A particle is subjected to two mutually perpendicular SHMs of the same frequency: $x = A \sin(\omega

Question

A particle is subjected to two mutually perpendicular SHMs of the same frequency: $x = A \sin(\omega t)$ and $y = A \sin(\omega t + \pi/2)$. What is the trajectory of the particle?

Professor PC Thomas Classes · Answer

Here $y = A \cos(\omega t)$. Squaring and adding gives $x^2 + y^2 = A^2(\sin^2\omega t + \cos^2\omega t) = A^2$. This is the equation of a circle.