= Solution
Convert the fixed position to $x=0.20\ \mathrm m$. The element's <displacement> is then $y(t)=0.27\sin(2.4-500t)\ \mathrm{mm}$. Taking time <derivatives> at fixed position gives its transverse <velocity> and <acceleration>:
$$
\boxed{v_y(t)=-135\cos(2.4-500t)\ \mathrm{mm\,s^{-1}},\qquad
a_y(t)=-67500\sin(2.4-500t)\ \mathrm{mm\,s^{-2}}.}
$$
Equivalently the prefactors in SI units are $-0.135\ \mathrm{m\,s^{-1}}$ and $-67.5\ \mathrm{m\,s^{-2}}$. The <phase> is measured in radians. At $t=4\ \mathrm s$,
$$
\boxed{y(4)=0.27\sin(-1997.6)\ \mathrm{mm}
\simeq0.118152\ \mathrm{mm}=1.18152\times10^{-4}\ \mathrm m.}
$$
These are the transverse motions of the material element, not the propagation speed of the <wave> along the string.
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