Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 4 - Review Exercises - Page 186: 57

Answer

The frequency of $0.5$ means that half a cycle is completed each second. At $t = 1.5~s$, the object is $4$ units below the equilibrium point. At $t = 2~s$, the object is at the equilibrium point. At $t = 3.25~s$, the object is $2.83$ units below the equilibrium point.

Work Step by Step

$s(t) = 4~sin~\pi t$ The frequency is the number of cycles per second. We can find the frequency: $f = \frac{\omega}{2\pi} = \frac{\pi}{2\pi} = 0.5$ The frequency of $0.5$ means that half a cycle is completed each second. The function $~~s(t) = 4~sin~\pi t~~$ oscillates between the values of $-4$ and $4$. The equilibrium point is the middle point which is 0. We can find the position at $t = 1.5~s$: $s(t) = 4~sin~(1.5\pi) = -4$ At $t = 1.5~s$, the object is $4$ units below the equilibrium point. We can find the position at $t = 2~s$: $s(t) = 4~sin~(2\pi) = 0$ At $t = 2~s$, the object is at the equilibrium point. We can find the position at $t = 3.25~s$: $s(t) = 4~sin~(3.25\pi) = -2.83$ At $t = 3.25~s$, the object is $2.83$ units below the equilibrium point.
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