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(A) Given the equations of motion:$x = 2 \sin \omega t \implies \sin \omega t = \frac{x}{2}$$y = 2 \sin \left( \omega t + \frac{\pi}{4} \right) = 2 \l

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an ellipse

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(A) Given the equations of motion:$x = 2 \sin \omega t \implies \sin \omega t = \frac{x}{2}$$y = 2 \sin \left( \omega t + \frac{\pi}{4} \right) = 2 \left( \sin \omega t \cos \frac{\pi}{4} + \cos \omega t \sin \frac{\pi}{4} \right)$Since $\sin \omega t = \frac{x}{2}$,then $\cos \omega t = \sqrt{1 - \sin^2 \omega t} = \sqrt{1 - \frac{x^2}{4}} = \frac{\sqrt{4 - x^2}}{2}$.Substituting these into the expression for $y$:$y = 2 \left( \frac{x}{2} \cdot \frac{1}{\sqrt{2}} + \frac{\sqrt{4 - x^2}}{2} \cdot \frac{1}{\sqrt{2}} \right)$$y = \frac{x}{\sqrt{2}} + \frac{\sqrt{4 - x^2}}{\sqrt{2}}$$\sqrt{2} y - x = \sqrt{4 - x^2}$Squaring both sides:$(\sqrt{2} y - x)^2 = 4 - x^2$$2y^2 + x^2 - 2\sqrt{2} xy = 4 - x^2$$2x^2 + 2y^2 - 2\sqrt{2} xy = 4$$x^2 + y^2 - \sqrt{2} xy = 2$This is the general equation of an ellipse ($Ax^2 + Bxy + Cy^2 = F$ where $B^2 - 4AC < 0$).

$A$ particle is subjected to two mutually perpendicular simple harmonic motions such that its $x$ and $y$ coordinates are given by:
$x = 2 \sin \omega t$
$y = 2 \sin \left( \omega t + \frac{\pi}{4} \right)$
The path of the particle will be:

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$A$ particle is subjected to two mutually perpendicular simple harmonic motions such that its $x$ and $y$ coordinates are given by:$x = 2 \sin \omega t$$y = 2 \sin \left( \omega t + \frac{\pi}{4} \right)$The path of the particle will be: