A simple pendulum `50cm` long is suspended from the roof of a acceleration in the horizontal direction with constant acceleration `sqrt(3) g m//s^(-1)`. The period of small oscillations of the pendulum about its equilibrium position is `(g = pi^(2) m//s^(2))` ltbRgt

Options

A. `1.0 sec`
B. `1.25 sec`
C. `1.53 sec`
D. `1.68 sec`

Answer

A

Explanation

With respect to the cart, equilibrium position of pendulum is shown

If displaced by small angle `theta` from this position ,the it will execute `SHM` about this equilibrium position time period of which is given by
`T = 2pi sqrt((L)/(g_(eff))) , g_(eff) = sqrt(g^(2) + (sqrt(3g))^(2))`
`rArr g_(eff) = 2g rArr T = 1.0 second`