Answer
Both particles are moving in the same direction.
Work Step by Step
We can express the positions of the two particles as follows:
$x_1 = \frac{A}{2}~cos(\omega~t)=\frac{A}{2}~cos(\frac{2\pi}{1.5}~t) = \frac{A}{2}~cos(4.189~t)$
$x_2 = \frac{A}{2}~cos(\omega~t-\frac{\pi}{6}) = \frac{A}{2}~cos(\frac{2\pi}{1.5}~t-\frac{\pi}{6}) = \frac{A}{2}~cos(4.189~t - \frac{\pi}{6})$
We can express the velocities of the two particles as follows:
$v_1 = -\frac{A~\omega}{2}~sin(\omega~t)=-\frac{A~\omega}{2}~sin(\frac{2\pi}{1.5}~t) = -\frac{A~\omega}{2}~sin(4.189~t)$
$v_2 = -\frac{A~\omega}{2}~sin(\omega~t-\frac{\pi}{6}) = -\frac{A~\omega}{2}~sin(\frac{2\pi}{1.5}~t-\frac{\pi}{6}) = -\frac{A~\omega}{2}~sin(4.189~t - \frac{\pi}{6})$
In part (a), we found that the time is $t = 0.625~s$ at a time that is $0.50~s$ after he time when the lagging particle is at the most positive end of the path.
We can find the velocity of each particle:
$v_1 = -\frac{A~\omega}{2}~sin[(4.189)(0.625)] = -0.25~A~\omega \lt 0$
$v_2 = -\frac{A~\omega}{2}~sin[(4.189)(0.625) - \frac{\pi}{6})] = -0.43~A~\omega \lt 0$
We can see that both particles are moving in the negative direction.
Thus, both particles are moving in the same direction.
