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 \[T=2\pi \sqrt{\frac{l}{{{g}_{eff}}}}=2\pi \sqrt{\frac{l}{g\cos \theta }}\]You need to login to perform this action.
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