Question 54 of 194

$xandy$ displacements of a particle are given as $x\left(t\right)=a\sin ⁡\omega tandy\left(t\right)=a\sin ⁡2\omega t.$ Its trajectory will look like:

ABCD

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Answer: (C)

$∵x=A\sin ⁡\omega t\Rightarrow sin\omega t=\dfrac{x}{A}$

Also, $cos\omega t=\sqrt{1-{\sin }^{2}⁡\omega t}=\sqrt{1-\dfrac{{x}^{2}}{{A}^{2}}}$

$\Rightarrow \cos ⁡\omega t=\dfrac{\sqrt{{A}^{2}-{x}^{2}}}{A}$

As, $y=2A\sin ⁡\omega t\cos ⁡\omega t$

$\Rightarrow y=2A\dfrac{x}{A}\dfrac{\sqrt{{A}^{2}-{x}^{2}}}{A}$

$\Rightarrow y=\dfrac{2}{A}x\sqrt{{A}^{2}-{x}^{2}}$

$\Rightarrow y=0atx=0andx=\pm A$

Which is possible only in option (C)