Question

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. $2\pi \sqrt{\frac{1}{g\sin \alpha }}$
• C. $2\pi \sqrt{\frac{l}{g}}$
• D. $2\pi \sqrt{\frac{l}{g\tan \alpha }}$

Answer 1

The correct option is A $2\pi \sqrt{\frac{l}{g\cos \alpha }}$

We are given that the simple pendulum of length $l$ 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 $m\left(g\sin \theta \right)$ 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
$T=2\pi \sqrt{\frac{1}{g_{eff}}}=2\pi \sqrt{\frac{l}{g\cos \alpha }}$
l