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Engineering › Fluid Mechanics › Waves ›

Dispersion

Uses a linear dispersion relationship to compute wave-frequency from wave-number

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Contents

Interface
Dispersion K
Dispersion K2
Dispersion W
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Interface

C++

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Name	HTML Code
dispersion_k Uses a linear dispersion relationship to compute wave-frequency from wave-number	<script type="text/javascript" src="http://www.codecogs.com/ic_json-71&u=4"></script>
dispersion_k2 Uses a 2nd order dispersion relationship, to obtain wave-frequency from wave-number	<script type="text/javascript" src="http://www.codecogs.com/ic_json-71&u=5"></script>
dispersion_w Itartively solves for wave-number from wave-frequency.	<script type="text/javascript" src="http://www.codecogs.com/ic_json-71&u=6"></script>

Dispersion K

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doubledispersion_k(	double	`k`
	double	`depth` `= 0`
	double	`gravity` `= 9.8066`	)

This function solves the linear dispersion equation,

, to obtain wave-frequency from a given wave-number, using a very simple rearrangement

$w = \sqrt { g k tanh(k d) }$

(1)

where w is wave-frequency, k is wave-number and g is gravity. In deep water (represented with d<=0), this solution reduces to

$w = \sqrt { g k }$

(2)

The opposite of this function is dispersion_w

Parameters

k	( $2\;\mathrm{\pi/m}$ ) is wave-number. [rad/m]
depth	the depth of the water to mean sea level. A value of zero or less corresponds to deep water. [m]
gravity	(default 9.8066). [m/s²]

Returns

wave-frequency ( $2\;\mathrm{\pi/s}$ ). [rad/s]

Source Code

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Dispersion K2

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doubledispersion_k2(	double	`a`
	double	`k`
	double	`depth` `= 0`
	double	`gravity` `= 9.8066`	)

This function solves a 2nd order dispersion relationship to more accurate compute the relationship between wave-frequency and wave-number for the specified wave amplitude a. The solution is based on work by Dalrymple, who derived:

$w^2 = g k (1 + f_1 \epsilon^2 D) tanh(k d + f_2 \epsilon)$

(3)

where

In deep water(represented with d<=0), this solution reduces to

$w = \sqrt { g k }$

(8)

As in the linear dispersion relationship.

Example:

Example - Second order Wave Number to Wave Frequency

Problem

Compute the second order Wave Number from Wave Frequency in water of depth 2m.

Workings

#include <stdio.h>
#include <codecogs/engineering/fluid_mechanics/waves/dispersion.h>
using namespace Engineering::Fluid_Mechanics::Waves;
 
int main()
{
  double amp=3;  // 3 meters
  printf("   k         w ");
  for(double k=0.01; k<1;k+=0.1)
  {
    printf("\n %.6lf", k);
    double w=dispersion_k2(amp,k,2);
    printf("  %.3lf", w);
  }
}

Parameters

a	amplitude of component with wavenumber k. [m]
k	wave-number of component ( $2\;\mathrm{\pi/m}$ ). [rad/m]
depth	the depth of the water to mean sea level. A value of zero or less corresponds to deep water. [m]
gravity	(default $9.8066\;\mathrm{m/s^2}$ ). [m/s²]

Returns

wave-frequency ( $2\;\mathrm{\pi/s}$ ). [rad/s]

Source Code

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Dispersion W

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doubledispersion_w(	double	`w`
	double	`depth` `= 0`
	double	`gravity` `= 9.8066`	)

Finds the wave-number associated with a particular wave-frequency, using this 1st order dispersion equation,

to obtain wave-number from a given wave-frequency.

In Deep water, this equations reduces to

, which can obviously solved directly.

For shallow water, an iterative approach must be used. For each iteration, we calculate the residual error:

$\epsilon = \frac{w^2}{g} - k tanh(k d)$

(9)

where w is wave-frequency, k is wave-number and g is gravity.

We then seek to minimise ε using the first derivative $\partial \epsilon / \partial k$ .

This convergence of this error function is fairly rapid, with poorest conversion, 8 iterations in water 2m deep (for 6dp precision) occurring when w is small i.e. w<=0.1. When w=1.5 only 4 iterations are needed, while w>3.5 needs only 2 iterations. Shallow water naturally requires more iteration. If your doing many calculation in either shallow water or very long wave periods (>30s), then you might want to consider fine tuning this function.

This relationship can me use to move from the wave-period ( $2\pi / w$ ) to the wave-length ( $2\pi/k$ ) of a water wave. This is also shown graphically in the following figure:

$\graph w=0:4, depth=1:5:3, gravity=9.8066$

The opposite of this function is dispersion_k

Example:

[metric]

Example - First order conversion of Wave Frequency to Wave Number

Problem

Compute Wave Frequency from Wave Number in water of depth 2m.

Workings

#include <stdio.h>
#include <codecogs/engineering/fluid_mechanics/waves/dispersion.h>
using namespace Engineering::Fluid_Mechanics::Waves;
 
int main()
{
  printf("   k         w  recalculated k");
  for(double k=0.01; k<1;k+=0.1)
  {
    printf("\n %.6lf", k);
    double w=dispersion_k(k,2);
    double k2=dispersion_w(w,2);
    printf("  %.3lf  %.6lf", w, k2);
  }
  return 0;
}

Solution

k         w  recalculated k
0.010000  0.044  0.010000
0.110000  0.483  0.110000
0.210000  0.904  0.210000
0.310000  1.294  0.310000
0.410000  1.648  0.410000
0.510000  1.962  0.510000
0.610000  2.241  0.610000
0.710000  2.489  0.710000
0.810000  2.710  0.810000
0.910000  2.910  0.910000

Example 1

Parameters

w	( $2\;\mathrm{\pi/s}$ ) is wave-frequency, usually written using $\omega$ (omega).
depth	(m) define the depth of the water to mean sea level. A value of zero or less corresponds to deep water.
gravity	(default $9.8066\;\mathrm{m/s^2}$ ).

Returns

wave-number ( $2\;\mathrm{\pi/m}$ ).

Source Code

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Page Comments

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double	`dispersion_k` (double k, double depth=0, double gravity=9.8066) Uses a linear dispersion relationship to compute wave-frequency from wave-number
double	`dispersion_k2` (double a, double k, double depth=0, double gravity=9.8066) Uses a 2nd order dispersion relationship, to obtain wave-frequency from wave-number
double	`dispersion_w` (double w, double depth=0, double gravity=9.8066) Itartively solves for wave-number from wave-frequency.