The time period of a simple pendulum inside a starionary lift is 2 second. What would be its period, when the lift moves upwards with an acceleration `(g)/(4)` ?

Options

A. 2 sec
B. `(4)/(sqrt(5))` sec
C. `sqrt(5)` sec
D. 4 sec

Answer

B

Explanation

`g'=g+(g)/(4)=(5g)/(4)`
`therefore (T_(2))/(T_(1))=sqrt((4L)/(5g)xx(g)/(L))=sqrt((4)/(5))=(2)/(sqrt(5))`
`therefore T_(2)=(2xx2)/(sqrt(5))=(4)/(sqrt(5))`