The acceleration due to gravity on the surface of moon is `1.7 m s^(-2)`. What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5 s ? (g on the surface of earth is 9.8 m `s^(-2)`)

Explanation

Suppose the time period of a simple pendulum on earth `T_(e ) = 3.5s` and on moon is `T_m` and length l of simple pendulum is same.
Time period of simple pendulum, `T= 2pi sqrt((l)/(g))`, here `2pi` and l are constant
`therefore T propto (1)/(sqrt(g))`
`therefore (T_m)/(T_e)= sqrt((g_e)/(g_m))`
`therefore T_(m)= T_(e ) xx sqrt((g_e)/(g_m))`
`therefore T_(m)= 3.5xx sqrt((9.8)/(1.7)) = 3.5xx sqrt(5.76)`
`= 3.5xx 2.4 " "` [Here `T_(m) lt T_(e )`, hence pendulum clock works very slowly]
`= 8.4s`.