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

The acceleration due to gravity on the surface of the moon is 1.7 ms^(

1.	The acceleration due to gravity on the surface of the moon is 1.7 ms^(-2). What is the time period of a simple pendulum on the surface of moon if its time period on the surface earth is 3.5 s. ? (g on the surface earth 9.8 ms^(-2))
Answer» SOLUTION :Here, `g_m = 1.7 MS^(-2) , g_e = 9.8 ms^(-2) , T_e = 3.5 s^(-1)` As `T_e = 2pi SQRT(L/(g_e)) and T_m = 2pi sqrt(l/g_m)` `therefore (T_m)/(T_e) =sqrt((g_e)/(g_m)) ` (or) `T_m = T_e sqrt((g_e)/(g_m)) = 3.5 sqrt((9.8)/(1.7)) =8.4 s`