Q. A particle of mass 10 g is executing simple harmonic motion with an amplitude of 0.5 m and periodic time of $(\pi/5)$ s. The maximum value of the force acting on the particle is
A) 25 N
B) 5 N
C) 2.5 N
D) 0.5 N
Explanation
(d) $F_{\max} = m\omega^{2}a = m\frac{4\pi^{2}}{T^{2}}a$
$= \frac{10}{1000} \times \frac{4 \times (\pi)^{2}}{(\pi/5)} \times 0.5 = 0.5\ N$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.