Съдържание

IPhO 1985, Теория и експеримент — Задача 1. Problem 1

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от IPhO 1985

Problems and solutions of the 16th IPhO Portorož, Slovenia, (Former Yugoslavia), 1985 · 10 т.

Условие

A young radio amateur maintains a radio link with two girls living in two towns. He positions an aerial array such that when the girl living in town A receives a maximum signal, the girl living in town B receives no signal and vice versa. The array is built from two vertical rod aerials transmitting with equal intensities uniformly in all directions in the horizontal plane.

a) Find the parameters of the array, i. e. the distance between the rods, its orientation and the phase shift between the electrical signals supplied to the rods, such that the distance between the rods is minimum.

b) Find the numerical solution if the boy has a radio station transmitting at 27 MHz and builds up the aerial array at Portorož. Using the map he has found that the angles between the north and the direction of A (Koper) and of B (small town of Buje in Istria) are $72^{\circ}$ and $157^{\circ}$ , respectively.

Отговори

Решение

a) Let the electrical signals supplied to rods 1 and 2 be $E_1 = E_0\cos\omega t$ and $E_2 = E_0\cos(\omega t + \delta)$ , respectively. The condition for a maximum signal in direction $\vartheta_A$ (Fig. 4) is:

\frac{2\pi a}{\lambda}\sin\vartheta_A - \delta = 2\pi N

and the condition for a minimum signal in direction $\vartheta_B$ :

\frac{2\pi a}{\lambda}\sin\vartheta_B - \delta = 2\pi N' + \pi

(2p.)

where $N$ and $N'$ are arbitrary integers. In addition, $\vartheta_A - \vartheta_B = \varphi$ , where $\varphi$ is given. The problem can now be formulated as follows: Find the parameters $a$ , $\vartheta_A$ , $\vartheta_B$ , $\delta$ , $N$ , and $N'$ satisfying the above equations such, that $a$ is minimum.

Two rods marked 1 (top) and 2 (bottom) a distance a apart on a vertical line; from each rod a parallel ray leaves upward to the right at angle ϑ to the horizontal line through rod 1; a perpendicular from rod 1 to the lower ray makes the same angle ϑ with the line joining the rods. — Fig. 4

We first eliminate $\delta$ by subtracting the second equation from the first one:

a\sin\vartheta_A - a\sin\vartheta_B = \lambda(N - N' - \tfrac{1}{2})\,.

Using the sine addition theorem and the relation $\vartheta_B = \vartheta_A - \varphi$ :

2a\cos(\vartheta_A - \tfrac{1}{2}\varphi)\sin\tfrac{1}{2}\varphi = \lambda(N - N' - \tfrac{1}{2})

a = \frac{\lambda(N - N' - \tfrac{1}{2})}{2\cos(\vartheta_A - \tfrac{1}{2}\varphi)\sin\tfrac{1}{2}\varphi}\,.

The minimum of $a$ is obtained for the greatest possible value of the denominator, i. e.:

\cos(\vartheta_A - \tfrac{1}{2}\varphi) = 1\,,\qquad \vartheta_A = \tfrac{1}{2}\varphi\,,

and the minimum value of the numerator, i. e.:

N - N' = 1\,.

The solution is therefore:

a = \frac{\lambda}{4\sin\frac{1}{2}\varphi}\,,\quad \vartheta_A = \tfrac{1}{2}\varphi\,,\quad \vartheta_B = -\tfrac{1}{2}\varphi \quad\text{and}\quad \delta = \tfrac{1}{2}\pi - 2\pi N\,.

(6p.)

(N = 0 can be assumed throughout without loosing any physically relevant solution.)

b) The wavelength $\lambda = c/\nu = 11.1\ \mathrm{m}$ , and the angle between directions A and B, $\varphi = 157^{\circ} - 72^{\circ} = 85^{\circ}$ . The minimum distance between the rods is $a = 4.1\ \mathrm{m}$ , while the direction of the symmetry line of the rods is $72^{\circ} + 42.5^{\circ} = 114.5^{\circ}$ measured from the north.

(2 p.)

Оригинал в Архива: ipho85.pdf