JEE MAIN - Physics (2022 - 28th June Morning Shift - No. 3)
Motion of a particle in x-y plane is described by a set of following equations $$x = 4\sin \left( {{\pi \over 2} - \omega t} \right)\,m$$ and $$y = 4\sin (\omega t)\,m$$. The path of the particle will be :
circular
helical
parabolic
elliptical
Explanation
$$x = 4\sin \left( {{\pi \over 2} - \omega t} \right)$$
$$ = 4\cos (\omega t)$$
$$y = 4\sin (\omega t)$$
$$ \Rightarrow {x^2} + {y^2} = {4^2}$$
$$\Rightarrow$$ The particle is moving in a circular motion with radius of 4 m.
Comments (0)
