question_answer

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 a, is given by [IIT-JEE (Screening) 2000]

A)            \[2\pi \sqrt{\frac{L}{g\cos \alpha }}\]                            

B)            \[2\pi \sqrt{\frac{L}{g\sin \alpha }}\]

C)            \[2\pi \sqrt{\frac{L}{g}}\]

D)            \[2\pi \sqrt{\frac{L}{g\tan \alpha }}\]

Correct Answer: A

Solution :

                   See the following force diagram.                                                     Vehicle is moving down the frictionless inclined surface so, its acceleration is\[g\sin \theta \]. Since vehicle is accelerating, a pseudo force \[m(g\sin \theta )\] will act on bob of pendulum which cancel the \[\sin \theta \] component of weight of the bob. Hence net force on the bob is Fnet \[=mg\cos \theta \] or net acceleration of the bob is \[{{g}_{eff}}=g\cos \theta \]                    \[\therefore \] Time period \[T=2\pi \sqrt{\frac{l}{{{g}_{eff}}}}=2\pi \sqrt{\frac{l}{g\cos \theta }}\]