Question

The time period of a simple pendulum of length l in a lift descending with acceleration g/3

Answer 1

A=s/t2
3=I/t2
t2=I/3
t=rootover of I/3

Answer 2

Answer:

T =

$2\pi \times \sqrt{3l \div 2g}$

Explanation:

The effective accelaration of a lift descending with accelaration g/3 =

g - g/3 = 2g/3

Time period of a simple pendulum, T = 2π√ 3l/2g