Q. A particle executing simple harmonic motion with amplitude of $0.1\ m.$ At a certain instant when its displacement is $0.02\ m,$ its acceleration is $0.5\ m/s^{2}.$ The maximum velocity of the particle is (in m/s)
A) 0.01
B) 0.05
C) 0.5
D) 0.25
Explanation
(c) Acceleration $A = \omega^{2}y \Rightarrow \omega = \sqrt{\frac{A}{y}} = \sqrt{\frac{0.5}{0.02}} = 5$
Maximum velocity $v_{\max} = a\omega = 0.1 \times 5 = 0.5$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.