Q. A body of mass $500\ g$ is attached to a horizontal spring of spring constant $8\pi^{2}\ N\ m^{- 1}.$ If the body is pulled to a distance of $10\ cm$ from its mean position, then its frequency of oscillation is
A) $2\ Hz$
B) $4\ Hz$
C) $8\ Hz$
D) $0.5\ Hz$
Explanation
(a) Here, Mass of the body, $m = 500g = 500 \times 10^{- 3}kg$
Spring constant, $k = 8\pi^{2}\ N\ m^{- 1}$
The frequency of oscillation is
$v = \frac{1}{2\pi}\sqrt{\frac{k}{m}} = \frac{1}{2\pi}\sqrt{\frac{8\pi^{2}\ {N\ m}^{- 1}}{500 \times 10^{- 3}\ kg}} = 2\ Hz$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.