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

Explanation

Here, `g_(m) = 1.7 ms^(-2), g_(e) = 9.8 ms^(-2), T_(m) = ? , T_(e) = 3.5 s^(-1)`
Since, `T_(e) = 2pi sqrt((1)/(8e))` and `T_(m) = 2pi sqrt((1)/(gm))`
`:. (T_(m))/(T_(e)) = sqrt((g_(e))/(g_(m))) rArr T_(m) = T_(e) = sqrt((g_(e))/(g_(m)))`
`3.5 sqrt((9.8)/(1.7)) = 8.4 s`.