The x and y displacements of a particle are g | vedclass

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(A) Given the equations of motion for the particle:$x = a \sin \omega t \Rightarrow \sin \omega t = \frac{x}{a}$$y = a \sin 2 \omega t$Using the trigo

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(A) Given the equations of motion for the particle:$x = a \sin \omega t \Rightarrow \sin \omega t = \frac{x}{a}$$y = a \sin 2 \omega t$Using the trigonometric identity $\sin 2 	heta = 2 \sin 	heta \cos 	heta$,we can write:$y = a (2 \sin \omega t \cos \omega t)$Since $\cos \omega t = \sqrt{1 - \sin^2 \omega t} = \sqrt{1 - (\frac{x}{a})^2} = \frac{\sqrt{a^2 - x^2}}{a}$,we substitute this into the equation for $y$:$y = 2a (\frac{x}{a}) (\frac{\sqrt{a^2 - x^2}}{a})$$y = \frac{2x}{a} \sqrt{a^2 - x^2}$Squaring both sides gives $y^2 = \frac{4x^2}{a^2} (a^2 - x^2)$,which represents a figure-eight curve (Lissajous figure) symmetric about the $x$ and $y$ axes. This corresponds to the shape shown in option $A$.

The $x$ and $y$ displacements of a particle are given as $x(t) = a \sin \omega t$ and $y(t) = a \sin 2 \omega t$. Its trajectory will look like:

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