Q. The displacement of tow particles executing SHM are represented equations$\ y_{1} = \ 2\sin(10t + \theta),y_{2} = 3cos\ 10t$. The phase difference between the velocity of these particles is
A) $\theta$
B) $- \theta$
C) $\theta + \pi/2$
D) $\theta - \pi/2$
Explanation
(d) $v_{1} = \frac{dy_{1}}{dt} = 2 \times 10\cos{(10t + \theta)}$;
$v_{2} = - 3 \times 10\sin{10t = 30\cos{(10 + \pi/2})}$
$\therefore$ Phase difference$= (10t + \theta) - (10t + \pi/2)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.