Kvant 1970, No. 9 · Pages

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Let’s consider an infinite
decimal…

S. G. Gindikin

4,92617231248,9
7,16214301695743,0
1,7

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Let us limit ourselves only to the additions concluded between 0 and 1, that is, expressions of the form

$$0, \alpha_1 \alpha_2 \dots \alpha_k \dots,$$

(1) where $\alpha_k$, $1 \leq k < \infty$, are some numbers 0, 1, …, 8, 9. In the theory of real numbers, there is quite a lot of trouble with the fact that any additional number with a “tail” of nines (all signs, starting from a certain place, are equal to 9) is equal to some additional number with a “tail” of zeros (for example, 0.099… = 0.100…). To avoid them, we will not consider dobies that contain nine in their entries. So, we will consider dobes of the form (1), that is, sequences of the number $0, \alpha_1 \alpha_2 \dots \alpha_k \dots$, where $\alpha_k$ can be equal to 0, 1, 2, 3, 4, 5, 6, 7, 8*). We must know that even among such great ones there are “immense ones.”

How can you prove this?

Let us first consider a similar situation, relative, however, to finite sets (for better understanding we describe it in a humorous form). The director of the cinema learned that there were spectators at the screening, and the seller had ninety tickets at the box office. You probably won’t be surprised that from a comparison of these two facts he concluded that there were free riders in the hall. He simply noticed that there were more spectators than tickets, and there could not be enough tickets for all the spectators. Please note that this reasoning only makes it possible to prove the existence of a ticketless spectator, and removing a specific “hare” requires additional research (for example, checking tickets at the exit).

In all the years that have passed centuries, G. Cantor **) came up with a wonderful way to carry out similar reasoning for the free-rider age G. Cantor **) came up with a wonderful way to carry out similar reasoning for the free-ride age G. Cantor (1845-1918) - the creator of the theory of free-rider sets.

*) The perceptive reader will certainly notice when reading the further text that all considerations would only be meaningful if we limited them to only two numbers, for example, 0 and 1.

**) G. Cantor (1845-1918) - creator of the theory of free-riding sets.

It turns out that we have arrived at a contradiction? It is pleasant to come to a contradiction if you have previously assumed the opposite of what you want to prove. But, as you probably remember, we did not make such an assumption. How can this be?

The only choice that can be made is that our concept of a “comprehensive” dobby, as a dobby admitting of a finite description, is incorrect and has led to a contradiction. To make sure of this, let’s give another example. It will no longer concern infinite dobbies, but natural numbers. Each such number can be described by a finite text (for example, by naming it), so there is no need to talk about “immenseness”.

However, it will find numbers that are not described in our universal library, that is, numbers for which there is no description in Russian, containing no more than a million characters. The latter follows from the fact that the number of volumes in a universal library is finite. Among the natural numbers that have not received a description, there are the smallest ones. We have proven the following statement:

There is the smallest natural number that cannot be specified in Russian text containing no more than one million characters.

The highlighted phrase contains this thirty-five characters. But this is thirty-five less than a million, and therefore we can be congratulated on one more contradiction: the above phrase unambiguously describes the mentioned number!

This is a real criticism of the concept of “verbal portrait”. Do you think the descriptions can also be considered phrases:

“α is an infinite tenth dobby, the signs of which with numbers divisible by the number of steps taken by Napoleon on the day of the Battle of Privato coincide with the corresponding signs of the number l, and the remaining signs are equal to zero.”

“To determine the k-th sign of dobby δ, you need to count the number of goals scored in all football mats played in the k-th year of the new era (not according to Moscow time), and take the last digit.”

“The fifth sign of γ is equal to 1 if in the notation √2 of the infinite decimal addition the digit 7 occurs an infinite number of times, otherwise it is equal to 3; all other signs of γ are equal to 2.”

I tried very hard to come up with absurd descriptions that would nevertheless define the only infinite doob. It is unlikely that we will agree to accept them as “verbal portraits” of Dobei. And the point is not even their absurdity, the important thing is that they do not provide a factual description of these dobes. We did not even exclude the possibility that the description could use already established information from the past (α), facts from the future (δ). In addition, the final text may require that an infinite number of prodedurs be sold to determine a single decimal place (γ). The same K. Prutkov provides against such errors:

“If everything that was past was present, and the present continued to exist at odds with the future, who would be able to figure out: where is the cause and where are the consequences?”

The main inconvenience, which, in fact, led us to a contradiction, is that the “verbal portrait” could use information about the set of all “verbal portraits”, in which he himself should be included as one of the elements.

The fact that this situation, which in logic bears the name of a “vicious circle,” is a source of contradiction, was known from the time of the semi-legendary sage Epimendas, who lived in the 6th century BC. e., who, being critical (that is, living on the island of Crete), declared: “All critical people are liars.” For simplicity, we will classify all people as liars, who always tell lies, and right people, who always tell lies.

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Electrical resistance is a quantum phenomenon

D. A. Frank-Kamenetsky

The resistance of a conductor to the flow of electric current is not at all as simple a thing as it seems at first glance. As we will see, we cannot do without quantum here either. Let’s first see what can be done without its help.

Classical theory

We are used to the fact that “classic” is a similar word. They say: Streltsov scored a “classic goal.” Instead of “classic,” you could say “cool” or even “iron” in more modern jargon. A physicist has a completely different attitude towards this word. In modern physics, classical are called theories created by the classes of science of the past, before the emergence of quantum physics and the theory of relativity. Most often, the word “classical” is used by physicists as the opposite of the word “quantum”. The classical theory is different in that it does not involve quanti. Such theories are very simple; to modern physicists they seem somewhat naive. They often bring a big role, but, as a rule, they do not fully resolve the issue. In classical theory, there is always something missing for a modern physicist.

In many books, including textbooks (and even very well), it is written that the electrical resistance of a metal originates from modern electricians,

carrying current, with atoms*) of the crystal lattice. Is this concept of resistance correct?

It is best not to say either “yes” or “no”, but to say that it expresses the classical theory of electrical resistance. Everything we just said about classical theories in general applies to it. But in this case, the classical theory turns out to be very weak: it cannot explain many of the main features of electrical resistance. In the classical theory, one thing is certainly true: resistance comes from the fact that electrons transfer part of their energy and momentum (momentum) to the crystal lattice. But exactly how this transmission originates is not a simple and very interesting question.

Temperature dependence

The resistance of pure metals increases greatly with temperature. For many of them it is applied in proportion to the absolute temperature. At low temperatures, this simple dependence breaks down.

*) It would be more accurate to say - with ponies, but this does not exist for us here. The mechanism of electrical resistance is that electrons hitting the atoms of the crystal lattice cause them to vibrate—electrical energy is converted into thermal energy.

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Chali results are not so flat.

The energy of a phonan, like any quantum, is equal to $hv$, where $h$ is Planck’s constant, and $v$ is the frequency of the wave.

The number of phonans is $\frac{E}{hv}$, where $E$ is the energy of thermal motion, proportional to the absolute temperature $T$. Boltzmann’s constant $k$ can be taken as the proportionality set. So, the number of phonans is $N = \frac{kT}{hv}$. We have obtained a very important result. The number of phonans in a solid is proportional to the absolute temperature. Therefore, if the resistance is proportional to the constant power, Boltzmann’s constant $k$ can be taken. So, the number of phonans is $N = \frac{kT}{hv}$. We have obtained a very important result. The number of phonans in a solid is proportional to the absolute temperature. Therefore, if the resistance is proportional to the constant power, Boltzmann’s constant $k$ can be taken. So, the number of phonans is $N = \frac{kT}{hv}$. We have obtained a very important result.

