Figure shows that two respective particles P' and Q' in SHM along with their corresponding particles in circular motion. Let P moves in upward direction when crossing Q' at A/2 as shown in Fig. (a), at this instant phase of P is 

${\mathrm{\varphi }}_1={\sin }^{-1}\left(\frac{1}{2}\right)=\frac{\mathrm{\pi }}{6}$                  …(i)

Similarly, as shown in Fig. (b) we can take particle Q' is moving in downward direction (opposite P) at A/2, this implies its circular motion particle is in second quadrant thus its phase angle is

${\mathrm{\varphi }}_2=\mathrm{\pi }-{\sin }^{-1}\left(\frac{1}{2}\right)=\mathrm{\pi }-\frac{\mathrm{\pi }}{6}=\frac{5\mathrm{\pi }}{6}$              …(ii)

As both are oscillating at same angular frequency their phase difference remains constant which can be given from Eqs. (i) and (ii) as

$\mathrm{\Delta \varphi }={\mathrm{\varphi }}_2-{\mathrm{\varphi }}_1=\frac{5\mathrm{\pi }}{6}-\frac{\mathrm{\pi }}{6}=\frac{2\mathrm{\pi }}{3}\mathrm{rad}={120}^{\circ }$