Chapter 4 - Graphs of the Circular Functions - Section 4.5 Harmonic Motion - 4.5 Exercises - Page 179: 1b

$s(1)=2$ Therefore, at t=1, the weight is at the same position it was at t=0. Since the weight is at its initial position again, it means that one complete oscillation has occurred. At this position, the weight is changing its motion from up to down.

First, lets find $s(1)$ by substituting $t=1$ in the equation: $s(t)=2\cos (4\pi t)$ $s(1)=2\cos (4\pi \times1)$ $s(1)=2\cos (4\pi)$ $s(1)=2(1)$ $s(1)=2$ Therefore, at t=1, the weight is at the same position it was at t=0. Since the weight is at its initial position again, it means that one complete oscillation has occurred. At this position, the weight is changing its motion from up to down.