But this result is valid only at sufficiently high temperatures. Indeed, according to the meaning of quantum theory, the number of phonans cannot be less than one. Let’s see at what temperature it is equal to unity. This temperature is called the Debye temperature and is denoted in Greek by the letter $\theta$. Equating $N$ to one, we find: $\theta = \frac{hv}{k}$. Debye temperature (or Debye temperature) plays a very important role in solid state physics. Using this quantity of interest to us, the number of phonans is expressed quite simply: $N = \frac{T}{\theta}$. At temperatures below the base, this number becomes less than one. This means that we can only talk about the average number of phonans, and the calculation is subject to conditions. But the quantities $\frac{T}{\theta}$ still remain measures of the number of phonans in a solid, and, consequently, the number of only electrins with phonans. Is

*) This is the number of phonans per lattice vibration, more precisely, per one vibrational degree of freedom, or, as physicists like to say, per one oscillator.

Fig. 8. Generalized temperature dependence curve of resistivity of pure metals.

Based on these reports, the following hypothesis was put forward: the temperature dependence of the electrical resistance of pure metals is expressed by a universal function of the measure of the number of phonans $\frac{T}{\theta}$. Mathematically, this hypothesis is written as follows: $\frac{R}{R_0} = F\left(\frac{T}{\theta}\right)$, where $R$ is the resistance at temperature $T$, $R_0$ is at temperature $\theta$, and $F$ is a function, the form of which can be found from experience. What is important is what the $F$ function promises for different pure metals. How well this hypothesis is justified experimentally can be seen from figure 8. This graph was constructed at one time by the American physicist Bardeen, who later became famous for his work on the theory of superconductivity.

It is interesting and useful to relate the Debye temperature $\theta$ to the speed of sound $c$. Let us remember that matter consists of atoms and the wave does not propagate continuously, but jumps from atom to atom. During the oscillation period $t$ the wave cannot travel a path less than the distance between neighboring atoms $a$.

Otherwise, the wave would “hang” in the void between two atoms. The path traveled by a sound wave in one oscillation is equal to $c\tau$. So, the atomic structure of matter imposes a limitation on the periods of sound vibrations: $c\tau \geq a$. Frequency $v$ is the number of periods per unit time, hence $v\tau=1$. Thus, the frequencies of sound vibrations are limited:

$$v \leq \frac{c}{a}.$$

During thermal motion, sound waves (in other words, phonons) with different frequencies are excited, as they say, a whole spectrum of frequencies. But the phonon spectrum has a limiting frequency. Waves with higher frequencies would not have time to jump from one atom to the next and therefore cannot be excited. Detailed calculations show that the Debye temperature can be determined in the same way as we did, if only the limiting frequency of the background spectrum in a solid is taken as the frequency $v$. This is what we will do.

All we have to do is estimate the distance between the atoms $a$. If the number of atoms per unit volume is $n$, then $n=\frac{1}{a^3}$.

From here and from formula (*) it turns out:

$$\frac{v^3}{c^3} \leq n.$$

A more accurate calculation, which we cannot provide here, is given by the additional factor $4\pi$. Thus, we obtain a formula connecting the limiting frequency $v$ with the speed of sound $c$:

$$4\pi \frac{v^3}{c^3} = n.$$

This very important formula is closely related to the Rayleigh-Jeans formula, which was discussed in the article by Ya. A. Smorodinsky («Kvant» No. 1). Expressing the frequency $v$ from formula (**) and substituting it into the expression for the Debye temperature, we obtain that

$$\theta = \frac{hc}{k} \sqrt[3]{\frac{n}{4\pi}}.$$

So, we got the connection between the

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Attention: in
equation - parameter!

A. Ya. Margulis,
A. G. Mordkovich,
B. A. Radunsky

Instead of a preface
I remember such a case. The student was asked to solve a seemingly harmless equation:
$(a^2 - 5a + 6)x = a^2 - 4$. (1)
Without thinking twice, the student continued the following calculations:
$$x = \frac{a^2 - 4}{a^2 - 5a + 6} = \frac{(a - 2)(a + 2)}{(a - 2)(a - 3)}; x = \frac{a + 2}{a - 3}.$$ (2)
He clearly did not take into account that the given equation contains the parameter $a$ and that an equation with a parameter is, in essence, a set of equations. So our equation for $a = 0$ takes the form $6x = -4$, for $a = 1$ it takes the form $2x = -3$, etc. In other words, having given the parameter a specific numerical value, we select one from the set of equations - corresponding to the selected value of the parameter. But then the solution depends on the parameter.
We have already said that for $a = 0$ equation (1) takes the form $6x = -4$. The root of this equation is the value $x = -\frac{2}{3}$. But according to formula (2) with $a = 0$ we get $x = -\frac{2}{3}$.
For $a = 1$, equation (1) takes the form $2x = -3$, hence $x = -\frac{3}{2}$. But according to formula (2) with $a = 1$ we get $x = -\frac{3}{2}$.
If such a situation occurred for any active knowledge $a$, then it could be argued that formula (2) determines the solution to equation (1). But that’s not true.
Let’s put $a = 2$. Equation (1) becomes $0 \cdot x = 0$. The root of this equation is any real number, while formula (2) gives in this case $x = -4$.
For $a = 3$, equation (1) takes the form $0 \cdot x = 5$. This equation has no solutions. However, the expression $\frac{a + 2}{a - 3}$ does not make sense with $a = 3$.
Thus, it is impossible to assert that $x = \frac{a + 2}{a - 3}$ for any value of the parameter.
In fact, it was necessary to reason as follows. Before dividing both sides of equation (1) by the coefficient for the unknown, let’s see if it could happen that for some values ​​of the parameter this coefficient will become zero. We note that the trinomial $a^2 - 5a + 6$ vanishes at $a = 2$, $a = 3$. This means that we need to consider these cases separately, which we have already done. If $a \neq 2$, $a \neq 3$, then we can divide both sides of the equation by the coefficient of $x$, we get $x = \frac{a + 2}{a - 3}$. Combining the results obtained, we write down
ANSWER: if $a = 2$, then $x$ is any real number;
if $a = 3$, then there are no solutions;
if $a \neq 2$, $a \neq 3$, then $x = \frac{a + 2}{a - 3}$.
This simple example shows that when solving an equation with pa-

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where
$$x_{1,2} = \frac{1-a \pm \sqrt{2a-3a^2}}{2}.$$

Both found roots for all admissible values ​​of the parameter $a$, (that is, for all $a$ satisfying the inequalities $0 \leq a \leq \frac{1}{2}$ - see figure 1) are real numbers. Indeed, the discriminant $D = 2a-3a^2 \geq 0$ for $0 \leq a \leq \frac{2}{3}$; that means, even more so $D \geq 0$ with $0 \leq a \leq \frac{1}{2}$.

Now we need to find out whether the found values ​​satisfy the conditions $x \geq 0$, $x+a \leq \frac{1}{2}$.

Let us first check that condition $x+a \leq \frac{1}{2}$ is met. We have
$$x_1+a = \frac{1-a-\sqrt{2a-3a^2}}{2} + a = \frac{1+a-\sqrt{2a-3a^2}}{2}.$$
Solving inequality
$$\frac{1+a-\sqrt{2a-3a^2}}{2} \leq \frac{1}{2},$$
we get $0 \leq a \leq \frac{1}{2}$. This means $x_1+a \leq \frac{1}{2}$ for all admissible values ​​of the parameter $a$.

