Two simple harmonic motions are represented by the equations

$y_1 = 0.1 sin (100 \pi t + \frac{\pi}{3})$ and $y_2 = 0.1 cos \pi t$.

The phase difference of the velocity of particle 1, with respect to the velocity of particle 2 is ;

Given equations of wave are :

$\begin{array}{l} y_{1}=0.1 \sin (100 \pi t+\pi / 3) \\ y_{2}=0.1 \cos \pi t=0.1 \sin (\pi / 2+\pi t)=0.1 \sin (\pi t+\pi / 2) \end{array}$

Now, to find the velocity of the particle, differentiate both equations with respect to time.

$\mathrm{dy}_{1} / \mathrm{dt}=\mathrm{v}_{1}=0.1 \times 100 \pi \cos (100 \pi t+\pi / 3)=10 \pi \cos (100 \pi t+\pi / 3)$

similarly for 2nd equation,

$\mathrm{dy}_{2} / \mathrm{dt}=\mathrm{v}_{2}=0.1 \times \pi \cos (\pi \mathrm{t}+\Pi / 2)=0.1 \pi \cos (\pi \mathrm{t}+\pi / 2)$

We know,

if equation $x=A \sin (\omega t+\Phi)$ is given then, at $t=0$ phase of motion is $\Phi$

Similarly at $t=0$ phase of 1 st particle velocity is $\pi / 3$ at $t=0$, phase of velocity of 2 nd particle is $\pi / 2$

Now, phase difference $=$ phase of 1 st particle at $t=0$ - phase of 2 nd particle at $\mathrm{t}=0$

$=\pi / 3-\pi / 2=-\pi / 6$