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 (prop), is given by.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))`

The period of oscillation of a simple pendulum of length (L) suspended

1.	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 (prop), is given by.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» Correct Answer - A<br>See the following force diagram<br> <br>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`<br> `:.` Time period `T = 2 pi sqrt((1)/(g_(eff))) = 2pi sqrt((1)/( g cos theta))`

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