Next,
$$x_2+a = \frac{1-a+\sqrt{2a-3a^2}}{2} + a = \frac{1+a+\sqrt{2a-3a^2}}{2}.$$
Inequality
$$\frac{1+a+\sqrt{2a-3a^2}}{2} \leq \frac{1}{2}$$

is executed only at $a = 0$. But for $a = 0$ we have $D = 0$ and $x_1=x_2 = \frac{1}{2}$.

The proof of the non-negativity of both found roots is left to the reader.

Answer: if $0 \leq a \leq \frac{1}{2}$, then $x = \frac{1-a-\sqrt{2a-3a^2}}{2}$;

if $a < 0$ or $a > \frac{1}{2}$, then there are no solutions.

Note that when solving the equation considered, drying of the ODZ turns out to be useful. The question naturally arises: when should such clarifications be made?

If we carry out a transformation that results in an equation equivalent to the previous one, then in the case of ODZ there is, of course, no need. Otherwise it is useful. This situation arises, for example, when squaring (or to any even power) both parts of an equation, which is what we encountered in the example considered.

Some examples
Example 1*). Solve the equation
$$\frac{\sqrt{a+x}}{a} + \frac{\sqrt{a+x}}{x} = \sqrt{x}.$$
Solution. The ODZ is determined by the equations $x > 0$, $a+x \geq 0$, $a \neq 0$ (from here, by the way, it immediately follows that at $a = 0$ the equation has no solution). Let’s transform the equation to the form
$$(a+x)^{3/2} = ax^{3/2}.$$
Let’s dry the ODZ: from the non-negativity part of the last equation and the equations $x > 0$, $a \neq 0$ it follows that $a > 0$.

Next we get: $a+x = a^{2/3}x$, from where $x (a^{2/3}-1) = a$.
With $a = 1$ there are very/no solutions. If $a \neq 1$, then
$$x = \frac{a}{a^{2/3}-1}.$$
Let’s find out whether, with everyone’s consideration,
*) No. 206 from the “Collection of problems in elementary mathematics.” Authors: N. P. Antonov, M. Ya. Vygodsky, V. V. Nikitin, A. I. Sankin. In the given problem the example is solved/herno.

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Given the values ​​of the parameter $a$ ($a > 0, a \neq 1$), condition $x > 0$ is satisfied (from which it obviously follows that the second condition $a + x \geq 0$ is satisfied).

The inequality $\frac{a}{a^{2/3} - 1} > 0$ holds only for $a > 1$ (taking into account that $a > 0$). Thus, not for all admissible values ​​of $a$ the found root turns out to be suitable.

Answer: if $a > 1$, then $x = \frac{a}{a^{2/3} - 1}$; if $a \leq 1$, then there is no solution.

Example 2: Solve the equation
$$\log_9 x + \log_9 \frac{2 - x}{2} = \log_9 \log_9 p.$$

Solution. The ODZ is determined by the inequalities $0 < x < 2$, $p > 1$ (at figure 2 this area is colored crimson). Solving the equation, we consistently obtain
$$\frac{x(2 - x)}{2} = \log_9 p,$$
$$x^2 - 2x + 2\log_9 p = 0,$$
$$x_{1,2} = 1 \pm \sqrt{1 - 2\log_9 p}.$$

Condition $1 - 2\log_9 p \geq 0$ is satisfied for $p \leq 3$, which, taking into account the ODZ, gives $1 < p \leq 3$. With such $p$ the roots will be real numbers. At $p > 1$ the inequality $\log_9 p > 0$ is satisfied, and then $1 - 2\log_9 p < 1$. It follows that for $1 < p \leq 3$ the inequality $0 < x_{1,2} < 2$ holds.

Answer: if $1 < p \leq 3$, then $x_{1,2} = 1 \pm \sqrt{1 - 2\log_9 p}$; for the remaining $p$ there are no solutions.

Comment. From equation (6) we obtain $p = 9^{\frac{x(2 - x)}{2}}$. The graph of this function, plotted on figure 2, clearly shows that the condition $0 < x < 2$ is satisfied at $1 < p \leq 3$.

Example 3: Solve the equation
$$\frac{a + \sin x}{a \cos x + 1} = \frac{a + \cos x}{a \sin x + 1}.$$

Solution. Let us first consider the case $a = 0$. The given equation takes the form $\sin x = \cos x$, hence $x = \frac{\pi}{4} + \pi n$ ($n = 0, \pm 1, \ldots$).

Let now $a \neq 0$. Then the ODZ is determined by the equations
$$\cos x \neq -\frac{1}{a}; \sin x \neq -\frac{1}{a}.$$

Let us transform the given equation to the form $(\sin x - \cos x)[a^2 + a(\sin x + \cos x) + 1] = 0$. The last property equation is to the following set of equations:
$$a^2 + a(\sin x + \cos x) + 1 = 0,$$
$$\sin x - \cos x = 0.$$

Solving the first equation of the set (7), we obtain:
$$a\sqrt{2}\sin(x + \frac{\pi}{4}) = -a^2 - 1$$
and onwards
$$\sin\left(x + \frac{\pi}{4}\right) = -\frac{a^2 + 1}{a\sqrt{2}}.$$

The last equation has no solution, since $\left| -\frac{a^2 + 1}{a\sqrt{2}} \right| > 1$ for any $a \neq 0$. Indeed, suppose that $\frac{a^2 + 1}{|a|\sqrt{2}} \leq 1$; transforming this inequality, we get
$$a^2 + 1 \leq |a|\sqrt{2}, (a^2 + 1)^2 \leq 2a^2,$$
$$a^4 + 1 \leq 0.$$

The last inequality is incorrect.

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The solutions to the second population equation (7) are the values ​​$x = \frac{\pi}{4} + \pi k$ ($k = 0, \pm 1, \ldots$). It remains to check whether, for all values ​​of $a \neq 0$, the found values ​​of $x$ are admissible.

If $k = 2n$, then $\sin x = \cos x = \frac{\sqrt{2}}{2}$ and the condition $\frac{\sqrt{2}}{2} \neq -\frac{1}{a}$ must be satisfied, that is, $a \neq -\sqrt{2}$.

If $k = 2n + 1$, then $\sin x = \cos x = -\frac{\sqrt{2}}{2}$.

From the condition $-\frac{\sqrt{2}}{2} \neq -\frac{1}{a}$ we obtain: $a \neq \sqrt{2}$.

Combining all the results, write

Answer: if $a \neq \pm \sqrt{2}$, then $x = \frac{\pi}{4} + \pi k$;

if $a = \sqrt{2}$, then $x = \frac{\pi}{4} + 2\pi n$;

if $a = -\sqrt{2}$, then $x = \frac{\pi}{4} + \pi (2n + 1)$ ($k,n = 0, \pm 1, \pm 2; \ldots$).

Example 4. Solve equation $\log^2_{\sqrt{2}} \cos x - 2a \log_{\sqrt{2}} \cos x = a^2 - 8$.

(8)

Solution. Putting $y = \log_{\sqrt{2}} \cos x$, we transform the given equation into a quadratic equation $y^2 - 2ay - (a^2 - 8) = 0$,

whence $y_{1,2} = a \pm \sqrt{2(a^2 - 4)}$.

