The time period of a simple pendulum measured inside a stationary lift is found to be T . If the lift starts accelerating upwards with an acceleration `g //3`, the time period is

Options

A. `(T)/(3)`
B. `(sqrt(3))/(2)(T)`
C. 3 T
D. `sqrt((3)/(2))T`

Answer

B

Explanation

`T=2pi sqrt((l)/(g))` Original period
When it moves up, `a=(g)/(3) g' =g+(g)/(3)=(4g)/(3)`
`therefore T'=2pi sqrt((3l)/(4g))=(sqrt(3))/(2) (2pi sqrt((l)/(g))) therefore T'=(sqrt(3))/(2)T`