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 :

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 \left( {\omega t + \frac{\pi }{4}} \right)$

The path of the particle will be :

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Solution

Hence simwt $=\frac{x}{2}$

$\cos \omega t=\frac{\sqrt{4-x^{2}}}{2}$

$y=2 \sin \omega t+\frac{2}{\sqrt{2}} \cos \omega$ ltbgt $=\sqrt{2} \times \frac{x}{2}+\sqrt{2} \frac{\sqrt{4-x^{2}}}{2}$

$=\frac{x}{\sqrt{2}}+\left(\frac{\sqrt{4-x^{2}}}{\sqrt{2}}\right)$

or $\sqrt{2} y-x=\sqrt{4-x^{2}}$

or $2 y^{2}+x^{2}-x y=4-x^{2}$

or $2 y^{2}+2 x^{2}-x y=4$

or $x^{2}+y^{2}-\sqrt{2} x y=2$

This is an equation of ellipse.

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