# Q & A: Speed of waves in a tray of water

Q:
Hi, Im doing my physics coursework for my physics GCSE, but Im having trouble making it scientific... Its an experiment involving filling a tray with different depths of water, then dropping it so a wave is created. We timed how fast the waves went in different depths. Ive got my results, and it goes faster in deeper water, but I have no idea why! Please can you help me?
Thanks very much
- Oli (age 15)
London, England
A:
The calculation of the speed of surface waves on a fluid is a little lengthy to do in full generality for all fluid heights and all wavelengths.  When you say that you "dropped it" do you mean you dropped the whole tray full of water on the floor, or did you drop something into the tray to make some waves? I would set up something with a card or a board on one end, shaken side to side so as to make waves of a desired wavelength -- dropping something in the tray results in a complicated mixture of waves of different wavelengths, which travel at different speeds.

I found a full expression for the speed of a wave in an incompressible fluid with no viscosity (there are no such fluids, but water is pretty close to this approximation for small trayfuls at normal temperatures and pressures). The speed of surface waves is:

speed = sqrt( ((g*lambda)/(2*pi))*tanh(2*pi*h/lambda) )

from Fetter and Walecka, "Theoretical Mechanics of Particles and Continua", McGraw-Hill, 1980. Here, g is the acceleration due to gravity, 9.81 m/s**2, lambda is the wavelength of the waves under study, and h is the depth of the fluid. For very shallow fluids (compared to the wavelength), the speed increases proportionally to the square root of the depth, and for very deep fluids, the speed increases with the square root of the wavelength.

In words why this is the case -- for a shallow fluid, the motion of the fluid is mostly side-to-side, and in order to accumulate more fluid in one place (to make the crest of the wave), each little bit of fluid must travel a little farther than it would have to in deeper water. When a wave passes, the bits of fluid, if you could watch one at a time, travel in ellipses. For shallow water, the ellipses are stretched out horizontally, and in very deep water, they are very nearly circular. So for a wave of the same height (top to bottom of the ellipse), the bits of water must travel farther in the shallow tray than the deep tray. Because the waves of the same height in shallow and deep water exert the same pressure differences due to gravity to get the water moving (although the motion is different due to the fact that the bottom is there), similar forces push and pull on the water. To get the water moving farther and faster with the same force takes a longer time for each push, and hence a slower speed for the wave, in the shallow water.

Note: this speed formula assumes that the waves are small -- for waves whose heights are comparable to the depth of the tray, you will get even more complicated behavior -- the most spectacular of these is the formation of "breakers" where the waves will curl over and crash as they do on beaches, making for good surfing.

Tom

(published on 10/22/2007)

## Follow-Up #1: Water waves and depth

Q:
What is the effect of amount of water on the water waves?
- Janu Verma (age 21)
Jammu, J&K, India
A:

We've answered this before, but possibly in a confusing manner.  Briefly, the deeper the water, the faster a surface wave will travel, and the lower the height will be.  As waves come ashore on the ocean, they slow down and get taller, preserving the amount of energy in the wave.  The wave speed and variation with depth also depends on the wavelength.  Ocean tsunamis travel at about 600 mph in deep water, and may only be a few inches to a couple of feet high, making them very difficult to detect before they crash ashore taveling much more slowly but being much taller.

Tom

(published on 10/22/2007)

## Follow-Up #2: Speed of surface water waves

Q:
Could the speed of the wave ever be equal to the square root of the gravity times height?
- Ananya (age 21)
Bombay, Mahastra, India
A:
If you look at the equation shown in the original question/answer (#2223):
speed = sqrt( ((g*lambda)/(2*pi))*tanh(2*pi*h/lambda) )  and note that the function tanh(x) for small x is proportional to x, then you arrive at the answer you suggest.

LeeH

(published on 04/08/2010)

## Follow-Up #3: speed of shallow water waves

Q:
How is the proof obtained for the shallow water eqn? How does the speed equation go to the eqn "c = (sqrt)((g*d)) where c is the speed, g is gravity and d is the depth of water? I dont understand the function of tanh really clearly.
- Talha (age 19)
Jeddah
A:

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)