Now we need to solve a set of logarithmic equations

$$\log_{\sqrt{2}} \cos x = a + \sqrt{2(a^2 - 4)}. \tag{9}$$

$$\log_{\sqrt{2}} \cos x = a - \sqrt{2(a^2 - 4)}. \tag{10}$$

We note that for $a^2 - 4 < 0$ the right-hand sides of equations (9) and (10) do not make sense. This means that, for $-2 < a < 2$, both equations have no solutions. Consider the case $a^2 - 4 \geq 0$, which is possible with $a \geq 2$ or with $a \leq -2$.

Let’s start with $a \geq 2$. Since $\cos x \leq 1$ and $\sqrt{2} > 1$, then $\log_{\sqrt{2}} \cos x \leq 0$.

This means that equations (9) and (10) will have solutions only for such values ​​of the parameter $a$ for which the right-hand sides of the equation are non-positive. It is easy to see that for $a \geq 2$ the right-hand side of equation (9) is positive, therefore, equation (9) has no solutions.

Let us find out at what values ​​of the parameter $a$ the right-hand side of equation (10) will be non-positive. Solving the inequality $a - \sqrt{2(a^2 - 4)} \leq 0$ (taking into account the condition $a \geq 2$), we obtain $a \geq \sqrt{8}$. Thus, if $2 \leq a < \sqrt{8}$, then equation (10) has no solutions (and since equation (9) also has no solutions, we conclude from this that equation (8) in this case also has no solutions).

If $a \geq \sqrt{8}$, then from equation (10) we obtain

$$x = \pm \arccos(\sqrt{2})^{a - \sqrt{2(a^2 - 8)}} + 2\pi n. \tag{11}$$

Let us now consider the case $a \leq -2$. For these parameter values, the right side of equation (10) is different, that is, this equation has solutions (they are already written above by formula (11)). Let us find out at what values ​​of the parameter $a$ the right-hand side of equation (9) will be non-positive. Solving the inequality $a + \sqrt{2(a^2 - 4)} \leq 0$ (taking into account the condition $a \leq -2$), we obtain $a \geq -\sqrt{8}$. Thus, if $a < -\sqrt{8}$, then equation (9) has no solutions, if $-\sqrt{8} \leq a \leq -2$, then equation (9) has solutions, namely:

$$x = \pm \arccos(\sqrt{2})^{a + \sqrt{2(a^2 - 8)}} + 2\pi n.$$

Combining all the results obtained, we write down

Answer: if $-2 < a < \sqrt{8}$, then there are no solutions;

if $a \geq \sqrt{8}$ or $a < -\sqrt{8}$ then

$$x = \pm \arccos(\sqrt{2})^{a - \sqrt{2(a^2 - 8)}} + 2\pi n;$$

if $-\sqrt{8} \leq a \leq -2$, then

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Fig. 3.

$$x = \pm \arccos(\sqrt{2})^{a \pm \sqrt{2}a^2 - 8} + 2\pi n.$$

Geometry helps algebra

When solving an equation with a parameter, various geometric considerations may be useful for scales. Consider, for example, the equation

$$\sqrt{x(2a-x)} = 1 - x \text{ при } a > 0.$$

Which solution of the equation can be interpreted as the abscissa of the point of intersection of the straight line $y = 1 - x$ with the semicircle

$$y = \sqrt{x(2a-x)}, \text{ или}$$
$$(x - a)^2 + y^2 = a^2; \text{ у} \geq 0$$

radius $a$, centered at point $(a, 0)$ (Fig. 3). The position of the semicircle and its radius change as the parameter $a$ changes. On figure 3 it is clearly visible that for some values of $a$ the semicircle does not intersect the line $y = 1 - x$ (semicircles lying under the line), for other values $a$ the semicircle intersects the line at two points (semicircles lying in the middle head zone), for third values $a$ the semicircle intersects the line at one point (semicircles lying in the red zone or above), and for some values of $a$ the semicircle touches the line (semicircle separating the head and brown zones).

Let us consider the case when the semicircle touches the line $y = 1 - x$. From geometric considerations we conclude that the distance of the center of this semicircle from the point of intersection of the straight line $y = 1 - x$ with the abscissa axis is equal to $a\sqrt{2}$. Hence $a + a\sqrt{2} = 1$, that is, the case of tangency occurs at $a = \sqrt{2} - 1$. This means that at $a < \sqrt{2} - 1$ the equation has no solution (blue zone), but at $a = \sqrt{2} - 1$ it has a unique solution, which is found from geometric considerations: $x = a + \frac{a\sqrt{2}}{2}$ (see Fig. 3), from which we obtain $x = \frac{\sqrt{2}}{2}$.

The semicircle passing at figure 3 through the point $(1, 0)$ corresponds to the value of the parameter $a = \frac{1}{2}$. This means that for $\sqrt{2} - 1 < a \leqslant \frac{1}{2}$ the equation has two solutions:

$$x_1 = \frac{a + 1 - \sqrt{a^2 + 2a - 1}}{2},$$
$$x_2 = \frac{a + 1 + \sqrt{a^2 + 2a - 1}}{2}$$

(the roots are found using an algebraic solution to the given equation). Finally, if $a \geqslant \frac{1}{2}$, then the equation has a unique solution $x = x_1$ (since $x_1 < x_2$).

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ELLIPSE
HYPERBOLA
PARABOLA

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Fig. 3. Deformations with $k=2$ and $k=\frac{1}{2}$. Notice the change in roof pitch.

the plane collapses. When $k=1$ all points of the plane remain in place; the latter case is called the mode-destiny transformation.

Let some figure $\omega$ (that is, some set of points) be given on the plane Π. We will be interested in the set $\Omega$ of all points obtained from $\omega$ as a result of deformation (see Fig. 3, 4).

What is Elipis? Elipis elements

An elipis is a figure called

Fig. 4. Deformation of the image of a soccer ball with coefficients $k=\frac{1}{2}$ (bottom left) and $k=2$ (right).

obtained by the deformation of the circle relative to its diameter (Fig. 5).

Let $a$ be the radius of the circle; then its diameter $cd$, lying on the axis, will transform into itself $(CD=2a)$, and the diameter $eh$ perpendicular to it will transform into the segment $EH$ $(EH=2ka)$. Let us denote the length of the segment $OE$ $(=OH)$ by $b$. Then $ka=b$.

The segments $CD$ and $EH$ are called the axes of the ellipse; the numbers $a$ and $b$ will be called the semi-axes of the elipis. If $k>1$ (stretch), then $b>a$ (Fig. 5).

Fig. 5. Elipis - circumferential deformation

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Fig. 6. This way you can build an ellipse.

If $k < 1$ (compression), then $b < a$ (Fig. 6). With $k = 1$ (identity transformation) the circle turns into a circle. In order not to violate the generality of the definition of an ellipse, we agree to consider a circle to be a special case of an ellipse ($b = a$). The larger of the segments $CD$, $EH$ is called the major axis of the ellipse, the smaller - the minor axis. So, if $k > 1$, then $a$ is the semiminor axis and $b$ is the major one, and if $k < 1$, then vice versa.

In both cases $k = \frac{b}{a}$.

Point $o$—the center of the circle—will transform into itself ($o \rightarrow O$) during deformation. O is called the center of the ellipse, the points $C$, $D$, $E$ and $H$ are its vertices.

The square $pqrs$, circumscribed around the circle (the “mother” of the ellipse), one side of which is parallel and the other perpendicular to the deformation axis, is transformed into a rectangle $PQRS$ from the sides $2a$ and $2b$, circumscribed around the ellipse; it is called characteristic of our ellipse, since it completely determines it. The sides of the Characteristic rectangle touch the ellipse at its vertices.

How to make an ellipse thicker?

