Q. This time period of a particle undergoing SHM is 16 s. It starts motion from the mean position. After 2 s, its velocity is $0.4\ ms^{- 1}.$ The amplitude is
A) 1.44 m
B) 0.72 m
C) 2.88 m
D) 0.36 m
Explanation
(a) Velocity, $v = r\omega\cos{\omega t};$
$0.4 = r \times \frac{2\pi}{16}\cos\frac{2\pi}{16} \times 2 = r \times \frac{2\pi}{16} \times \frac{1}{\sqrt{2}}$
or $r = \frac{0.4 \times 16 \times \sqrt{2}}{2\pi} = \frac{3.2\sqrt{2}}{\pi} = 1.44\ m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.