$$x$$ and $$y$$ are 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:

Using the trigonometric identity $$\sin 2\theta = 2 \sin \theta \cos \theta$$, we can rewrite $$y(t)$$:

$$y = a(2 \sin \omega t \cos \omega t)$$

$$\sin \omega t = \frac{x}{a}$$

Using $$\cos \theta = \sqrt{1 - \sin^2 \theta}$$: $$\cos \omega t = \sqrt{1 - \left(\frac{x}{a}\right)^2}$$

$$y = 2a \left(\frac{x}{a}\right) \sqrt{1 - \frac{x^2}{a^2}}$$

$$y = 2x \sqrt{1 - \frac{x^2}{a^2}}$$

$$y^2 = 4x^2 \left(1 - \frac{x^2}{a^2}\right)$$

$$y^2 = 4x^2 - \frac{4x^4}{a^2}$$

When $$x = 0 \implies y = 0$$. (Passes through the origin). When $$x = \pm a \implies y = 0$$

As $$\omega t$$ goes from $$0$$ to $$\pi/2$$, $$x$$ moves from $$0$$ to $$a$$.

In the same interval, $$2\omega t$$ goes from $$0$$ to $$\pi$$, meaning $$y$$ completes a half-sine wave ($$0 \rightarrow a \rightarrow 0$$). This forms a loop in the first and fourth quadrants. As $$\omega t$$ continues to $$\pi$$, $$x$$ returns to $$0$$ and $$y$$ goes from $$0 \rightarrow -a \rightarrow 0$$.

Shape: This specific frequency ratio of $$1:2$$ with zero phase difference results in a horizontal figure-eight shape.