 Solution

Fourier modes of the harmonic potential decay as $e^{|k|y}$; the two surface conditions give ${\eta }_{tt}+|k{|}^3\eta =0$, hence ${\omega }^2=|k{|}^3$. The initial transform is $F(k)=-i\sqrt{\pi },k{e}^{-k^2/4}/2$. At $x=Vt$, the contributing stationary wavenumber is $k_0=4V^2/9$, givingDifferent observer speeds select different group-velocity wavenumbers; fast observers see exponentially weaker, higher-frequency ripples.