In part (D), we found
[tex]2\sin(4\pi t)+5\cos(4\pi t)=\sqrt{29}\sin\left(4\pi t+\tan^{-1}\left(\dfrac52\right)\right)[/tex]
so the phase [tex]\phi[/tex] is [tex]\tan^{-1}\left(\frac52\right)\approx1.19\,\rm rad[/tex], which falls between 0 and [tex]\frac\pi2[/tex]. This means the weight is somewhere between the maximum positive position (where [tex]\phi[/tex] would be 0) and the equilibrium position (where [tex]\phi[/tex] would be [tex]\frac\pi2[/tex]), and would be traveling in the negative direction.
