The formulas for the time period \( T \) of a simple pendulum are given by:

\[ T_e = 2\pi \sqrt{\frac{L}{g_e}} \]
\[ T_m = 2\pi \sqrt{\frac{L}{g_m}} \]
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