Verified Solution • 2025–26 Session moderate

QJEEM22Ph1a10: JEE Main 2022 Physics - Class 11 JEEM22Ph1a10 Solution: Motion of a particle in x-y plane is described by a set of…

Motion of a particle in x-y plane is described by a set of following equations x=4sin⁡(π2−ωt) mx = 4\sin \left( {{\pi \over 2} - \omega t} \right)\,m and y=4sin⁡(ωt) my = 4\sin (\omega t)\,m . The path of the particle will be :

Multiple Choice Options

  • A circular
  • B helical
  • C parabolic
  • D elliptical

Step-by-Step Verified Solution

x=4sin⁡(π2−ωt)x = 4\sin \left( {{\pi \over 2} - \omega t} \right)

=4cos⁡(ωt)= 4\cos (\omega t)

y=4sin⁡(ωt)y = 4\sin (\omega t)

⇒x2+y2=42\Rightarrow {x^2} + {y^2} = {4^2}

⇒\Rightarrow The particle is moving in a circular motion with radius of 4 m.

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