$x$ 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
JEE MAIN 2015, Diffcult
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$\frac{x}{a}=\sin \omega t$
$\Rightarrow \cos \omega t=\sqrt{1-\left(\frac{x_{2}}{a^{2}}\right)}$
$\frac{y}{a}=\sin 2 \omega t$
$=2 \sin \omega t \times \cos \omega t$
$=2 \frac{x}{a} \times \sqrt{1-\left(\frac{x^{2}}{a^{2}}\right)}$
$ \Rightarrow y = \frac{{2x}}{{{a^2}}}\sqrt {\left( {a - x} \right)\left( {a + x} \right)} $
Hence trajectory of particle will look like as $(\mathrm{c})$
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