Skip to content
Skip to main content

About this free course

Download this course

Share this free course

Engineering: environmental fluids
Engineering: environmental fluids

Start this free course now. Just create an account and sign in. Enrol and complete the course for a free statement of participation or digital badge if available.

2.2 Wave speed, amplitude and displacement

The interaction of a wind over the surface of water to produce waves is complex. On the surface of deep water the wave speed c sub w of typical waves can be modelled as

c sub w equals Square root of cap l times g divided by two pi
Equation label: (Equation 12)

where cap l is the wavelength from peak to peak between two waves following each other and g is the acceleration due to gravity. The peak is the highest point of the wave, and is also known as the crest. Thus the wave speed is greater for longer wavelengths. Consequently a swell may comprise long and fast waves, which can also be very high if the initiating wind speed itself is both high and sustained for a significant time. The height, cap h, of a wave is taken to be the distance from the lowest level of the surface to the top of the wave. The lowest level is called the trough. The shape of a wave – its cross section or side view – depends upon its height and wavelength. At lower heights, it tends to be sinusoidal, so the amplitude of this type of wave will be half the height from trough to crest.

Wave speeds

The action of wind over deep water is to create a disturbance on the surface layers of the water. This disturbance takes the form of a wave which travels more or less in the same direction as the wind but at a speed which is given by

c sub w equals Square root of cap l times g divided by two pi

where c sub w is the wave speed, cap l is the wavelength and g is the acceleration due to gravity.

Note that the wave is a disturbance which moves along and through the water – the water itself does not move along, except in the case of a tsunami.

Higher waves tend to have a narrower crest and a wider and shallower trough, as sketched in Figure 12.

At a value of cap h divided by cap l equals one divided by seven the crests become more pronounced and sharp-edged in profile and the top edges break into foaming white water (white horses). This foaming dissipates energy, which effectively stops further growth in height, meaning that the ratio cap h divided by cap l stays at a maximum value of one divided by seven.

Described image
Figure 12 Cross sections of wave profiles

The breaking of waves is of course most evident in shallow water at the beach. Once the depth of water has reduced to about cap l divided by two, the shape of the wave profile alters again. If the slope of the beach is small (e.g. less than about 1 in 30 or 3.3%) the wave will break progressively as it rolls in, and the water itself does travel along in this case. If the slope is much greater, the wave is effectively slowed down and cannot adjust; instead, it becomes unstable, growing in height and then breaking by plunging over in a dramatic fashion. This still contains a lot of energy and can impart high forces on anything in its path. Even a non-breaking wave can cause large forces owing to the energy it contains, as the speed causes drag forces on anything it flows past.

A typical wave can be modelled quite easily. Figure 13 shows a wave of sinusoidal form with a relatively small surface displacement amplitude cap h divided by two in comparison to the wavelength and depth d of the water. The sea bed is assumed to be flat and smooth (with negligible friction), and there is a steady series of waves flowing to the right with speed c sub w. The wave depth is cap h, which in this model will be twice the amplitude. The wave periodic time, cap t, is the time taken for one complete wavelength to pass through and is given by

cap t sub equals cap l divided by c sub w
Equation label: (Equation 13)
Described image
Figure 13 Cross section of a sinusoidal wave profile

As mentioned above, each water particle will move in an approximately circular orbit as the wave disturbance passes through. Figure 14 shows the shapes of an individual water particle orbits for shallow, intermediate depth and deep water in schematic form; the relative sizes are not to scale. In shallow water, the orbit is elliptical in cross section and reaches to the sea bed. In the intermediate depth, the orbit is more circular, and in deep water the orbit is completely circular and does not extend to the sea bed.

Described image
Figure 14 Wave water particle orbits (not to scale)

Generally at a water depth equivalent to half the wave length, d equals cap l divided by two, the amplitude of wave motion is barely 4% of that at the surface. This forms a useful rule of thumb in defining a ‘deep water’ wave. Assuming a deep water situation, the wave motion of an individual particle of water as a wave passes is near enough circular. Taking a stationary reference axis set at the flat sea level, as the depth increases with z, the wave motion amplitude reduces by a factor of e super left parenthesis two pi z divided by cap l right parenthesis, where numerically z will be negative.

