Two particles move parallel to the x-axis about the origin with same amplitude 'a' and frequency $\omega$. At a certain instant they are found at a distance a/3 from the origin on opposite sides but their velocities are in the same direction. What is the phase difference between the two? 

$\text{ Let }{x}_1=a\sin \omega t\text{ and }{x}_2=a\sin (\omega t+\delta )\text{ be two SHMs. }$

$\begin{array}{l}\frac{a}{3}=a\sin \omega t\text{ and }-\frac{a}{3}=a\sin (\omega t+\delta )\\ \sin \omega t=1/3\text{ and }\sin (\omega t+\delta )=-1/3\end{array}$

$\text{ Eliminating }t,\frac{1}{3}\cos \delta +\sqrt{1-\frac{1}{9}}\sin \delta =-\frac{1}{3}$

$\begin{array}{l}9{\cos }^2\delta +2\cos \delta -7=0\\ \cos \delta =-1\text{ or }\frac{7}{9}\\ i.e., \delta ={180}^{\circ }\text{ or }{\cos }^{-1}\left(\frac{7}{9}\right)\end{array}$

If we put 180^o, we find that v₁ and v₂ are of opposite signs.
Hence $\delta$ = 180^o is not applicable.

$\therefore \delta ={\cos }^{-1}\left(\frac{7}{9}\right)$