The time period of a simple pendulum inside a stationary lift is `sqrt(5)` s. What will be the time period when the lift moves upward with an acceleration `(g)/(4)`?

Options

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

Answer

C

Explanation

In this case `g'=g+(g)/(4)=(5g)/(4)`
`T=2pi sqrt((l)/(g)) and T'=2pi sqrt((4l)/(5g))`
`therefore T'=(2)/(sqrt(5))xx2pi sqrt((l)/(g))=(2)/(sqrt(5))xxT`
`=(2)/(sqrt(5))xxsqrt(5)=2sec`