The displacement equation of particle executing $SHM$ is
$x=a \cos (\omega t+\phi)$(i)
Velocity, $v=\frac{d x}{d t}=-a \omega \sin (\omega t+\phi)$ (ii)
Acceleration,
$A=\frac{d v}{d t}=-a \omega^2 \cos (\omega t+\phi)$ (iii)
Fig. (i) is a plot of Eq. (i) with $\phi=0$. Fig. (ii) shows Eq. (ii) also with $\phi=0$. Fig. (iii) is a plot of Eq. (iii). It should be noted that in the figures the curve of $v$ is shifted (to the left) from the curve of $x$ by one-quarter period $\left(\frac{1}{4} T\right)$. Similarly, the acceleration curve of $A$ is shifted (to the left) by $\frac{1}{4} T$ relative to thevelocity curve of $v$. This implies that velocity is $90^{\circ}(0.5 \pi)$ out of phase with the displacement and the acceleration is $90^{\circ}(0.5 \pi)$ out of phase with the velocity but $180^{\circ}(\pi)$ out of phase with displacement.