If the axes of the ellipse are given, then the easiest way is to measure the ellipse by hand by first constructing a characteristic rectangle and then inscribing an oval curve into it. But this method, with little practice, can confidently draw an almost real ellipse). We already know that there are many oval curves with the same axes. You can accurately construct how many corner points of the ellipse. The most common and simple way of such construction is shown in figure 6.

On this figure two concentric circles have radii $a$ and $b$. Figure out this construction yourself and prove its legitimacy, that is, prove that on figure $6$ $m_0M = m_0m \cdot \frac{b}{a}$.

By making this construction for several points $m_1$, $m_2$, $m_3$, … of a circle of radius $a$, we can get how many corner points of our ellipse. All that remains is to connect them with a smooth line by hand or using a pattern.

But it is interesting to measure an ellipse with a continuous movement, that is, without lifting the pencil from the paper, just as we draw a circle with a compass. This can be done with a special device - an ellipsoid. We will talk about it below when we get acquainted with the equation of the ellipse. But first we need to find out what an equation of lines is.

Equation of lines *)

One of the replacement methods discovered in the 17th century is the coordinate method, which allows one to translate geometric concepts into algebraic language. This method makes it possible to establish a correspondence between lines lying in the plane on which the coordinate system is installed and equations containing two letters $x$ and $y$.

An equation of a certain line is such an equation for $x$ and $y$, which turns to the same place when the values ​​$(x_0, y_0)$ are substituted into it in that and only in that

*) You can read more about the equation of lines in any textbook on analytical geometry.

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Fig. 7. $m(x_{окр}, y_{окр})$ is a point on a circle, $M(x_0, y_0)$ is a point on an ellipse.

case when the point with coordinates $(x_0, y_0)$ lies on this line.

We can also say that a line defined by an equation is the locus of points whose coordinates satisfy this equation.

For example, the equation $y=x$ is the equation of the bisector of the I and III coordinate angles, and the equation of a circle of radius $a$ with the center at the origin will be written as follows: $x^2+y^2=a^2$.

It can be easily obtained from the Pythagorean theorem.

Ellipse equation

An ellipse with semi-axes $a$ and $b$ is obtained from a circle of radius $a$ by deformation with a coefficient $k=\frac{b}{a}$ (Fig. 7). Let us establish coordinate axes on the plane where both of these lines are located; Let’s direct the $OX$ axis along the deformation axis, and the $OY$ axis through the center of the circle (also the center of the ellipse) perpendicular to the $OX$ axis. Let $m$ be any point on the circle, and $M$ be the point of the ellipse at which $m$ passes.

the coordinates of the point $m$ are denoted by $(x_{окр}, y_{окр})$, and the coordinates of the point $M$ by $(x_0, y_0)$. Obviously, by definition of deformation

$$x_0=x_{окр}, y_0=k \cdot y_{окр},$$

(*)

Fig. 8. Point $M$ describes an ellipse.

where

$$x_{окр}=x_0, y_{окр}=\frac{y_0}{k}.$$

If we substitute the coordinates of any of its points $(x_{окр}, y_{окр})$ into the equation of the circle $x^2+y^2=a^2$, then the equality will be true

$$x^2_{окр}+y^2_{окр}=a^2,$$

which means, replacing $x_{окр}$ and $y_{окр}$ according to formula (*), we see that the equality will also be true

$$x^2_{9}+\frac{y^2_{9}}{k^2}=a^2$$

for any point $M$ $(x_0, y_0)$.

If the point $M'(X', Y')$ does not lie on the ellipse, then the point $m'(x', y')$, which fell into $M'$ as a result of deformation, did not lie on the circle, and the equation $(x')^2+(y')^2=a^2$ is incorrect; This means that equation $(X')^2+\frac{(Y')^2}{k^2}=a^2$ is also false.

From the definition of the equation of a line we obtain the following equation of the ellipse

$$x^2+\frac{y^2}{k^2}=a^2.$$

Noticing that $ka=b$, we can reduce this equation to the form

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.$$

This is the equation of the ellipse in the most convenient (so-called canonical) form.

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Let us consider the following geometric problem (Fig. 8).

The segment $PQ$ moves in a plane so that one of its ends $Q$ slides along the straight line $OX$, and the other $P$ slides along the straight line $OY$ perpendicular to it. What line does the point $M$ of this segment $(MP=a, MQ=b)$ cover?

Let us prove that this curve is an ellipse. Let’s apply the coordinate method (Fig. 9) - for the coordinate axes used along which the segment slides. Let the coordinates of the point be $M-x$ and $y$ (they depend on the angle $PQO$). From the similarity of triangles $QNM$ and $MLP$ we have $\frac{NM}{LP} = \frac{QM}{MP}$, or $\frac{y}{\sqrt{a^2-x^2}} = \frac{b}{a}$.

Equation (1) cannot yet be called the equation of the desired trajectory: it is compiled in relation to figure 9, where the segment lies in the quarter and both coordinates of the point $M$ are positive. Only under this condition can we assume that $ON=x$, and $NM=y$; if the segment is in the II, III or IV quarter at figure 8, then you need to count $ON=\pm x$, $NM=\pm y$, choosing the necessary signs each time.

Fig. 10. Ellipsograph.

However, equation (1) is also suitable for the second quarter, since $x$ is squared and changing the sign of $x$ will not change this equation. For III and IV quarters, equation (1) should be replaced with this:

$$-\frac{y}{\sqrt{a^2-x^2}} = \frac{b}{a}.$$

The pair of equations (1) and (2) can be replaced by a single equation

$$\frac{y^2}{a^2-x^2} = \frac{b^2}{a^2},$$

which is easily reduced to the form

$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$

The required curve is an ellipse.

The design of the ellipsograph is based on the result obtained. A cross is placed on the paper or drawing board, in which there are two mutually perpendicular slots (Fig. 10). Three couplings are connected to the straight bar, which can be secured at any point on the bar. The outer muffs $P$ and $Q$ can go in the slots of the cross, and the middle one $M$ is equipped with a pencil, which, when moving the bar, draws an ellipse. Having secured the muffs along the given half-axes $PM=a$, $MQ=b$, we can draw an ellipse with these half-axes.

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Inclined sections of a circular cylinder - ellipses

By intersecting a round cylinder with a plane that is neither perpendicular nor parallel to its axis, we obtain more or less elongated ovals, depending on the angle of inclination of the section. This means that every housewife cuts the sausage into slices for guests and often (to “save”) making the corner sharp enough so that the oval has a larger area.

Let us prove that these sections are not just ovals, but real ellipses. To do this, it is enough to establish that the oval is obtained from a circle (the base of the cylinder) by stretching; coefficient $k$ is equal to $\frac{1}{\cos \alpha}$, where $\alpha$ is the angle between the planes of the section and the base of the cylinder.

Let us draw (Fig. 11) at the base of the cylinder two mutually perpendicular diameters: $cd$ - parallel and $eh$ - perpendicular to the line of intersection of both planes, and draw through them the planes $cdDC$ and $ehHE$, parallel to the axis of the cylinder, as well as the plane $M_0Mmm_0$, parallel to $ehHE$ passing through the point $M$ of the section. Knowing that the length of the projection of a segment onto a certain line is equal to the length of the segment itself,

multiplied by the cosine of the angle between it and its projection, we get

$$\frac{M_0M}{m_0m} = \frac{OE}{oe} = \frac{b}{a} = \frac{1}{\cos \alpha} = k.$$

If we now put both figures $CEDH$ and $cedh$ on the same plane so that $CD$ is aligned with $cd$, then they will take the same position as on figure 5, with $M_0M = m_0m \cdot \frac{1}{\cos \alpha} = m_0m \cdot k$, where $k > 1$.