In other words, if cap a sub s is the surface amplitude and cap a sub z is the amplitude at depth z (where z is a negative number) then

cap a sub z equals cap a sub s times e super left parenthesis two times pi times z divided by cap l right parenthesis full stop
Equation label: (Equation 14)

If the radius of the circular motion is r, the particle speed will be given by u equals omega times r (as with any circular motion), and its acceleration will be

a equals omega squared times r
Equation label: (Equation 15)

where omega is the radian circular frequency. Again, as with all circular orbiting motion

omega equals two pi f equals two pi divided by cap t
Equation label: (Equation 16)

and

cap l equals c sub w times cap t

which is a rearranged version of Equation 13.

Deep water wave study

For a wave in deep water of wavelength 200.0 m and height 6.0 m, calculate:

  • a.the wave speed
  • b.the periodic time
  • c.the displacement amplitude at the surface
  • d.the displacement amplitude at 50.0 m depth
  • e.the maximum horizontal acceleration at 25.0 m depth.

Give your answers to 3 significant figures.

Solution

  • a.Using Equation 12:

    equation sequence part 1 c sub w equals part 2 Square root of cap l times g divided by two pi equals part 3 Square root of 200.0 m prefix multiplication of 9.81 m s super negative two divided by two pi equals part 4 17.67 times ellipsis m s super negative one equals 17.7 m s super negative one left parenthesis to three s full stop f full stop right parenthesis full stop
  • b.Using Equation 13;

    equation sequence part 1 cap t equals part 2 cap l divided by c sub w equals part 3 200.0 m divided by 17.67 times ellipsis m s super negative one equals part 4 11.3 s left parenthesis to three s full stop f right parenthesis full stop
  • c.The amplitude at the surface is

    equation sequence part 1 cap a sub s equals part 2 cap h divided by two equals part 3 six m divided by two equals part 4 three m full stop
  • d.To find the displacement amplitude, Equation 14 is used:

    cap a sub z equals cap a sub s times e super left parenthesis two times pi times z divided by cap l right parenthesis full stop

    First calculate the factor at 50 m depth:

    equation sequence part 1 e super left parenthesis two times pi times z divided by cap l right parenthesis equals part 2 e super left parenthesis two times pi of negative 50 times m divided by 200 times m right parenthesis equals part 3 e super negative 1.571 times ellipsis equals part 4 0.2078 horizontal ellipsis full stop

    This gives a displacement amplitude

    cap a sub cap z equals three times m prefix multiplication of 0.2078 times ellipsis equals 0.624 times m left parenthesis to three s full stop f full stop right parenthesis full stop
  • e.First calculate the displacement amplitude at 25 m depth, since this is also the radius of the particle’s circular motion:

    equation sequence part 1 e super left parenthesis two times pi times z divided by cap l right parenthesis equals part 2 e super left parenthesis two times pi of negative 25 m divided by 200 m right parenthesis equals part 3 e super negative 0.785 times ellipsis equals part 4 0.4559 times ellipsis

    so the radius of circular motion (or displacement amplitude) is

    equation sequence part 1 r equals part 2 cap a sub s times e super left parenthesis two times pi times z divided by cap l right parenthesis equals part 3 three m prefix multiplication of 0.4559 times ellipsis equals 1.367 times ellipsis m full stop

    Now, using Equation 16:

    equation sequence part 1 omega equals part 2 two times pi divided by cap t equals part 3 two times pi divided by 11.3 times ellipsis s equals part 4 0.5551 times ellipsis times s super negative one full stop

    Therefore, from Equation 15, the acceleration amplitude is

    equation sequence part 1 a equals part 2 omega squared times r equals part 3 left parenthesis 0.5551 times ellipsis s super negative one right parenthesis squared multiplication 1.367 times ellipsis m equals 0.422 m s super negative two left parenthesis to three s full stop f full stop right parenthesis full stop

Activity 1

The wave as described above has decayed such that its height is reduced by 50%. Determine the same parameters, noting the changes in values:

  • a.the wave speed
  • b.the periodic time
  • c.the displacement amplitude at the surface
  • d.the displacement amplitude at 50.0 m depth
  • e.the maximum horizontal acceleration at 25.0 m depth.

