= First turning point of an initially stationary coupled oscillator
{title2=$\tan z_*=-z_*$}
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 $\eta=\delta/\bar\omega\ll1$, writing $z=\bar\omega t$ gives $z\cos z+\sin z+O(\eta^2)=0$. Thus $t_*=2.028757838\ldots/\bar\omega+O(\eta^2/\bar\omega)$, using the first root of $\tan z=-z$ in $(\pi/2,\pi)$. The envelope instead first reaches its maximum at $\pi/(2\delta)$.
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