This proves our statement.

It is interesting that for each ellipse you can specify a position such that it turns out to be a cross section of a round cylinder, the radius of which is equal to the small area of ​​the ellipse. To do this, you first need to take a perpendicular section of the cylinder and rotate the plane of this section around one of the diameters of the resulting circle by an angle whose cosine is equal to $\frac{a}{b}$.

This is always possible since here.

Mating ellipse diameters

An interesting property of the ellipse immediately follows from the last result: the midpoints of all parallel chords of the ellipse form a straight segment passing through the center of the ellipse (Fig. 12). This property is very indo

36

Fig. 17. This is how an ellipse is obtained in practice.

equality $MF_1 + MF_2 = const$.

Thus, the sum of the distances from any point of the ellipse to its two foci is a constant value.

This is the focal property of the ellipse. An ellipse is often defined as a “geometric location” (in modern language - as a set) of points lying in the same plane for which the sum of the distances to two points (foci) is a constant value.

If we choose the end of its major axis as the point $M$ of the ellipse (for example, the point $C$ on figure 15), then by the focal property $CF_1 + CF_2 = const$. Due to symmetry $F_1C = F_2D$; this means that this constant is equal to $CF_2 + F_2D$, that is, the major axis $CD = 2a$. If we take the end of the minor axis, for example, the point $E$, as the point $M$, then $EF_1 + EF_2 = 2a$, and since due to symmetry $EF_1 = EF_2$, then each of these segments is equal to $a$. This proves the statement made at the beginning of this section, and provides a way to construct the foci of an ellipse along its axes (Fig. 15).

Ellipsograph for gardeners

The well-known construction of an ellipse by continuous movement using a thread ring is based on the focal property of an ellipse. Let’s put a sheet of paper on a lined board and stick two canile pins on it. Having thrown a thread ring onto both cops (the perimeter of the ring should be greater than twice the distance between the cops), pull the thread with a sharp pencil so that you get a triangle ($\Delta F_1F_2M$). If you move the pencil along the paper so that the thread remains taut, the triangle will be deformed, but its perimeter remains constant. The length of the side $F_1F_2$ also remains constant, which means $F_1M + F_2M = const$ and the pencil will describe the ellipse.

We have a new, very simple ellipsograph. This method has been successfully used with elliptical flower terminals. Instead of kopki, two stakes are driven into the ground, a ring is made of a thick rope, and an ellipse is drawn on the ground not with a pencil, but with a stick (Fig. 17).

We did not explain the origin of the name “ellipse” (in Greek “ellipsis” means “disadvantage” - what does the disadvantage have to do with it?), We did not talk about many other very important properties of this curve - properties common to all finite sections: ellipse, hyperbola and parabola. These issues will be covered in future issues «Kvant».

Spatial figure in this stat only give a general idea of ​​their day. When performing figure, preference was given to clarity, rather than building a correct spatial figure on a plane *).

The laws of representation of spatial figures on a plane will be discussed in one of the upcoming issues of the magazine.

*) Try to prove that the hyperbole given on page 26 is, strictly speaking, incorrectly depicted.

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LABORATORY “QUANTA”
BALL
INSTEAD OF
LENSES
G. I. Kosourov

In this and subsequent issues, the section “Kvant’s Problem” is composed of problems proposed at the All-Union Optimization in Physics and Mathematics in 1970. After each problem, the class in which it was offered is indicated in parentheses.

F48. A rope is wound around the armature shaft of the dynamo, to which a weight is attached. As the weight descends, it rotates the anchor. When the armature was sufficiently untwisted, a load resistance was connected to the terminals of the machine. Draw on a graph the dependence of the speed of rotation of the anchor on the time from the moment the load begins to move. (X)

G. I. Kosourov

F49. The U-shaped tube is filled with water. Air has been removed from one elbow; the air pressure in the other leg at temperature $t = 20^\circ\text{C}$ is equal to atmospheric pressure. Both ends of the tube are sealed. The difference between the water levels in the elbows is 15 m. What will be the difference in water levels in the elbows if the tube is heated to $100^\circ\text{C}$? (IX-X)

B. B. Bukhovtsev

F50. On a smooth horizontal table lie two identical cubes with masses $m$, connected by a spring of stiffness $k$ and length $l_0$ in an unstretched state. A force $F$, constant in magnitude and direction, will suddenly begin to act on the left cube (Fig. 1). Find the minimum and maximum distance

Fig. 1.

Fig. 2.

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M6. There is a clock in front of you. How many hand positions are there that cannot be used to determine the time unless you know which hand is the hour hand and which hand is the minute hand?
(It is believed that the position of each of the arrows can be determined accurately, but following how the arrows move cannot.)

Answer. There are 132 such positions. In other words, there are 66 pairs of hand arrangements such that in each of these pairs one arrangement is obtained from the other by replacing the hour hand with the minute hand and the minute hand with the hour hand (as with Fig. 1); Of course, such positions are not considered when the directions of the hour and minute hands coincide: in these cases, the time is determined unambiguously.

Solution and e. We got many different solutions to this problem. Let’s list some of them. Here’s the most common one.

Let’s assume that the dial is divided into 12 parts. Let’s consider one of the positions of the arrows that are discussed in the conditions. Let one arrow occupy the position $x$, the other - $y$; here $0 \leq x < 12$, $0 \leq y < 12$. If the first hand is considered to be the hour hand, and the second hand is considered to be the minute hand, then the equality must be fulfilled

$$x = a + \frac{5y}{60},$$

(1)

Fig. 1.

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Fig. 2.

where $a$ is an integer number of hours, $0 \leqslant a \leqslant 11$.
Similarly, if the first hand is the minute hand and the second hand is the hour hand, then
$$y = b + \frac{5x}{60},$$

where $b$ is an integer, $0 \leqslant b \leqslant 11$.
The system of two equalities (1) and (2) can be replaced by equivalent ones:

$$\begin{cases} 12x = 12a + y, \\ 12y = 12b + x, \\ 143x = 12(b + 12a), \end{cases}$$

$$\begin{cases} x = 12 \cdot \frac{b + 12a}{143}, \\ y = 12 \cdot \frac{a + 12b}{143}. \end{cases}$$

Here $0 \leqslant a \leqslant 11, 0 \leqslant b \leqslant 11$ are integers. Every pair $(x, y)$ satisfying this system is a solution to it. There are only $12^2 = 144$. It is clear that the case when $x = y$, that is, when $a = b$, must be excluded. 132 remain.

Two of them - the corresponding $a = 2, b = 5$ and $a = 5, b = 2$ - are depicted on figure 1 (it is clear that the replacement of $a$ by $b$ and $b$ by $a$ corresponds to the replacement of arrows).

A very clear solution was sent by 10th grade student S. Pozdnyakov (e. Vilnius). Here’s his reasoning.

Let us set the position of the minute hand $y$ as a function of the position of the hour hand $x$. The graph of this function $x \rightarrow y$ is shown in figure 2 in blue. This graph is the set of all positions of the hands $(x, y)$ that can be found on the dial of a watch. On the area of ​​interest there are pairs $(x, y)$ for which the pair $(y, x)$ also belongs to this set. But the point $(y, x)$ is a point symmetrical to $(x, y)$ with respect to the bisector of the angle $xOy$. On figure 2 the set of points symmetrical to the systems of points is marked in red *).

