Question

Help please precalc will give brainliest

Answer 1

In part (D), we found

$2\sin(4\pi t)+5\cos(4\pi t)=\sqrt{29}\sin\left(4\pi t+\tan^{-1}\left(\dfrac52\right)\right)$

so the phase $\phi$ is $\tan^{-1}\left(\frac52\right)\approx1.19\,\rm rad$, which falls between 0 and $\frac\pi2$. This means the weight is somewhere between the maximum positive position (where $\phi$ would be 0) and the equilibrium position (where $\phi$ would be $\frac\pi2$), and would be traveling in the negative direction.