IPhO 1995 · Theory problem 2

Sound Propagation

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Introduction

The speed of propagation of sound in the ocean varies with depth, temperature and salinity. Figure 1(a) below shows the variation of sound speed \(c\) with depth \(z\) for a case where a minimum speed value \(c_0\) occurs midway betweeen the ocean surface and the sea bed. Note that for convenience \(z=0\) at the depth of this sound speed minimum, \(z=z_S\) at the surface and \(z=-z_b\) at the sea bed. Above \(z=0\), \(c\) is given by

\[ c=c_0+bz\;. \]

Below \(z=0\), \(c\) is given by

\[ c=c_0-bz\;. \]

In each case \(b=\left|\dfrac{dc}{dz}\right|\), that is, \(b\) is the magnitude of the sound speed gradient with depth; \(b\) is assumed constant.

Figure 1

Figure 1 (a)

Figure 2

Figure 1 (b)

Figure 1(b) shows a section of the \(z\)–\(x\) plane through the ocean, where \(x\) is a horizontal direction. The variation of \(c\) with respect to \(z\) is shown in figure 1(a). At the position \(z=0\), \(x=0\), a sound source \(S\) is located. A ‘sound ray’ is emitted from \(S\) at an angle \(\theta_0\) as shown. Because of the variation of \(c\) with \(z\), the ray will be refracted.

(a) (6 marks)

Show that the trajectory of the ray, leaving the source \(S\) and constrained to the \(z\)–\(x\) plane forms an arc of a circle with radius \(R\) where

\[ R=\frac{c_0}{b\sin\theta_0} \quad\text{for}\quad 0\leq\theta_0<\frac{\pi}{2}\;. \]

(b) (3 marks)

Derive an expression involving \(z_S\), \(c_0\) and \(b\) to give the smallest value of the angle \(\theta_0\) for upwardly directed rays which can be transmitted without the sound wave reflecting from the sea surface.

(c) (4 marks)

Figure 1(b) shows the position of a sound receiver \(H\) which is located at the position \(z=0\), \(x=X\). Derive an expression involving \(b\), \(X\) and \(c_0\) to give the series of angles \(\theta_0\) required for the sound ray emerging from \(S\) to reach the receiver \(H\). Assume that \(z_S\) and \(z_b\) are sufficiently large to remove the possibility of reflection from sea surface or sea bed.

(d) (2 marks)

Calculate the smallest four values of \(\theta_0\) for refracted rays from \(S\) to reach \(H\) when

  • \(X=10000\ \mathrm{m}\)
  • \(c_0=1500\ \mathrm{m\,s^{-1}}\)
  • \(b=0.02000\ \mathrm{s^{-1}}\)

(e) (5 marks)

Derive an expression to give the time taken for sound to travel from \(S\) to \(H\) following the ray path associated with the smallest value of angle \(\theta_0\), as determined in part (c). Calculate the value of this transit time for the conditions given in part (d). The following result may be of assistance:

\[ \int\frac{dx}{\sin x}=\ln\tan\left(\frac{x}{2}\right) \]

Calculate the time taken for the direct ray to travel from \(S\) to \(H\) along \(z=0\). Which of the two rys will arrive first, the ray for which \(\theta_0=\pi/2\), or the ray with the smallest value of \(\theta_0\) as calculated for part (d)?

Official English text of the problem.

Answers

  • (a) Each segment lying entirely above or below \(z=0\) is a circular arc with \[ \boxed{R=\frac{c_0}{b\sin\theta_0}}. \] The case \(\theta_0=0\) is the straight vertical ray, corresponding to \(R\to\infty\).

  • (b) The limiting angle is \[ \boxed{\theta_{0,\min}=\sin^{-1}\!\left(\frac{c_0}{c_0+bz_S}\right)}. \] Equality gives a ray tangent to the surface; larger angles turn below it.

  • (c) For initially upward, refracted rays, \[ \boxed{\theta_{0,n}=\tan^{-1}\!\left(\frac{2nc_0}{bX}\right)}, \qquad n=1,2,3,\ldots, \] where \(n\) counts the circular arcs between source and receiver. The direct ray has \(\theta_0=\pi/2\).

