Question

The magnitude of displacement of a particle moving in a circle of radius $a$ with constant angular speed $\omega$ varies with time $t$ is

• A. $2a\cos \omega t$
• B. $2a\cos \frac{\omega t}{2}$
• C. $2a\sin \omega t$
• D. $2a\sin \frac{\omega t}{2}$

Answer 1

The correct option is D $2a\sin \frac{\omega t}{2}$

Draw a diagram of particle performing circular motion, showing angular displacement $\theta$ in time interval $t$


Find displacement in $x$ direction and in $y$ direction

Particle starting from point $P$ and going till point $Q$ as shown in figure
$PR$ is displacement in $x$ direction and $RQ$ is displacement in $y$ direction

$PR=OP-OR=a-a\cos \theta$
$RQ=a\sin \theta$

In time $t$ particle has rotated an angle, $\theta =\omega t$

Calculate linear displacement
$s=PQ=\sqrt{Q{R}^2+P{R}^2}$
$s=\sqrt{{\left(a\sin \omega t\right)}^2+{(a-a\cos \omega t)}^2}$
$s=2a\sin \frac{wt}{2}$

Final Answer: (b)