Solve the problem.
-A weight suspended from a spring is vibrating vertically with up being the positive direction. The function $f ( t ) = 10 \sin \left( \frac { 3 \pi t } { 4 } - \frac { \pi } { 4 } \right)$ represents the distance in centimeters of the weight from its rest position as a function of time t, where t is measured in seconds. Find the smallest positive value of t for which the displacement of the weight above its rest position is 5 cm.
Round answer to three decimal places, if necessary.
A) 1.586
B) 2.293
C) 0.556
D) 0.222

Question

Quizplus · Accepted Answer

To find the smallest positive value of $t$ for which the displacement is 5 cm, set the function equal to 5 and solve for $t$:$$5 = 10 \sin\left(\frac{3\pi t}{4} - \frac{\pi}{4}\right)$$Divide both sides by 10:$$0.5 = \sin\left(\frac{3\pi t}{4} - \frac{\pi}{4}\right)$$Using the inverse sine function:$$\sin^{-1}(0.5) = \frac{3\pi t}{4} - \frac{\pi}{4}$$Since $\sin^{-1}(0.5) = \frac{\pi}{6}$ or $\frac{5\pi}{6}$ (but we use the first value for the smallest positive $t$):$$\frac{\pi}{6} = \frac{3\pi t}{4} - \frac{\pi}{4}$$Solving for $t$:$$\frac{\pi}{6} + \frac{\pi}{4} = \frac{3\pi t}{4}$$$$\frac{\pi}{6} + \frac{\pi}{4} = \frac{2\pi + 3\pi}{12} = \frac{5\pi}{12}$$$$\frac{5\pi}{12} = \frac{3\pi t}{4}$$$$t = \frac{5\pi}{12} \cdot \frac{4}{3\pi} = \frac{5}{9}$$$$t \approx 0.556$$

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