## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

$f = 0.5~Hz$ $A = 10~cm$
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.