${y}_{1}={A}_{1} \sin {k}({x}-v {t}), {y}_{2}={A}_{2} \sin {k}\left({x}-{vt}+{x}_{0}\right) .$ Given amplitudes ${A}_{1}=12\, {mm}$ and ${A}_{2}=5\, {mm}$ ${x}_{0}=3.5\, {cm}$ and wave number ${k}=6.28\, {cm}^{-1}$. The amplitude of resulting wave will be $......\,{mm}$

${y_1} = 0.05\,\cos \,\left( {0.50\,\pi x - 100\,\pi t} \right)$
${y_2} = 0.05\,\cos \,\left( {0.46\,\pi x - 92\,\pi t} \right)$
then velocity is..... $m/s$