#### Answer

(a) $\phi = \frac{5\pi}{3}$
(b) $v = 0.136~m/s$
(c) $v_{max} = 0.157~m/s$

#### Work Step by Step

(a) The general equation for the motion is:
$x(t) = A~cos(\omega~t+\phi)$
At $t = 0$:
$x = A~cos(\phi) = \frac{A}{2}$
$cos(\phi) = \frac{1}{2}$
$\phi = arccos(\frac{1}{2})$
$\phi = \frac{\pi}{3}, \frac{5\pi}{3}$
Since the basic cos-curve is shifted to the left $\frac{5\pi}{3}$, the phase constant is $\phi = \frac{5\pi}{3}$
(b) From the graph, we can see that one cycle is completed in 4 seconds. Therefore, the period $T = 4~s$. We can find the angular frequency.
$T = \frac{2\pi}{\omega}$
$\omega = \frac{2\pi}{T}$
$\omega = \frac{2\pi}{4~s}$
$\omega = 1.57~rad/s$
From the graph, we can see that the amplitude is 0.10 meters. We can find the velocity at t = 0:
$v(t) = -A~\omega~sin(\omega~t+\phi)$
$v = -A~\omega~sin(\phi)$
$v = -(0.10~m)(1.57~rad/s)~sin(\frac{5\pi}{3})$
$v = 0.136~m/s$
(c) We can find the maximum speed as:
$v_{max} = A~\omega$
$v_{max} = (0.10~m)(1.57~rad/s)$
$v_{max} = 0.157~m/s$