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

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

1.	The acceleration due to gravity on the surface of moon is 1.7 ms^(-2) . What is the time period of a simple pendulum on the moon if its time period on the earth is 3.5 s ? (g on earth = 9.8 ms^(-2))
Answer» Solution :`T=2pisqrt(l/g)` Period on the moon = T On EARTH, `3.5 =2pisqrt(l/(9.8))""...(1)` On moon , `T. = 2PI SQRT(l/(1.7))""...(2)` EQUATION (2) `DIV` (1) `(T.)/(3.5) = sqrt((9.8)/(1.7))=2.4` `T. = 3.5 xx2.4 =8.4 s`