A pendulum suspended from the roof of an elevator at rest has a time period `T_(1)`, when the elevator moves up with an acceleration `a` its time period becomes `T_(2)`, when the elevator moves down with an acceleration `a`, its time period becomes `T_(3)`, then

Options

A. `T_(1) = sqrt((T_(3)T_(2)))`
B. `T_(3) = sqrt((T_(1)^(2)T_(2)^(2)))`
C. `T_(1) = (sqrt(2) T_(2)T_(3))/(sqrt(T_(2)^(2)+T_(3)^(2)))`
D. `T_(1) = (T_(2)^(2))/(T_(3))`

Answer

C

Explanation

`T_(1) = 2pi sqrt((l)/(g)), T_(2) = 2pi sqrt((l)/(g + a)), T_(3) = 2 pi sqrt((l)/(g-a))`
`(2)/(T_(1)^(2)) = (1)/(T_(2)^(2)) + (1)/(T_(3)^(2))" "rArr T_(1) = (sqrt(2) T_(2)T_(3))/(sqrt(T_(2)^(2) + T_(3)^(2)))`