What is the phase difference between two simple harmonic motions represented by x1=A sin(omegat +pi/6) and x2=A cos omegat A. pi/6 B. pi/3 C. pi/2 D. (2pi)/3 ?

$B$>
$A \cos \omega t$ can be written as $A \sin \left(\frac{\pi}{2} + \omega t\right)$
So, $\partial \phi = \left(\frac{\pi}{2} + \omega t - \omega t - \frac{\pi}{6}\right) = \frac{\pi}{3}$