The variation of velocity of a particle executing SHM with time is shown in figure. The velocity of the particle when a phase change of $\pi /6$ takes place from the instant it is at one of the extreme positions will be 

From the graph T = (5 -1) = 4 s

(distance between the two adjacent crests shown in the figure)

$\begin{array}{l}\mathrm{and} v_{max}=5\mathrm{m}/\mathrm{s};\omega A=5\mathrm{m}/\mathrm{s}\\ \left(\frac{2\pi }{T}\right)A=5\Rightarrow A=\frac{5T}{2\pi }=\frac{5\times 4}{2\pi }=\frac{10}{\pi }\mathrm{m}\\ \mathrm{Also}, \omega =\frac{2\pi }{4}=\frac{\pi }{2}\mathrm{rad}/\mathrm{s}\end{array}$

The equation of velocity can be written as

$v=5\sin \left(\frac{\pi }{2}t\right)\mathrm{m}/\mathrm{s}$

$\text{ At extreme position, }v=0;\sin \left(\frac{\pi }{2}t\right)=0\text{ or }t=2\mathrm{s}$

Phase of the particle velocity at that instant corresponding
to the above equation = $\pi$.
Therefore, when a phase change of $\pi$/6 takes place, the
resulting phase = $\pi$+$\pi$/6.

$\begin{array}{c}v=5\sin \left(\pi +\frac{\pi }{6}\right)=-5\sin \frac{\pi }{6}=-5\left(\frac{1}{2}\right)\\ =2.5\mathrm{m}/\mathrm{s}\text{ (numerically) }\end{array}$