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A particle moves in $x-y$ plane according to rule $\mathrm{x}=\mathrm{a} \sin \omega \mathrm{t}$ and $\mathrm{y}=\mathrm{a} \cos \omega \mathrm{t}$. The particle follows
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Verified Answer
The correct answer is:
a circular path
$$
\begin{aligned}
& \mathrm{x}=\mathrm{a} \sin \omega \mathrm{t} \Rightarrow \frac{\mathrm{x}}{\mathrm{a}}=\sin \omega \mathrm{t} \\
& \mathrm{x}=\mathrm{a} \cos \omega \mathrm{t} \Rightarrow \frac{\mathrm{x}}{\mathrm{a}^2}=\cos \omega \mathrm{t} \\
& \therefore \quad \frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1 \\
& \text { or } \quad \mathrm{x}^2+\mathrm{y}^2=\mathrm{a}^2
\end{aligned}
$$
This is equation of circle, so the particle follows a circular path.
\begin{aligned}
& \mathrm{x}=\mathrm{a} \sin \omega \mathrm{t} \Rightarrow \frac{\mathrm{x}}{\mathrm{a}}=\sin \omega \mathrm{t} \\
& \mathrm{x}=\mathrm{a} \cos \omega \mathrm{t} \Rightarrow \frac{\mathrm{x}}{\mathrm{a}^2}=\cos \omega \mathrm{t} \\
& \therefore \quad \frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1 \\
& \text { or } \quad \mathrm{x}^2+\mathrm{y}^2=\mathrm{a}^2
\end{aligned}
$$
This is equation of circle, so the particle follows a circular path.
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