 First turning point of an initially stationary coupled oscillator

For two equal masses with equal wall springs released with displacements $(d,0)$, their second displacement is $x_2=d\sin (\bar{\omega }t)\sin (\delta t)$, where $\bar{\omega }$ is the mean of the two positive normal mode frequencies and $2\delta$ their difference. The first nonzero time with ${\dot{x}}_2=0$ is a fast turning point, not the slow maximum of its beat envelope. With $\eta =\delta /\bar{\omega }\ll 1$, writing $z=\bar{\omega }t$ gives $z\cos z+\sin z+O({\eta }^2)=0$. Thus $t_{\ast }=2.028757838\dots /\bar{\omega }+O({\eta }^2/\bar{\omega })$, using the first root of $\tan z=-z$ in $(\pi /2,\pi )$. The envelope instead first reaches its maximum at $\pi /(2\delta )$.