Answer
$f = 0.5~Hz$
$A = 10~cm$
Work Step by Step
From the graph, we can see that one cycle is completed every 2 seconds. Therefore, the period $T = 2~s$.
$f = \frac{1}{T} = \frac{1}{2~s} = 0.5~Hz$
The motion moves between -10 cm and 10 cm. Therefore, the oscillation amplitude is 10 cm.