It remains to find the number of intersections of the blue and red segments (not counting the points lying on the bisector $x = y$). It is easy to see that there are 132 of them, i.e. 66 pairs of points, symmetrical with respect to the bisector.

And finally, one more thing, no mathematical apparatus is required.

Let’s put next to our clock (on the right) another, imaginary one, which runs exactly 12 times faster. Let both clocks be at the same time when they show 12 o’clock; then the hour hand of the right clock always coincides with the minute hand of the left one. It is clear that the “indistinguishable” positions of the hands that interest us are exactly those when the hour hand of the left coincides with the minute hand of the right, fast (think why!). How many times will this happen? Of the 144 revolutions that the minute hand of the right watch will make during the time that the hour hand of the “normal” watch makes one revolution,

*) The author writes that the “red” set is “the graph of the function inverse to our function $x \rightarrow y$”. This is not entirely true: it is clear that our function does not have an inverse on the entire segment $0 \leqslant x \leqslant 12$, since here each $y$ corresponds to many (namely twelve) different $x$. One could say this: we will consider the function $x \rightarrow y$ on each segment $0 \leqslant x < 1, 1 \leqslant x < 2, \dots, 11 \leqslant x < 12$, take the inverse of each of these functions and construct a graph for each of them. Then we will really get a “red” set consisting of devenaudami graphs! (For the definition of the inverse function, see «Kvant» No. 1, pp. 32-33.)

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Each rotation there will be one match (including the starting point of the first rotation); Of these, we need to exclude 12 cases when all four arrows coincide, leaving 132.

M7. a, b, c are the sides of the triangle.

Prove that $\frac{a}{b+c-a} + \frac{b}{c+a-b} + \frac{c}{a+b-c} \geq 3$.

Solution. This problem has quite a few solutions. We will give one of the shortest ones. We only need the inequality well known from the school course
$$\frac{m}{n} + \frac{n}{m} \geq 2,$$
(1)

where $m > 0, n > 0$.

Let us introduce the following notation:
$$b + c - a = x,$$
$$c + a - b = y,$$
$$a + b - c = z.$$

Note that
$$x > 0, y > 0, z > 0;$$
(3)
this follows from the fact that $a, b, c$ are the sides of the triangle, and each side is less than the sum of the other two (unfortunately, from the letters sent to the editor, it is clear that most of those who solved the problem did not even think about where they use the fact that $a, b, c$ are the sides of the triangle).

From (2) it follows that
$$a = \frac{y+z}{2}, \quad b = \frac{x+z}{2}, \quad c = \frac{x+y}{2};$$

Posibly, the original inequality can be rewritten as follows:
$$\frac{y+z}{2x} + \frac{x+z}{2y} + \frac{x+y}{2z} \geq 3.$$

Transforming the left-hand side of (4) and using (1) and (3), we obtain
$$\frac{y+z}{2x} + \frac{x+z}{2y} + \frac{x+y}{2z} = = \frac{1}{2} \left[ \left( \frac{y}{x} + \frac{x}{y} \right) + \left( \frac{z}{x} + \frac{x}{z} \right) + \left( \frac{z}{y} + \frac{y}{z} \right) \right] \geq \frac{1}{2} (2 + 2 + 2) = 3,$$
which is what was required.

Note that a more general inequality can be proven similarly:
$$\frac{x_1}{x_2 + x_3 + \dots + x_n - x_1} + \frac{x_2}{x_3 + x_4 + \dots + x_n + x_1 - x_2} + \dots + \frac{x_{n-1}}{x_n + x_1 + x_2 + \dots + x_{n-2} - x_{n-1}} + \frac{x_n}{x_1 + x_2 + \dots + x_{n-1} - x_n} \geq \frac{n}{n-2},$$

where all the denominators are positive.

In conclusion, we note that one geometric fact directly follows from the proven inequality. Let $R$ be the radius of the circumscribed circle of the triangle, and $r$ the radius of the circle inscribed in it. It’s good to know that
$$R = \frac{abc}{4S}, \quad r = \frac{S}{p},$$
(5)

$$S = \sqrt{p(p-a)(p-b)(p-c)},$$

where $S$ is the area of ​​the triangle, and $p$ is its half-perimeter. From (2) it follows that
$$p - a = \frac{x}{2},$$
$$p - b = \frac{y}{2},$$
$$p - c = \frac{z}{2}.$$

Now, using (5) and (6), we get
$$\frac{R}{2r} = \frac{abcp}{8S^2} = \frac{abcp}{8p(p-a)(p-b)(p-c)} = = \frac{abc}{(a+b-c)(b+c-a)(c+a-b)} = = \frac{(x+y)(y+z)(z+x)}{8xyz} = = \frac{1}{4} \left[ 1 + \left( \frac{y+z}{2x} + \frac{x+z}{2y} + \frac{x+y}{2z} \right) \right].$$

Note that the left side of inequality (4) is in parentheses on the right side of the last equality. Using it, we get
$$\frac{R}{2r} \geq 1,$$
that is, $R \geq 2r$.

So, the radius of the circle circumscribed about the triangle is not less than the diameter of the circle inscribed in it. Try to prove this geometric number (which is much more difficult).

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Fig. 3

F7. A horizontal rod $O_1A$ of length $l$ rotates around a vertical axis $O_1$ (Fig. 3). A wheel of radius $r$ is mounted on the axis attached to the end of the rod $A$. The wheel axis is horizontal and makes an angle $\alpha$ with the rod $O_1A$. The wheel rotates on its axle without friction and rolls along the ground. There is more trenne between the wheel and the soil. How many revolutions will the wheel make when the rod $O_1A$ makes one revolution around the vertical axis?

Solution. The solution problem is easiest if you reason as an observable sitting at the end $A$ of the rod $O_1A$ would reason. If the angular velocity of rotation of the rod $O_1A$ relative to the ground is equal to $\Omega$, then the observer will say that in his coordinate system the rod is stationary, and the earth rotates around the axis $O_1$ with an angular velocity equal to $\Omega$, in the direction opposite to the direction of rotation of the rod relative to the earth. The speed of the point of the earth at which the wheel touches it is equal in magnitude to $v = \Omega l$ and is directed perpendicular to the rod $O_1A$. Let us decompose this speed into two components: $v_1 = v \cos \alpha$ and $v_2 = v \sin \alpha$, corresponding to the parallel and perpendicular planes of the wheel. It is clear that since the friction of the wheel on the soil is high, the point of the wheel at which it touches the ground moves relative to the point $A$, and therefore the axis of the wheel, at a speed $v_1$. This means that the wheel rotates around its axis with an angular velocity of $\omega = \frac{v_1}{r} = \Omega \frac{l}{r} \cos \alpha$.

Fig. 4.

The ratio of the rotation angles of the rod and the wheel around their axes is equal to the relative angular velocities of rotation of the rod and the wheel. Therefore, when the rod $O_1A$ makes one revolution around the axis $O_1$, the wheel will make $n = \frac{\omega}{\Omega} = \frac{l}{r} \cos \alpha$ revolutions around its axis.

F8. A long rod $AB$ with a threaded hole at the end is screwed onto a vertical screw (Fig. 4). The rod is released. The trenne between the screw and the rod is negligible. How will the rod move after it comes off the screw?

