If $y = A \sin(\omega t)$, then $v = \frac{dy}{dt} = A\omega \cos(\omega t)$ and $a = \frac{dv}{dt} = -A\omega^2 \sin(\omega t) = A\omega^2 \sin(\omega t + \pi)$. Thus, the phase difference between displacement ($y$) and acceleration ($a$) is $\pi$.