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))

The acceleration due to gravity on the surface of moon is 1.7 m s^(-2)

1.	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))
Answer» SOLUTION :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`.