  • (d) The four smallest angles are \[ \boxed{86.19^\circ,\quad88.09^\circ,\quad88.73^\circ,\quad89.05^\circ}. \]

  • (e) For the smallest angle, \(n=1\), \[ \boxed{ t_1=\frac{2}{b}\ln\!\left[ \sqrt{1+\left(\frac{bX}{2c_0}\right)^2} +\frac{bX}{2c_0}\right] \approx6.662\ \mathrm{s} }. \] The direct-ray time is \[ \boxed{t_{\mathrm{direct}}=\frac{X}{c_0}\approx6.667\ \mathrm{s}}. \] The refracted ray arrives first, by approximately \(\boxed{4.93\ \mathrm{ms}}\).

Solution

We use geometrical acoustics: the wavelength is sufficiently short for propagation to be described by rays in a stationary, locally isotropic medium. The sound speed depends only on \(z\).

(a)

Consider the medium as a succession of thin horizontal layers. Snell’s law, with the angle measured from the vertical normal to these layers, gives

\[ \frac{\sin\theta}{c(z)} =\frac{\sin\theta_0}{c_0} \equiv p, \]

where \(p\) is constant along the ray.

Let \(s\) be distance along the ray. For a ray travelling toward increasing \(x\),

\[ \frac{dx}{ds}=\sin\theta, \qquad \frac{dz}{ds}=\cos\theta. \]

In the upper region, \(c=c_0+bz\). Differentiating \(\sin\theta=pc\) along the ray gives

\[ \cos\theta\,\frac{d\theta}{ds} =pb\,\frac{dz}{ds} =pb\cos\theta. \]

Thus, with the value at the horizontal turning point obtained by continuity,

\[ \frac{d\theta}{ds}=pb =\frac{b\sin\theta_0}{c_0}. \]

The rate of change of the tangent direction with arc length is the curvature. Since it is constant, the trajectory is circular, with

\[ \boxed{R=\frac{1}{pb}=\frac{c_0}{b\sin\theta_0}}. \]

The circle can also be found explicitly. Using \(ds=R\,d\theta\) and the initial position \(x=z=0\),

\[ x=R(\cos\theta_0-\cos\theta), \qquad z=R(\sin\theta-\sin\theta_0). \]

Since \(R\sin\theta_0=c_0/b\), these equations imply

\[ \boxed{ \left(x-R\cos\theta_0\right)^2 +\left(z+\frac{c_0}{b}\right)^2=R^2 }. \]

The upper arc therefore has its centre at

\[ \left(R\cos\theta_0,-\frac{c_0}{b}\right). \]

Below \(z=0\), the gradient changes sign, so \(d\theta/ds=-pb\). The curvature reverses, while its magnitude remains \(1/R\). A ray crossing \(z=0\) consequently follows successive circular arcs of equal radius.

For \(\theta_0=0\), Snell’s law gives a vertical ray with zero curvature. The stated radius formula is understood in the limiting sense \(R\to\infty\).

(b)

At the highest point of an upward arc, the ray is horizontal, so \(\theta=\pi/2\). Snell’s law then gives

\[ \frac{1}{c_0+bz_{\max}} =\frac{\sin\theta_0}{c_0}, \]

and hence

\[ z_{\max} =\frac{c_0}{b}\left(\frac{1}{\sin\theta_0}-1\right). \]

The limiting ray that avoids crossing the sea surface satisfies \(z_{\max}=z_S\). Therefore,

\[ \sin\theta_{0,\min} =\frac{c_0}{c_0+bz_S}, \]

so

\[ \boxed{ \theta_{0,\min} =\sin^{-1}\!\left(\frac{c_0}{c_0+bz_S}\right) }. \]

Because increasing \(\theta_0\) decreases \(z_{\max}\), angles above this threshold turn below the surface. Equality is the grazing case; strict separation from the surface requires \(\theta_0>\theta_{0,\min}\).

(c)

During the first upper arc, the ray returns to \(z=0\) when

\[ \sin\theta=\sin\theta_0, \qquad \theta=\pi-\theta_0. \]

Its horizontal displacement at that crossing is

\[ D =R\left[\cos\theta_0-\cos(\pi-\theta_0)\right] =2R\cos\theta_0. \]

Using the radius from part (a),

\[ \boxed{D=\frac{2c_0}{b}\cot\theta_0}. \]

The speed is continuous at \(z=0\), so the direction of the ray is continuous there. The equal gradient magnitudes above and below \(z=0\) produce alternating upper and lower arcs, each spanning the same horizontal distance \(D\).

