The mass and the diameter of a planet are three times the repective value for the Earth. The period of oscillation of a simple pendulum on the Earth is 2 s. The period of oscillation of the same pendulum on the planet would be:

Answer

`03.46`

Explanation

Time period of simple pendulum, `T = 2pisqrt(l/g)`
`g_(P) = (GM_(p))/R_(p)^(2) = 4(GM_(p))/D_(P)^(2) = (4GM_(E))/D_(E)^(2) xx 3/3^(2) = g_( c)/3`
`rArr g_(P)/g_(e) = 1/3`
`therefore T_(e)/T_(P) = sqrt(g_(p)/g_(e)) = sqrt(1/3)`
`rArr T_(P) = sqrt(3) xx 2 = 3.46 s`