Question

The magnitude of displacement vector of a particle which is moving in a circle of radius $a$ with constant angular velocity $\omega $ as a function of time is

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

Answer 1

Answer: (B)

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

This means the magnitude of the displacement of the particle is
$ s = \textit{PQ} = \sqrt{\textit{QR}^{2} + \textit{PR}^{2}}$
$= \sqrt{\left(\textit{a sin} \omega t\right)^{2} + \left(a - \text{a cos} \omega t\right)^{2}}$
$s =2a \, \text{sin}\frac{\omega t}{2}$