What is the frequency of a second's pendulum in an elevatore moving up with an acceleration of `g//2`?

Explanation

For second pendulum, frequency `v=(1)/(2)s.`
When elevator is moving upwards with acceleration a, the effective acceleration due to gravity is
`f_(1)=g+a=g+g//2=3g//2`
As, `v=(1)/(2pi)sqrt((g)/(l))` so `v^(2)propg`
`:. (v_(1)^(2))/v^(2)=(g_(1))/(g)=(3g//2)/(g)=(3)/(2)`
`:. (v_(1))/(v)=sqrt((3)/(2))=1.225`
`v_(1)=1.225v=1.225xx(1)/(2)=0.612s^(-1)`