Question

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

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

Answer 1

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

Let the particle start the motion at point $P$, as shown in the below figure.


In time $t$, particle has rotated an angle, $\theta =\omega t$
So, the displacement is given by
$s=PQ=\sqrt{Q{R}^2+P{R}^2}=\sqrt{(a\sin \theta {)}^2+(a-a\cos \theta {)}^2}$
$=\sqrt{(a\sin \omega t{)}^2+(a-a\cos \omega t{)}^2}$
$=\sqrt{2a^2-2a^2\cos \omega t}$
$=\sqrt{2a^2\times 2{\sin }^2\left(\frac{\omega t}{2}\right)}$
$\Rightarrow s=2a\sin \frac{\omega t}{2}$