Answer
$24\;m/s$
Work Step by Step
Given wave Eq. is
$y=(0.10\;m)(\sin\frac{\pi x}{2})\sin12\pi t\;.................(1)$
From Eq. $1$, we obtain
$k=\frac{\pi}{2}\;rad/m$ and $\omega=12\pi\;rad/s$
Therefore, the speed of the waves on the rope is
$v=\frac{\omega}{k}$
or, $v=\frac{12\pi\times2}{\pi}\;m/s$
or, $\boxed{v=24\;m/s}$
