Q. A simple pendulum of length $l$ has been set up inside a railway wagon sliding down a frictionless inclined plane having an angle of inclination $\theta = 30^{0}$ with the horizontal. What will be its period of oscillation as recorded by an observer inside the wagon?
A) $2\pi\sqrt{\frac{2l}{\sqrt{3\ g}}}$
B) $2\pi\sqrt{2l/g}$
C) $2\pi\sqrt{l/g}$
D) $2\pi\sqrt{\frac{\sqrt{3l}}{2g}}$
Explanation
(a) On the inclined plane, the effective acceleration due to gravity
$g^{'} = g\cos{30{^\circ}}$
$= g \times \sqrt{3}/2$
$\therefore\ \ \ T = 2\pi\sqrt{\frac{l}{g^{'}}} = 2\pi\sqrt{\frac{2l}{\sqrt{3}g}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.