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 \[T=2\pi \sqrt{\frac{l}{{{g}_{eff}}}}=2\pi \sqrt{\frac{l}{g\,\cos \alpha }}\]You need to login to perform this action.
You will be redirected in
3 sec