Question

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

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

Answer 1

Answer: (A)

$x=4\sin \left(\frac{\pi }{2}-\omega t\right)y=4\cos (\omega t)$
$x=4\cos (\omega t)y=4\sin (\omega t)$
Eliminate ' $t$ ' to find relation between $x$ and $y$
$x^2+y^2=y^2{\cos }^2\omega t+y^2{\sin }^2\omega t=4^2$
$x^2+y^2=4^2$