The period of oscillation of a simple pendulum of length (L) suspended from the roof of a vehicle which moves without friction down an inclined plane of inclination `alpha`, is given by.

Options

A. `2pi sqrt((1)/(g cos alpha))`
B. `pi sqrt((1)/(g cos alpha))`
C. `(1)/(2pi) sqrt((1)/(g cos alpha))`
D. `(1)/(pi) sqrt((1)/(2g cos alpha))`

Answer

A

Explanation

Since the vehicle is moving down the plane with an acceleration (because each and every acceleration done system is acceleration with same accleration done the plane) the string will remain perpendicular to if plane.
When the bob is increased, it moves down to the loss of the verticle relative it vehicle with a acceleration
`g = sqrt((g^(2) - g^(2) sin^(2) alpha)) = g cos alpha`
The period of oscilation of teh perpendicular
`T = 2 pi sqrt((1)/(g')) = 2pi sqrt((1)/(g cos alpha))`.