The time period of a simple pendulum is T remaining at rest inside a lift. Find the time period of pendulum when lift starts to move up with an acceleration of g/4

Options

A. T
B. T/2
C. 2T/5
D. `2T//sqrt(5)`

Answer

D

Explanation

When the lift is at rest, the time period of the simple pendulum is `T =(2pi)sqrt((T)/(g))`
Now, if the lift is moving up with an acceleration `g/4` then the effective acceleration due to gravity
`g'=g+(g)/(4)=(5g)/(4)`
`therefore ` Time period `T'=2pisqrt((lk)/(g'))`
`=2pisqrt((l)/(5g//4))`
`=(2)/(sqrt(5))xx2pisqrt((l)/(g))=(2)/(sqrt(5))T`