Solution:
Superposition of two mutually perpendicular SHMs of equal amplitude and a phase difference of $\pi/2$ produces a circular trajectory.\n$x = A\sin(\omega t)$ and $y = A\sin(\omega t + \pi/2) = A\cos(\omega t)$ gives $x^2 + y^2 = A^2$.
Two S.H.M.s with same amplitude and time period, when acting together in perpendicular directions with a phase difference of $\pi/2$, give rise to: (1997)
Leave a Reply