Does a circular surface water wave decrease by 2$\pi$r?

Question

Does the energy of a wave in water decrease by (2$\pi$)r?

If that is right, why do we see waves at a considerable distance of almost same intensity of the near ones?

Answer 1

A surface wave spreading from a sinusoidal point source over a lossless and dispersion-free medium will indeed spread as $\propto \frac{1}{r}$. This follows easily from energy conservation: because the geometry has circular symmetry the instantaneous power $p(\rho,\phi)$may depend only on $\rho$ and not $\phi$ thus the total power integrated over circle of radius $r$ is constant is then $2\pi r p(r)=P_0$ and thus $p(r)=\frac{P_0}{2\pi r}$where $P_0$ is the power of the source. As is normally the case $p \propto a^2$ where $a$ is the amplitude of the wave, one has that $a(r) \propto \frac{1}{\sqrt{r}}$ (A similar same argument shows the $\propto \frac {1}{r^2}$ power spreading in the 3D spherically symmetric case.)

In the case of gravity waves the situation is complicated by the fact that the surface waves are dispersive. Consequently if instead of a periodic source we have an impulse like excitation, such as a stone dropped, the elementary waves of different frequency will travel with different velocity and the wavefront will break apart: the longer wavelengths travel faster and leave behind the smaller wavelength components. The result is that the wavefront amplitude is no more $\propto \frac{1}{\sqrt{r}}$ but it also depends on the bandwidth and frequency distribution of the excitation.