The time period of a simple pendulum of length L as measured in an elevator descending with acceleration g / 3 is

Options

A. `2pi sqrt((3L)/(2g))`
B. `pi sqrt((3L)/(g))`
C. `2 pi sqrt((3L)/(g))`
D. `2pi sqrt((2L)/(3g))`

Answer

A

Explanation

The effective acceleration in a lift descending with acceleration `(g)/(3)` is
`g_(eff) = g - (g)/(3) = (2g)/(3)`
Time period of simple pendulum
`T = 2pi sqrt((L)/(g_(eff))) = 2pi sqrt((L)/(2g//3)) = 2pi sqrt((3L)/(2g))`