Question

Turn object attached to a spring oscillates around the position and is represented by the function Y= 2 cos(2pi(x-0.02)),With time in x seconds. When does a height of 1 foot occur on the interval 0

Answer 1

$\begin{gathered} y=2\cos (2\pi(x-0.02)) \\ \end{gathered}$$y=1,_{\text{ }}0The cosine function is equal to 1 for:

[tex]\begin{gathered} \cos (x)=1 \\ x=2\pi n \\ n\in\Z \end{gathered}$

so:

$\begin{gathered} 2\pi x-0.04\pi=2\pi n+\frac{\pi}{3} \\ 2\pi x=2\pi n+\frac{\pi}{3}+0.04\pi \\ x=n+0.02+\frac{1}{6} \\ or \\ x=n+0.02+\frac{5}{6} \end{gathered}$

For 0

$\begin{gathered} x=0.187 \\ x=0.853 \end{gathered}$