The equation that models the height above or below equilibrium in inches, h, of a spring over time in seconds, t, is h = -15 cos(2pi/5 t). At which times will the spring be at a height of 8 in. above equilibrium? Select two of the following, make two selections.

Accepted Solution

Answer:1.7 seconds, and3.3 secondsStep-by-step explanation:We simply need to plug in 8 into h and solve for t:[tex]h=-15Cos(\frac{2\pi}{5}t)\\8=-15Cos(\frac{2\pi}{5}t)\\-\frac{8}{15}=Cos(\frac{2\pi}{5}t)\\\frac{2\pi}{5}t=Cos^{-1}(-\frac{8}{15})\\\frac{2\pi}{5}t=2.13\\t=\frac{2.13}{\frac{2\pi}{5}}\\t=1.70[/tex]Since cosine is negative in the 3rd quadrant as well, we need to figure out the 3rd quadrant equivalent of 2.13 radians. First, Ο€ - 2.13 radians = 1.01 radians.Then, we add 1.01 to Ο€ radians, so we get 4.15 radiansSolving from the last part, we have:[tex]t=\frac{4.15}{\frac{2\pi}{5}}\\t=3.3[/tex] also, t = 3.30 seconds*Note: we put the calculator mode in radians when solvingSo, t = 1.7 seconds & 3.30 seconds