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JEE Mains · Physics · STD 11 - Motion in plane
  • JEE Mains 2022
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  • MCQ

Motion of a particle in \(x - y\) plane is described by a set of following equations \(x=4 \sin \left(\frac{\pi}{2}-\omega t\right) m\) and \(y=4 \sin (\omega t) m\). The path of particle will be 

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Answer & Solution

Correct Answer

(A) circular

Step-by-step Solution

Detailed explanation

\(x=4 \sin \left(\frac{\pi}{2}-\omega t\right) y=4 \cos (\omega t)\)

\(x=4 \cos (\omega t) \quad y=4 \sin (\omega t)\)

Eliminate ' \(t\) ' to find relation between \(x\) and \(y\)

\(x^{2}+y^{2}=y^{2} \cos ^{2} \omega t+y^{2} \sin ^{2} \omega t=4^{2}\)

\(x^{2}+y^{2}=4^{2}\)

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