Question

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

• A. an ellipse
• B. a straight line
• C. a parabola
• D. a circle

Answer 1

The correct option is A an ellipse

$x=2\sin \omega t;y=2\sin (\omega t+\frac{\pi }{4})$
$\Rightarrow \sin \omega t=\frac{x}{2}$ .....$\left(I\right)$
$\Rightarrow \cos \omega t=\sqrt{1-{\sin }^2\omega t}=\sqrt{1-{\left(\frac{x}{2}\right)}^2}=\sqrt{1-\frac{x^2}{4}}$ .....$\left(II\right)$
$y=2\sin (\omega t+\frac{\pi }{4})\Rightarrow y=2\frac{(\sin \omega t+\cos \omega t)}{\sqrt{2}}=\sqrt{2}(\sin \omega t+\cos \omega t)$
using the values of from $\left(I\right)$ and $\left(II\right)$, we have
$y=\sqrt{2}(\frac{x}{2}+\sqrt{1-\frac{x^2}{4}})\Rightarrow \frac{y}{\sqrt{2}}-\frac{x}{2}=\sqrt{1-\frac{x^2}{4}}\Rightarrow \frac{y^2}{2}+\frac{x^2}{4}-\frac{xy}{\sqrt{2}}=1-\frac{x^2}{4}\Rightarrow \frac{x^2}{2}+\frac{y^2}{2}-\frac{xy}{\sqrt{2}}-1=0\Rightarrow x^2+y^2-\sqrt{2}xy-2=0$
which is an equation of an oblique ellipse.