Kinematics

JEE Main level, single correct, medium

165 of 308

Question 165 of 308

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(\dfrac{\pi }{2}-\omega t\right)m and y=4sin⁡(ωt)my=4\sin \left(\omega t\right)m. The path of the particle will be

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Answer: (A) circular

x=4sin⁡(π2−ωt)=4cos⁡(ωt)x=4\sin \left(\dfrac{\pi }{2}-\omega t\right)=4\cos \left(\omega t\right),
y=4sin⁡(ωt)y=4\sin \left(\omega t\right)
Clearly, x2+y2=(4cos⁡(ωt))2+(4sin⁡(ωt))2⇒x2+y2=16{x}^{2}+{y}^{2}={\left(4\cos \left(\omega t\right)\right)}^{2}+{\left(4\sin \left(\omega t\right)\right)}^{2} \Rightarrow {x}^{2}+{y}^{2}=16
Hence, the path covered by particle is a circle.

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