Solution. At the moment when the rod flies off the screw, the speed of its center of mass (center of gravity) will have both vertical and horizontal components. Since the only force acting on the rod from this moment will be the merging of gravity, the center of mass of the rod will move along a parabola - the trajectory of a body thrown at an angle to the horizon with an initial speed equal to the speed of movement of the center of mass of the rod at the moment when it flew off the propeller. But at the moment when the rod flew off the screw, different points of the rod had different horizontal velocities. Therefore, the rod will also rotate in a horizontal plane around the center of mass.

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F9. On the horizontal table there is a weight attached to the table using a long spring. At first the spring was not stretched. Then the weight was moved 20 cm from the equilibrium position and released. The weight began to oscillate along the spring. Due to friction, the amplitude of its oscillations over the period decreased by 7%. How many oscillations will the weight make before stopping? At what distance from the equilibrium position will it stop?

Solution. If a constant external force acts on the oscillating mat, then it will change the equilibrium position of the mat. For example, if we compare the oscillations of a load attached to a spring and moving without friction in a horizontal plane with the oscillations of the same system in a vertical plane, when the load along the line of its motion is acted upon by gravity, then we will see that in the second case the equilibrium position of the load will be further from the fixed end of the spring by a distance of $x = \frac{mg}{k}$, where $k$ is the spring stiffness coefficient, and $m$ is the mass cargo In this case, the period of oscillation of the load will be the same in both cases.

The action of a constant external force on the matnik does not change the period of oscillation of the matnik and does not cause damping of its oscillations, since the position of the action of the external force reduces the energy of the matnik, and the second positive period increases the energy of the matnik by the same amount. It’s another matter when, as in our case, the force acting on the mat is constant only during each period, and through the polariod in the opposite direction. In this case, the oscillations of the mat are damped, since when the load moves, the force acting on it is always directed in the direction opposite to the direction of its movement. Moreover, every polariod of oscillation of the load occurs around different equilibrium positions. When the load moves, the equilibrium position is expressed - this is the point $A$ (Fig. 5), since the friction force acting on the load is directed

Fig. 5.

to the left (compare with the oscillations of a load suspended on a spring), and when moving to the left - point $B$.

It is clear that points $A$ and $B$ are at the same distance from point $O$—the position of the load with an undeformed spring. This distance is found from the condition that the spring tension force $F = k \Delta x$ ($\Delta x$ is the spring deformation) and the friction force acting on the load are equal: $k \Delta x = F_{\text{тр}}$. From here

$$a = AO = OB = \Delta x = \frac{F_{\text{тр}}}{k}.$$

The oscillations of the load cannot be limited to the area between points $A$ and $B$, since in this area the spring tension force does not exceed the maximum possible friction force. Therefore, if the speed of the load at every point in this area turns out to be zero, then the load will remain: the tension force of the spring will be balanced by the friction force.

Let us denote the amplitude of the $n$-th oscillation of the load (the deviation of the load from the point $O$) by the letter $y_n$, if the load is deflected to the right from the point $O$, and $x_n$, if the load is deflected to the left from the point $O$. $y_0 = 20$ cm is the initial deflection of the load. According to conditions

$$\frac{y_0 - y_1}{y_0} = 0.07.$$

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Let us find the law of change in the amplitude of oscillations of the load. The change in the energy of the system when the load moves is equal to the work done by the friction force, therefore, using the formula for the energy of the deformation spring $W = \frac{k(\Delta x)^2}{2}$, we can write that

$$\frac{k y_n^2}{2} - \frac{k x_n^2}{2} = F_{\text{тр}}(y_n + x_n)$$

And

$$\frac{k x_n^2}{2} - \frac{k y_n^2}{2} = F_{\text{тр}}(x_n + y_n+1).$$

Dividing both sides of the first equation by $y_n + x_n$, we find that $y_n - x_n = 2 \frac{F_{\text{тр}}}{k}$. Similarly, from the second equation we obtain that $x_n - y_n+1 = 2 \frac{F_{\text{тр}}}{k}$. Adding the resulting equations, we find that

$$y_n - y_n+1 = 4 \frac{F_{\text{тр}}}{k}$$

(9)

But $\frac{F_{\text{тр}}}{k} = a$, therefore $y_0 - y_1 = y_1 - y_2 = \dots = y_n - y_{n+1} = 4 a$. (10)

The amplitude of the load oscillations during the period decreases by the same amount, equal to 4a. Substituting into formula (7) instead of the difference $y_0 - y_1$ the equal difference $y_n - y_{n+1}$, we obtain that $\frac{y_n - y_{n+1}}{y_0} = 0,07$, or that over the period the amplitude of the load’s oscillations decreases by $0,07 y_0 = 1,4$ cm. After $n$ vibrations it becomes equal to $y_n = y_0 - n \cdot 0,07 y_0$. In this case, the amplitude of oscillations of the load, as we have already said, cannot be less than $a$. Since, according to formulas (7) and (10), $4a = y_0 - y_1 = 0,07 y_0 = 1,4$ cm, then $a = 0,35$ cm.

From the condition $y_n \geq a$ or $20(1 - 0,07n) \geq 0,35$ we obtain that $n \leq 14,04$. $n$ is an integer. Therefore, the load will make 14 complete oscillations.

Since, $y_{14} = 20(1 - 0,07 \cdot 14) = 0,4$ cm, that is, $y_{14} > a$, and the load is outside the area between points $A$ and $B$. Therefore, the tension force acting on the load is greater than the friction force, and the load will perform another part of the oscillation, stopping at a distance $y$ from the point $O$. Moreover, the load cannot reach point $O$. When moving to the left, its equilibrium position is point $B$, and the load cannot deviate to point $B$ by a distance greater than $y_{14} - a \approx 0,5$ cm.

Let the distance traveled by the load before stopping be equal to $y_{14} - y$. The energy of the system at the point where the load stops is $\frac{k y_n^2}{2}$. Equating the energy difference between the system of work done by the friction force, we obtain equation $\frac{k y_{14}^2}{2} - \frac{k y_n^2}{2} = F_{\text{тр}}(y_{14} - y)$. Dividing this equation by the non-zero difference $y_{14} - y$, we obtain that $ky_{14} + ky = 2F_{\text{тр}}$. Hence $y = 2 \frac{F_{\text{тр}}}{k} - y_{14} = 2a - y_{14} = 0,3$ cm.

F10. How to construct a heater of the highest possible power from four thin wire spirals with resistances $10 \text{ om}$, $20 \text{ om}$, $30 \text{ om}$ and $40 \text{ om}$, designed to release a power of no more than $2 \text{ om}$ on each, if there is a current source with e. d.s. $20 \text{ b}$ and internal resistance $20 \text{ om}$?

Solved.e. Let us first prove that the greater the power released in the external circuit, the closer the load resistance is to the internal resistance of the source.

Current in a circuit consisting of a source connected in series with e. d.s. $\varepsilon$ and internal resistance $r$ and load with resistance $R$, is equal to $\frac{\varepsilon}{R + r}$. Therefore, the power allocated to the load is $W = \frac{\varepsilon^2}{(R + r)^2} R$.

Let’s draw a graph of the dependence of $W$ on the load resistance $R$. At $R = 0$, $W = 0$; with $R \rightarrow \infty$ you can beat $r$ compared to $R$ in a significant matter. Then $W \approx \frac{\varepsilon^2 R}{R^2} = \frac{\varepsilon^2}{R}$.

That is, for large $R$ ($R \gg r$), the graph of the dependence of $W$ on $R$ is a hyperbola, and the symptoms are approaching the $R$ axis. Since the function $W(R)$ is continuous, equal to zero at $R = 0$ and decreasing at $R \rightarrow \infty$, it must have a maximum. Let’s find him.