Water waves

From dimensional analysis, the speed of water waves can be determined as follows:

Only the gravitational constant 'g=10 m/s^2' matters. g has units of L/T^2, so the only way to get a velocity is $$\frac{1}{(2\pi)^{\frac{1}{2}}}g^{\frac{1}{2}}*\lambda^{\frac{1}{2}}$$ where $$\lambda$$ is the wavelength.

The problem is that I looked at a small brook which is normally still, but was moving a little from a breeze, and estimated the wavelength (no longer than a foot), and applied the formula. The result is that the speed of the wave comes out a lot faster than what I observed.

So my question is has anyone else observed something similar? Or does this formula only work for big ocean waves?

Answers and Replies

The equation you quoted in an approximation for deep water.

I think it is due to the depth of the water. In shallow water the wave speed is v ~ sqrtgd where d is the depth of water. Depth < lambda/20 in this case.

The formula of all depth is a little more complicated.

See http://hyperphysics.phy-astr.gsu.edu/Hbase/watwav.html#hwav

The equation you quoted in an approximation for deep water.

I think it is due to the depth of the water. In shallow water the wave speed is v ~ sqrtgd where d is the depth of water. Depth < lambda/20 in this case.

The formula of all depth is a little more complicated.

See http://hyperphysics.phy-astr.gsu.edu/Hbase/watwav.html#hwav

I guess dimensional analysis can only take you so far.

At least it makes sense that for shallow water, the speed is less. I guess this is due to friction from the floor under the sea. It's interesting that it doesn't depend on the type of floor. For solids, the type of material determines a coefficient of friction. For liquids, the layer touching the ground doesn't move at all (does anyone know why?), and the layers above it have a friction that's the viscosity of the liquid. However, the viscosity doesn't even enter the equation, so maybe it's not because of friction from the floor of the sea!