Answer
(a) The amplitude is 10 cm.
(b) $f = 0.25~Hz$
(c) $\phi = \frac{4\pi}{3}$
Work Step by Step
(a) The motion moves between -10 cm and 10 cm. Therefore, the amplitude is 10 cm.
(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 frequency.
$f = \frac{1}{T} = \frac{1}{4~s} = 0.25~Hz$
(c) The general equation for the motion is:
$x(t) = A~cos(\omega~t+\phi)$
At $t = 0$:
$x = A~cos(\phi) = \frac{-A}{2}$
$cos(\phi) = \frac{-1}{2}$
$\phi = arccos(\frac{-1}{2})$
$\phi = \frac{2\pi}{3}, \frac{4\pi}{3}$
Since the basic cos-curve is shifted to the left $\frac{4\pi}{3}$, the phase constant is $\phi = \frac{4\pi}{3}$.
