Directivity of a source plus its ghost

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Problem

An air gun is fired at a depth of 10 m. The waveform includes frequencies in the range , the amplitudes of the 10- and 80-Hz components being the same near the source. Compare their amplitudes for the wave plus ghost at considerable distance from the source in the directions , , , and to the vertical.

Background

Air guns are described in problem 7.7.

Solution

We take the velocity in water as 1.5 km/s so that the wavelengths are 150 m and 19 m for the 10-Hz and 80-Hz components. From equation (6.7a) the amplitude of the ghost is , where the depth of the source is . For the 10-Hz component, ; for the 80 Hz component, . The ratio of the amplitude of the 10-Hz component to that of the 80-Hz component is

Table 6.8a. Ratios of ghost amplitudes of 10- and 80- Hz components.
0.42 3.3 Amplitude ratio
0.42 3.3 –2.6
0.36 2.9 1.5
0.30 2.3 1.04
0.21 1.6 0.21
0.13
* For , , so the ratio is 0/0. However, when is slightly less than , the arguments of the sines are small and we can replace the sines with the angles; the cos factors cancel and the ratio is

Table 6.8a shows the results for the given values of . The values in columns 2 and 3 are in radians and the column headed “Amplitude ratio” is the sine of the values in column two divided by the sine of the values in column three.

Thus, the 10-Hz component is stronger than the 80-Hz component as the direction approaches the vertical. The minus sign in the first ratio is due to a phase reversal of the 80-Hz component.

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