The receiver is reached after an integer number \(n\) of these arcs:

\[ X=nD=\frac{2nc_0}{b}\cot\theta_0, \qquad n=1,2,3,\ldots. \]

Therefore,

\[ \boxed{ \theta_{0,n} =\tan^{-1}\!\left(\frac{2nc_0}{bX}\right) }. \]

These angles increase with \(n\), so the smallest angle corresponds to a single upper arc:

\[ \boxed{\theta_{0,1}=\tan^{-1}\!\left(\frac{2c_0}{bX}\right)}. \]

Symmetry also gives corresponding paths launched downward. The horizontal direct ray, \(\theta_0=\pi/2\), is an additional path.

(d)

The numerical factor in the angle formula is

\[ \frac{2c_0}{bX} =\frac{2(1500)}{(0.02000)(10000)} =15. \]

Thus \(\theta_{0,n}=\tan^{-1}(15n)\), giving

\(n\) \(\tan\theta_{0,n}\) \(\theta_{0,n}\)
1 15 \(86.19^\circ\)
2 30 \(88.09^\circ\)
3 45 \(88.73^\circ\)
4 60 \(89.05^\circ\)

(e)

Write the smallest launch angle as \(\theta_1=\theta_{0,1}\). This ray travels along one upper arc, with \(\theta\) increasing from \(\theta_1\) to \(\pi-\theta_1\).

The travel time is

\[ t_1=\int\frac{ds}{c}. \]

Along this arc,

\[ ds=R\,d\theta, \qquad c=\frac{c_0}{\sin\theta_1}\sin\theta =bR\sin\theta. \]

Consequently,

\[ t_1 =\frac{1}{b}\int_{\theta_1}^{\pi-\theta_1} \frac{d\theta}{\sin\theta} =\frac{1}{b} \left[\ln\tan\left(\frac{\theta}{2}\right)\right]_{\theta_1}^{\pi-\theta_1}. \]

Using \(\tan[(\pi-\theta_1)/2]=\cot(\theta_1/2)\),

\[ \boxed{t_1=\frac{2}{b}\ln\cot\left(\frac{\theta_1}{2}\right)}. \]

To express this in terms of the given quantities, define

\[ u=\frac{bX}{2c_0}=\cot\theta_1. \]

The half-angle identity gives

\[ \cot\left(\frac{\theta_1}{2}\right) =\frac{1+\cos\theta_1}{\sin\theta_1} =\csc\theta_1+\cot\theta_1 =\sqrt{1+u^2}+u. \]

Hence,

\[ \boxed{ t_1=\frac{2}{b} \ln\!\left[ \sqrt{1+\left(\frac{bX}{2c_0}\right)^2} +\frac{bX}{2c_0} \right] }. \]

For the numerical data,

\[ u=\frac{(0.02000)(10000)}{2(1500)}=\frac{1}{15}, \]

so

\[ t_1 =\frac{2}{0.02000} \ln\!\left[\sqrt{1+\frac{1}{225}}+\frac{1}{15}\right] \mathrm{s} =\boxed{6.662\ \mathrm{s}}. \]

The direct ray remains at \(z=0\), where its speed is \(c_0\). Its time is therefore

\[ t_{\mathrm{direct}} =\frac{X}{c_0} =\frac{10000}{1500}\ \mathrm{s} =\boxed{6.667\ \mathrm{s}}. \]

Using unrounded values, the difference is

\[ t_{\mathrm{direct}}-t_1 =0.00492842\ldots\ \mathrm{s} \approx\boxed{4.93\ \mathrm{ms}}. \]

Thus the refracted ray with the smallest launch angle arrives first. The increased sound speed away from \(z=0\) more than compensates for its extra path length.

This comparison also follows generally from

\[ t_1=\frac{2}{b}\operatorname{arsinh}u, \qquad t_{\mathrm{direct}}=\frac{2u}{b}, \]

because, for \(u>0\),

\[ \operatorname{arsinh}u =\int_0^u\frac{dv}{\sqrt{1+v^2}}<u. \]

The solution was checked against the official solution.