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IPhO 1995, Теория — Задача 2. Sound Propagation

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XXVI IPhO — Theoretical Questions · 20 т.

Условие

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 between 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.

Graph of sound speed c (horizontal axis) against depth z (vertical axis, marked z_S at the top, 0 and −z_b at the bottom): a V-shaped line with its vertex at c0 on z = 0, the upper branch labelled c=c0+bz and the lower branch c=c0−bz, between horizontal lines at z_S and −z_b. — Figure 1 (a)

Section of the z–x plane: vertical z axis with dashed lines at z_S and −z_b; source S at the origin, receiver H at x = X on the dashed x axis; a curved ray leaves S at angle θ0 to the vertical and bends over, with the local angle θ marked against a vertical line further along the ray. — 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) 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 \le \theta_0 < \frac{\pi}{2}\ .

[6 т.]

(b) 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. [3 т.]

(c) 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. [4 т.]

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

$X = 10000\ \mathrm{m}$
$c_0 = 1500\ \mathrm{ms^{-1}}$
$b = 0.02000\ \mathrm{s^{-1}}$ [2 т.]

(e) 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 rays 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)? [5 т.]

Отговори

Решение

(a)

Snell’s Law may be expressed as

\frac{\sin\theta}{\sin\theta_0} = \frac{c}{c_0}\ , \qquad (1)

where $c$ is the speed of sound.
Consider some element of ray path $ds$ and treat this as, locally, an arc of a circle of radius $R$ . Note that $R$ may take up any value between 0 and $\infty$ . Consider a ray component which is initially directed upward from $S$ .

Small element of the ray path: a chord ds at angle θ to the vertical, rising by dz, with two radii R to the centre of curvature enclosing the angle dθ.

In the diagram, $ds = Rd\theta$ , or $\dfrac{ds}{d\theta} = R$ .

From equation (1), for a small change in speed $dc$ ,

\cos\theta d\theta = \frac{\sin\theta_0}{c_0}dc

For the upwardly directed ray $c = c_0 + bz$ so $dc = bdz$ and

\frac{\sin\theta_0}{c_0}b\,dz = \cos\theta d\theta\ , \quad \text{hence} \quad dz = \frac{c_0}{\sin\theta_0}\frac{1}{b}\cos\theta\,d\theta\ .

We may also write (here treating $ds$ as straight) $dz = ds\cos\theta$ . So

ds = \frac{c_0}{\sin\theta_0}\frac{1}{b}\,d\theta

Hence

\frac{ds}{d\theta} = R = \frac{c_0}{\sin\theta_0}\frac{1}{b}\ .

This result strictly applies to the small arc segments $ds$ . Note that from equation (1), however, it also applies for all $\theta$ , i.e. for all points along the trajectory, which therefore forms an arc of a circle with radius $R$ until the ray enters the region $z < 0$ .

Ray leaving z = 0 at angle θ0 to the vertical and curving up to just below the surface z = z_S; the radius R from the centre of curvature makes angle θ0, and the vertical distance R sin θ0 below z = 0 is marked.

(b)

Here

\begin{aligned} z_s &= R - R\sin\theta_0 \\ &= R(1 - \sin\theta_0) \\ &= \frac{c_0}{b\sin\theta_0}(1 - \sin\theta_0)\ , \end{aligned}

from which

\theta_0 = \sin^{-1}\left[\frac{c_0}{bz_s + c_0}\right]\ .

Single circular arc from S at x = 0 to H at x = X, launched at angle θ0 to the vertical; two radii R meet at the centre of the circle below the line SH.

(c)

The simplest pathway between $S$ and $H$ is a single arc of a circle passing through $S$ and $H$ . For this pathway:

X = 2R\cos\theta_0 = \frac{2c_0\cos\theta_0}{b\sin\theta_0} = \frac{2c_0}{b}\cot\theta_0\ .

Hence

\cot\theta_0 = \frac{bX}{2c_0}\ .

The next possibility consists of two circular arcs linked as shown.

Two linked circular arcs from S (x = 0) to H (x = X): the ray first curves above the line SH, then below it, crossing it at the midpoint.

For this pathway:

\frac{X}{2} = 2R\cos\theta_0 = \frac{2c_0}{b}\cot\theta_0\ .

i.e.

\cot\theta_0 = \frac{bX}{4c_0}\ .

In general, for values of $\theta_0 < \frac{\pi}{2}$ , rays emerging from $S$ will reach $H$ in $n$ arcs for launch angles given by

\theta_0 = \cot^{-1}\left[\frac{bX}{2nc_0}\right] = \tan^{-1}\left[\frac{2nc_0}{bX}\right]

where $n = 1, 2, 3, 4, \ldots$
Note that when $n = \infty$ , $\theta_0 = \frac{\pi}{2}$ as expected for the axial ray.

(d)

With the values cited, the four smallest values of launch angle are

$n$	$\theta_0$ (degrees)
1	86.19
2	88.09
3	88.73
4	89.04

(e)

The ray path associated with the smallest launch angle consists of a single arc as shown:

Single arc from point 1 (S) through the top point 2 to point 3 (H).

We seek

\int_1^3 dt = \int_1^3 \frac{ds}{c}

Try first:

t_{12} = \int_1^2 \frac{ds}{c} = \int_{\theta_0}^{\pi/2} \frac{Rd\theta}{c}

Using

R = \frac{c}{b\sin\theta}

gives

t_{12} = \frac{1}{b}\int_{\theta_0}^{\pi/2} \frac{d\theta}{\sin\theta}

so that

t_{12} = \frac{1}{b}\left[\ln\tan\frac{\theta}{2}\right]_{\theta_0}^{\pi/2} = -\frac{1}{b}\ln\tan\frac{\theta_0}{2}

Noting that $t_{13} = 2t_{12}$ gives

t_{13} = -\frac{2}{b}\ln\tan\frac{\theta_0}{2}\ .

For the specified $b$ , this gives a transit time for the smallest value of launch angle cited in the answer to part (d), of

t_{13} = 6.6546\ \mathrm{s}

The axial ray will have travel time given by

t = \frac{X}{c_0}

For the conditions given,

t_{13} = 6.6666\ \mathrm{s}

thus this axial ray travels slower than the example cited for $n = 1$ , thus the $n = 1$ ray will arrive first.

Оригинал в Архива: th.pdf · официални решения: th_sol.pdf