Two vectors $\overrightarrow{A} \mathrm{a}\mathrm{n}\mathrm{d} \overrightarrow{B}$ are defined as $\overrightarrow{A}=a\widehat{i} and \overrightarrow{B}=a\left(\cos \omega t \widehat{i}+\sin \omega t\widehat{j}\right)$, where a is a constant and $\omega =\frac{\pi }{6} rad s^{-1}.$ If $\left|\overrightarrow{A}+\overrightarrow{B}\right|=\sqrt{3}\left|\overrightarrow{A}-\overrightarrow{B}\right|$ at time $t=\tau$ for the first time, the value of $\tau$ in seconds, is_______

Solution

$\left|\overrightarrow{A}+\overrightarrow{B}\right|=2 a\cos \frac{\omega t}{2}$
$\left|\overrightarrow{A}-\overrightarrow{B}\right|=2a\sin \frac{\omega t}{2}$
So $2a\cos \frac{\omega t}{2}=\sqrt{3}\left(2a\sin \frac{\omega t}{2}\right)$
$\tan \frac{\omega t}{2}=\frac{1}{\sqrt{3}}$
$\frac{\omega t}{2}=\frac{\pi }{6}\Rightarrow \omega t=\frac{\pi }{3}$
$\frac{\pi }{6}t=\frac{\pi }{3}$
$t=2.00 sec$

Hence, the correct answer is 2.