(published on 10/22/2007)
(published on 10/22/2007)
(published on 04/08/2010)
It's questions like this one that make working on this site fun, because you give us a chance to relearn old physics in intuitive ways. Here I assume that it's not so much a precise proof that you need but rather a convincing argument for the form of the equation for the wave speed c. Let's try this.
Think of a sinusoidal wave with wavelength λ in water of depth h much less than λ. The wave has height w, much less than any other lengths. The water has density ρ. The initial static wave pattern can be viewed as the sum of two waves traveling opposite directions. When they've each gone λ/4 in their opposite ways, the water becomes level. All the initial gravitational potential energy has now been converted to kinetic energy, since we're treating the water as incompressible and having no viscosity. The initial energy density, per area, is around ρgw2 (not worrying about factors of two, etc.) because the amount of water lifted from the low points up by a height w is proportional to w. When that energy is converted to kinetic energy, the water velocity is about uniform throughout the depth, because we're considering waves in the shallow limit. So the energy density per area is around ρhv2, which must be the same as the initial potential energy density. So we have about (vh/w)2=gh.
How to we get c from the water velocity v? Water flowing over the depth h in time ~T, a wave period, must carry enough fluid a distance about λ to flatten out the disturbance of height w. So T is about λw/vh or about λ/(gh)1/2 , from our earlier result. That makes c=λ/T about (gh)1/2 . Apparently if one keeps track of all the numerical factors that result is actually right without any extra factors of 2 etc.
Probably there's a shorter argument, but this should do for now.
Mike W.
ps. If you want all the gory details check out
(published on 01/02/2015)