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(B) Given the equations of motion:$x = a \sin \omega t \implies \frac{x}{a} = \sin \omega t$  $(i)$$y = a \cos \omega t \implies \frac{y}{a} = \cos \o

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a circular path

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(B) Given the equations of motion:$x = a \sin \omega t \implies \frac{x}{a} = \sin \omega t$  $(i)$$y = a \cos \omega t \implies \frac{y}{a} = \cos \omega t$  $(ii)$Squaring and adding equations $(i)$ and $(ii)$,we get:$\left(\frac{x}{a}ight)^2 + \left(\frac{y}{a}ight)^2 = \sin^2 \omega t + \cos^2 \omega t$Since $\sin^2 	heta + \cos^2 	heta = 1$,we have:$\frac{x^2}{a^2} + \frac{y^2}{a^2} = 1$$x^2 + y^2 = a^2$This is the standard equation of a circle with radius $a$ centered at the origin. Therefore,the particle follows a circular path.

$A$ particle moves in the $x-y$ plane according to the rules $x = a \sin \omega t$ and $y = a \cos \omega t$. The particle follows:

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