Give your answers to 3 significant figures.

Answer

  • a.Using Equation 12,

    equation sequence part 1 c sub w equals part 2 Square root of cap l times g divided by two pi equals part 3 Square root of 200.0 m prefix multiplication of 9.81 m s super negative two divided by two pi equals part 4 17.67 times ellipsis m s super negative one equals 17.7 m s super negative one left parenthesis to three s full stop f full stop right parenthesis full stop

    The value is the same.

  • b.Using Equation 13,

    equation sequence part 1 cap t equals part 2 cap l divided by c sub w equals part 3 200.0 m divided by 17.67 times ellipsis m s super negative one equals part 4 11.31 times ellipsis s equals 11.3 s left parenthesis to three s full stop f full stop right parenthesis full stop

    The value is the same.

  • c.The amplitude at the surface is

    equation sequence part 1 cap a sub s equals part 2 cap h divided by two equals part 3 three m divided by two equals part 4 1.5 m full stop

    The surface amplitude has reduced by 50%.

  • d.To find the displacement amplitude Equation 14 is used:

    cap a sub z equals cap a sub s times e super left parenthesis two times pi times z divided by cap l right parenthesis full stop

    The factor at 50 m depth will be the same:

    equation sequence part 1 e super left parenthesis two times pi times z divided by cap l right parenthesis equals part 2 e super left parenthesis two times pi of negative 50 times m divided by 200 times m right parenthesis equals part 3 e super negative 1.570 times ellipsis equals part 4 0.2078 times ellipsis full stop

    This gives a displacement amplitude of

    cap a sub z equals 1.5 times m prefix multiplication of 0.2078 times ellipsis equals 0.3118 times ellipsis times m equals 0.312 times m left parenthesis to three s full stop f full stop right parenthesis full stop

    This shows a reduction of 50%.

  • e.First calculate the displacement amplitude at 25 m depth, since this is also the radius of the particle’s circular motion. The factor will be the same at

    equation sequence part 1 e super left parenthesis two times pi times z divided by cap l right parenthesis equals part 2 e super left parenthesis two times pi of negative 25 m divided by 200 m right parenthesis equals part 3 e super negative 0.785 times ellipsis equals part 4 0.4559 times ellipsis full stop

    The radius of circular motion (or displacement amplitude) is

    equation sequence part 1 r equals part 2 cap a sub s times e super left parenthesis two times pi times z divided by cap l right parenthesis equals part 3 1.5 m prefix multiplication of 0.4559 times ellipsis equals 0.683 times ellipsis m full stop

    This is a reduction of 50%. Now, using Equation 16, angular velocity will be the same at

    equation sequence part 1 omega equals part 2 two times pi divided by cap t equals part 3 two times pi divided by 11.3 times ellipsis s equals part 4 0.555 times ellipsis times s super negative one full stop

    The acceleration amplitude is

    equation sequence part 1 a equals part 2 omega squared times r equals part 3 left parenthesis 0.555 times ellipsis s super negative one right parenthesis squared multiplication 0.683 times ellipsis m equals 0.211 m s super negative two left parenthesis to three s full stop f full stop right parenthesis full stop

    This is a reduction of 50%.

Wave models

For typical wind-provoked waves over deep water the following relationships can be used to model the wave properties:

The speed of the wave is

c sub w equals Square root of cap l times g divided by two pi

where cap l is the wavelength and g is the acceleration due to gravity.

The periodic time for one complete wave to pass by is

cap t sub equals cap l divided by c sub w

and the wavelength in terms of velocity and periodic time is cap l equals c sub w times cap t.

The frequency of the waves in relation to the periodic time is

omega equals two pi f equals two pi divided by cap t

where omega is the radian circular frequency and f is the frequency in Hertz or cycles per second.

At a depth z (a negative numerical value) the amplitude of a wave is

cap a sub z equals cap a sub s times e super two times pi times z divided by cap l

where cap a sub s is the amplitude at the surface.