Q. The acceleration of a particle performing SHM is $12\ cms^{- 2}$ at a distance of 3 cm from the mean position. Its time period is
A) 2.0 s
B) 3.14 s
C) 0.5 s
D) 1.0 s
Explanation
(b) Acceleration, $a = - \omega^{2}y = \frac{- 4\pi^{2}}{T^{2}}y$
or $T = \left( \frac{4\pi^{2}y}{a} \right)^{1/2} = 2\pi\sqrt{\frac{y}{a}}$
$= 2 \times \frac{22}{7} \times \sqrt{\frac{3}{12}} = 3.14$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.