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. `sqrt(2) sec`
C. 1.53 sec
D. 1.68 sec

Answer

A

Explanation

With respect to the cart, equilibrium position of the pendulum is shown.
If displaced by small angle `theta` from this positon, then it will execute SHM about this equilibrium position, time period of which is given by

`T=2pisqrt(L/(g_(eff))) , g_(eff) =sqrt(g^(2)+(sqrt(3g)^(2)))rArr g_(eff) =2grArr T=1.0` second