Answer
(a) A = 0.120 m
(b) T = 1.60 s
(c) f = 0.625 Hz
Work Step by Step
(a) The object moves a distance of 0.120 m on either side of the equilibrium position. Therefore the amplitude is 0.120 meters.
(b) In 0.800 seconds, the object completes half of a cycle. Therefore, the period $T$ is 1.60 seconds.
(c) $f = \frac{1}{T} = \frac{1}{1.60~s} = 0.625~Hz$
