Q. A particle executes SHM in a line $4\ cm$ long. Its velocity when passing through the centre of line is $12\ cm/s.$ The period will be
A) $2.047\ s$
B) $1.047\ s$
C) $3.047\ s$
D) $0.047\ s$
Explanation
(b) Length of the line = Distance between extreme positions of oscillation $= 4\ cm$
So, Amplitude $a = 2\ cm$
also $v_{\max} = 12\ cm/s$
$\because v_{\max} = \omega a = \frac{2\pi}{T}a$
$\Rightarrow T = \frac{2\pi a}{v_{\max}} = \frac{2 \times 3.14 \times 2}{12} = 1.047\ s$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.