Question

A particle moves according to the equation $x=a\sin \omega t,y=a\cos \omega t$. Then it's resultant motion is

• A. uniform circular motion and $\omega$ is clockwise.
• B. uniform circular motion and $\omega$ is anticlockwise.
• C. nonuniform circular motion with $\omega$ clockwise.
• D. nonuniform circular motion with $\omega$ anticlockwise.

Answer 1

Answer: (A)

uniform circular motion and $\omega$ is clockwise.

Given $x=a\sin \omega t$ and $y=a\cos \omega t$
so, $x^2+y^2=a^2({\sin }^2\omega t+{\cos }^2\omega t)=a^2\Rightarrow$ its a equation of circle.
here $T=\frac{2\pi }{\omega }$ and the angular velocity does not vary with time the particle. So it is uniform motion.
now $v_x=\frac{dx}{dt}=a\omega \cos \omega t$ and $v_y=\frac{dy}{dt}=-a\omega \sin \omega t$
$a_x=\frac{d^{2}x}{d{t}^2}=-a{\omega }^2\sin \omega t$ and $a_y=\frac{d^{2}y}{d{t}^2}=-a{\omega }^2\cos \omega t$
Thus it is a clock-wise direction.