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JEE PHYSICS MCQ: WORK-ENERGY-POWER

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Analyzing +4 | -1
Motion of a particle in x−yx - y plane is described by a set of following equations x=4sin⁡(π2−ωt)mx = 4\sin \left( {\frac{\pi }{2} - \omega t} \right)m and y=4sin⁡(ωt)m.y = 4\sin \left( {\omega t} \right)m. The path of the particle will be
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Correct option: A
Circular
Option A is correct because the given equations describe a circular path with a radius of 4 m.

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