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 \[\alpha ,\]is given by               

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 :

                 We are given that the simple pendulum of length I is hanging from the roof of a vehicle which is moving down the frictionless inclined plane. 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 we can say that the effective acceleration would be equal to \[{{g}_{eff}}=g\,\cos \alpha \] Now the time period of oscillation is given by \[T=2\pi \sqrt{\frac{l}{{{g}_{eff}}}}=2\pi \sqrt{\frac{l}{g\,\cos \alpha }}\]