Motion of a particle in $$x-y$$ plane is described by a set of following equations $$x = 4\sin\left(\frac{\pi}{2} - \omega t\right)$$ m and $$y = 4\sin(\omega t)$$ m. The path of the particle will be

We are given the equations of motion: $$x = 4\sin\left(\frac{\pi}{2} - \omega t\right)$$ and $$y = 4\sin(\omega t)$$. Using the identity $$\sin\left(\frac{\pi}{2} - \theta\right) = \cos\theta$$, this simplifies to $$x = 4\cos(\omega t)$$, so we have $$x = 4\cos(\omega t)$$ and $$y = 4\sin(\omega t)$$.

Squaring and adding both equations gives:

$$x^2 + y^2 = 16\cos^2(\omega t) + 16\sin^2(\omega t)$$

$$x^2 + y^2 = 16(\cos^2(\omega t) + \sin^2(\omega t))$$

$$x^2 + y^2 = 16$$

This is the equation of a circle with radius 4 m centered at the origin. Therefore, the path of the particle is circular. The correct answer is Option A.