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((L)/(g cos alpha))`
B. `2pi sqrt((L)/(g sin alpha))`
C. `2pi sqrt((L)/(g))`
D. `2pi sqrt((L)/(g tan alpha))`

Answer

A

Explanation

See the following force diagram

Vehicle is moving down the frictionless inclined surfacwe so, its acceleration is `g sin theta` Since vehicle is acceleration apseudo force `m(g sintheta)` will act on bob of pendulum which cancels the `sintheta ` component of weight of the bob.Hence net force on the bob is `F_(net) = mg cos theta ` or net acceleration of the bob is `g_(eff) = g cos theta`
`:.` Time period `T = 2 pi sqrt((1)/(g_(eff))) = 2pi sqrt((1)/( g cos theta))`