Answer
a. $t=k\pi$
b. $t=\frac{2\pi}{3},\frac{4\pi}{3}$, maximum $\frac{3\sqrt 3}{2}$
Work Step by Step
a. Given $s_1=2sin(t), s_2=sin(2t)=2sin(t)cos(t), t\gt0$, the two masses pass each other when $s_1=s_2$ or $2sin(t)=2sin(t)cos(t)$; thus $sin(t)(1-cos(t))=0$ and we have $t=k\pi$, where $k$ is a positive integer.
b. Define $f(t)=s_1-s_2$; we need to find extrema of $f$ and compare their absolute values. Take derivative to get $f'(t)=2cos(t)-2cos(2t)=2cos(t)-2(2cos^2t-1)=-2(2cos^2t-cos(t)-1)$. Let $f'=0$ we can find $cos(t)=-\frac{1}{2},1$ which gives $t=0,\frac{2\pi}{3},\frac{4\pi}{3}$ in the interval of $[0,2\pi]$. $|f(0)|=0, |f(\frac{2\pi}{3})|=\frac{3\sqrt 3}{2},|f(\frac{4\pi}{3})|=\frac{3\sqrt 3}{2}$.
Thus at $t=\frac{2\pi}{3},\frac{4\pi}{3}$ the vertical distance reaches a maximum of $\frac{3\sqrt 3}{2}$ (as $|f(0)|\lt |f(\frac{2\pi}{3})|$, the extreme is a maximum not a minimum).
