JEE Mains · Physics · STD 11 - 3.2 motion in plane
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
- A circular
- B helical
- C parabolic
- D elliptical
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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