Answer
The amplitude is 4 cm
The frequency is 2 Hz
The phase constant is $\frac{\pi}{6}$
Work Step by Step
The wave oscillates vertically between -4 cm and 4 cm. Therefore, the amplitude is 4 cm.
From the graph, we can see that the horizontal length of one complete oscillation is 12 cm. That is, the wavelength is 12 m. We can find the frequency as:
$f = \frac{v}{\lambda}$
$f = \frac{24~m/s}{12~m}$
$f = 2~Hz$
At t = 0, $D = \frac{A}{2}$ at the point x = 0. Therefore;
$D(t) = A~cos(\omega~t+\phi)$
$cos(\phi) = \frac{1}{2}$
$\phi = arccos(\frac{1}{2})$
$\phi = \frac{\pi}{6}, \frac{5\pi}{6}$
From the graph, we can see that the point x = 0 is moving downward at t = 0. Therefore, $\phi = \frac{\pi}{6}$.