Answer
The distance between the two particles is $~~0.18~A$
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 find the time when the lagging particle is at the most positive end of the path:
$x_2 = \frac{A}{2}~cos(4.189~t - \frac{\pi}{6}) = \frac{A}{2}$
$cos(4.189~t - \frac{\pi}{6}) = 1$
$4.189~t - \frac{\pi}{6} = 0$
$4.189~t = \frac{\pi}{6}$
$t = \frac{\pi}{(6)(4.189)}$
$t = 0.125~s$
The time $0.50~s$ later is $t = 0.625~s$
We can find the position of each particle:
$x_1 = \frac{A}{2}~cos[(4.189)(0.625)] = -0.433~A$
$x_2 = \frac{A}{2}~cos[(4.189)(0.625) - \frac{\pi}{6})] = -0.25~A$
We can find the distance between the two particles:
$\Delta x = (-0.25~A)-(-0.433~A) = 0.18~A$
The distance between the two particles is $~~0.18~A$
