Q. Two equations of two S.H.M. are $y = a\sin{(\omega t - \alpha)}$ and $y = b\cos{(\omega t - \alpha)}.$ The phase difference between the two is
A) $0{^\circ}$
B) $\alpha{^\circ}$
C) $90{^\circ}$
D) $180{^\circ}$
Explanation
(c) $y = a\sin{(\omega t - \alpha) = a\cos\left( \omega t - \alpha - \frac{\pi}{2} \right)}$
Another equation is given $y = b\cos{(\omega t - \alpha)}$
So, there exists a phase difference of $\frac{\pi}{2} = 90{^\circ}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.