Answer
(a) $\phi_0 = \frac{7\pi}{6}$
(b) At point 1:
$phase = \frac{5\pi}{6}$
At point 2:
$phase = \frac{3\pi}{2}$
At point 3:
$phase = \frac{\pi}{6}$
Work Step by Step
(a) The general equation for the velocity of the particle is:
$v(t) = -A~\omega~sin(\omega~t+\phi_0)$
We can find $\phi_0$ when $v(0) = \frac{v_{max}}{2}$
$\frac{v_{max}}{2} = -A~\omega~sin(\omega~t+\phi_0)$
$sin(\phi_0) = \frac{-1}{2}$
$\phi_0 = arcsin(\frac{-1}{2})$
$\phi_0 = \frac{7\pi}{6}, \frac{11\pi}{6}$
In the graph, we can see that the velocity is increasing just after $t=0$. Therefore $\phi_0 = \frac{7\pi}{6}$.
(b) We can find the phase at point 1:
$phase = \frac{7\pi}{6}-\frac{2\pi}{6} = \frac{5\pi}{6}$
We can find the phase at point 2:
$phase = \frac{7\pi}{6}+\frac{\pi}{3} = \frac{3\pi}{2}$
We can find the phase at point 3:
$phase = \frac{7\pi}{6}+\pi = \frac{13\pi}{6}= \frac{\pi}{6}$
