Q. A simple pendulum performs simple harmonic motion about $X = 0$ with an amplitude $A$ and time period $T.$ The speed of the pendulum at $X = \frac{A}{2}$ will be
A) $\frac{\pi A\sqrt{3}}{T}$
B) $\frac{\pi A}{T}$
C) $\frac{\pi A\sqrt{3}}{2T}$
D) $\frac{3\pi^{2}A}{T}$
Explanation
(a) Velocity of a particle executing S.H.M. is given by
$v = \omega\sqrt{a^{2} - x^{2}} = \frac{2\pi}{T}\sqrt{A^{2} - \frac{A^{2}}{4}} = \frac{2\pi}{T}\sqrt{\frac{3A^{2}}{4}} = \frac{\pi A\sqrt{3}}{T}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.