The time period of a pendulum on the surfacce of the moon is 5s. If it is a seconds pendulum on earth and the acceleratio due to gravity on the earth is `9.8 m s^(-2)`, find the acceleration due to gravity on the surface of the moon.

Explanation

`Ta sqrt((1)/(g))`
`(T_E )/( T_M) = sqrt((g_M)/(g_E))`
` 2/5 = sqrt((g_M)/(g))`
` (4)/(25 ) = (g_M)/(g)`
` g_M = 4/(25) xx9.8 ( :. g= 9.8 m s^(-2))`
`= 1.6 s^(-2)`