Answer
$A = \frac{g~\mu~\lambda^2}{(2\pi)^2~F}$
Work Step by Step
We can find an expression for $\omega$.
$\omega = 2\pi~f$
$\omega = 2\pi~(\frac{v}{\lambda})$
$\omega = 2\pi~\sqrt{\frac{F}{\mu}}~(\frac{1}{\lambda})$
The ant will become momentarily weightless if the magnitude of the vertical acceleration of the rope exceeds $g$ as the rope descends in the vertical plane. The maximum acceleration is given by the expression $\omega^2~A$. We can find the minimum wave amplitude.
$\omega^2~A = g$
$A = \frac{g}{\omega^2}$
$A = \frac{g}{[2\pi~\sqrt{\frac{F}{\mu}}~(\frac{1}{\lambda})]^2}$
$A = \frac{g~\mu~\lambda^2}{(2\pi)^2~F}$
