Given SHMs

$y_1=0.1\sin\left(100\pi t+\frac{\pi}{3}\right), \qquad y_2=0.1\cos(\pi t)$

Find velocities by differentiating

For particle 1: $v_1=\frac{dy_1}{dt}=0.1\cdot 100\pi \cos\left(100\pi t+\frac{\pi}{3}\right)$ $v_1=10\pi \cos\left(100\pi t+\frac{\pi}{3}\right)$

Write cosine as sine: $\cos\theta=\sin\left(\theta+\frac{\pi}{2}\right)$ So, $v_1=10\pi \sin\left(100\pi t+\frac{\pi}{3}+\frac{\pi}{2}\right)$ $v_1=10\pi \sin\left(100\pi t+\frac{5\pi}{6}\right)$

Hence phase of velocity of particle 1 is $\phi_{v_1}=\frac{5\pi}{6}$

For particle 2: $v_2=\frac{dy_2}{dt}=0.1\cdot(-\pi)\sin(\pi t)$ $v_2=-0.1\pi\sin(\pi t)$

Now, $-\sin\theta=\sin(\theta-\pi)$ therefore $v_2=0.1\pi\sin(\pi t-\pi)$

Hence phase of velocity of particle 2 is $\phi_{v_2}=-\pi$

Equivalently, we may also write $v_2=0.1\pi\sin(\pi t+\pi)$ since phase differing by $2\pi$ is same.

Phase difference of velocity of particle 1 with respect to velocity of particle 2

$\Delta\phi=\phi_{v_1}-\phi_{v_2}$

Using $\phi_{v_2}=\pi$ (equivalent to $-\pi$): $\Delta\phi=\frac{5\pi}{6}-\pi=-\frac\pi6$

Check options

$-\frac\pi6$ corresponds to Option B.

Final Answer

The phase difference of the velocities is $\boxed{-\frac{\pi}{6}}$