Q. Two SHMs are represented by the equations $y_{1} = 0.1\sin{(100\pi t + \pi/3}$ and $y_{2} = 0.1cos100\pi t.$ The phase difference of velocity of particle 2 with respect to the velocity of particle 1 is
A) $- \pi/3$
B) $\pi/6$
C) $- \pi/6$
D) $\pi/3$
Explanation
(b) $v_{1} = \frac{dy_{1}}{dt}$
$= 0.1 \times 100\pi\cos{(100\pi t + \pi/3)}$
$v_{2} = \frac{dy_{2}}{dt} = 0.1 \times 100\pi\sin{100\ \pi t}$
$= 0.1 \times 100\pi\cos{(100\pi t + \pi/2)}$
Phase difference between two velocities
$(100\pi t + \pi/2) - (100\pi t + \pi/3)$
$= \pi/2 - \pi/3 - \pi/6$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.