Explanation

(a) Given,
$y_{1} = 0.1\sin{(100\pi t + \frac{\pi}{3}})\ $
$\therefore\ \ \ \ \ \ \ \frac{dy_{1}}{dt} = v_{1} = 0.1 \times 100\pi\cos\left( 100\pi t + \frac{\pi}{3} \right)$
or $v_{1} = 10\pi\sin\left( 100\pi t + \frac{\pi}{3} + \frac{\pi}{2} \right)$
or $v_{1} = 10\pi\sin\left( 100\pi t + \frac{5\pi}{6} \right)$
and $y_{2} = 0.1\cos{\pi t}$
$\therefore\ \ \ \ \frac{dy_{2}}{dt} = v_{2} = - 0.1\sin{\pi t = 0.1\sin{(\pi t + \pi)}}$
Hence, phase difference
$\mathrm{\Delta}\varnothing = \varnothing_{1} - \varnothing_{2} = \left( 100\pi t + \frac{5\pi}{6} \right) - (\pi t + \pi)$
$= \ \frac{5\pi}{6} - \pi$ (at $t = 0)$
= $- \frac{\pi}{6}$