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

Acceleration due to garavity on the surface of moon,`g=1.7ms^(2)`
Acceleration due to garavity on the surface of earth, `g=9.8ms^(-2)`
Time period of a simple pendulum on earth, T=3.5s
`T=2pisqrt((I)/(g))`
Where ,
// is the lenth of the pendulum
` thereforel=(T^(2))/((2pi)^(2))xxg=((3.5)^(2))/(4xx(3.14)^(2)xx9.8m)`
The length of the pendulum remains constant.
`T'=2pisqrt((I)/(g))`
`2pisqrt((((3.5)^(2))/(4xx(3.14)^(2))xx9.8)/(1.7))=8.4`s
Hence , the time period of the simple pendulum on the surface of moon is 8.4s.