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 on earth, `T=3.5s`,`g=9.8ms^(-2)` For simple pendulum `T=2pisqrtl/g`
`3.5 = 2pisqrtl/9.8` on moon, `g=1.7ms^(-2)` and if `T` is time period then
`T.=2pisqrt((l/1.7)`
Dividing eqn, (ii) by eqn. (i), we get
`T/3.5=sqrt((9.8/1.7)) or T = sqrt((9.8/1.7))**3.5=8.4s`