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

 

Option: 1

circular
 


Option: 2

helical
 


Option: 3

parabolic
 


Option: 4

elliptical


Answers (1)

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

\mathrm{x=4 \cos \omega t} \\              .........(1)

\mathrm{y=4 \sin \omega t}              ..........(2)

Square both equation and add

\mathrm{x^{2}+y^{2}=4\left[\cos ^{2} \omega t+\sin ^{2} \omega t\right]} \\

\mathrm{x^{2}+y^{2}=4}

this is equation of circle of radius 2m and centre at origin.

Hence correct option is 1

Posted by

Kshitij

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