Q. The bob of a simple pendulum of length L is released at time t = 0 from a position of small angular displacement. Its linear displacement at time t is given by
A) $X = a\ \sin 2\pi\sqrt{\frac{L}{g\ }} \times t$
B) $X = a\ \cos 2\pi\sqrt{\frac{g}{L\ }} \times t$
C) $X = a\ sin\sqrt{\frac{g}{L\ }} \times t$
D) $X = a\ cos\sqrt{\frac{g}{L\ }} \times t$
Explanation
(d) Time period, $T = 2\pi\sqrt{\frac{L}{g}}$
And $\omega = \frac{2\pi}{T} = \sqrt{\frac{g}{L}}$
Displacement, $x = a\cos{\omega t} = a\cos{\sqrt{\frac{g}{L}}t}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.