A plane wave is a wave with a flat front. In this case, the rays are parallel.

A plane wave is excited in the vicinity of an oscillating plane or if a small section of the wavefront of a point source is considered. The area of ​​this area can be the larger, the farther it is from the emitter.

The rays covering the section of the plane of the considered wave front form a "pipe". The amplitude of sound pressure in a plane wave does not decrease with distance from the source, since there is no spreading of energy outside the walls of this pipe. In practice, this corresponds to highly directional radiation, such as radiation from electrostatic panels. large area, horn emitters.

Signals in various points beams of a plane wave differ in the phase of the oscillations. If the sound pressure in a certain section of the plane wave front is sinusoidal, then it can be represented in an exponential form p zv = p tv-exp (icot). On distance G along the beam, it will lag behind the source of oscillations:

where g/s sv is the time it takes a wave to travel from a source to a point at a distance G along the beam k \u003d (o / c zb \u003d 2g/D - wave number, which determines the phase shift between the signals in the fronts of a plane wave, located at a distance G.

Real sound waves more complex than sinusoidal, however, the calculations carried out for sinusoidal waves are also valid for non-sinusoidal signals, if the frequency is not considered as a constant, i.e. consider a complex signal in the frequency domain. This is possible as long as the wave propagation processes remain linear.

A wave whose front is a sphere is called a spherical wave. The rays in this case coincide with the radii of the sphere. A spherical wave is formed in two cases.

• 1. The dimensions of the source are much smaller than the wavelength, and the distance to the source allows us to consider it as a point. Such a source is called a point source.
• 2. The source is a pulsating sphere.

In both cases, it is assumed that there are no re-reflections of the wave, i.e. only the direct wave is considered. There are no purely spherical waves in the field of interest of electroacoustics; this is the same abstraction as a plane wave. In the region of medium-high frequencies, the configuration and dimensions of the sources do not allow us to consider them as either a point or a sphere. And in the low-frequency region, at least the floor begins to have a direct influence. The only wave that is close to spherical is formed in a dampened chamber with small dimensions of the emitter. But consideration of this abstraction makes it possible to understand some important aspects of the propagation of sound waves.

At large distances from the emitter, the spherical wave degenerates into a plane wave.

On distance G from the emitter, the sound pressure can be

presented in the form r sv= -^-exp(/ (co? t - to? G)), where p-Jr- amplitude

sound pressure at a distance of 1 m from the center of the sphere. The decrease in sound pressure with distance from the center of the sphere is associated with the spreading of power over an ever larger area - 4 pg 2 . The total power flowing through the entire area of ​​the wavefront does not change, so the power per unit area decreases in proportion to the square of the distance. And the pressure is proportional to the square root of the power, so it decreases in proportion to the actual distance. The need for normalization to pressure at some fixed distance (1 mV this case) is related to the same fact that pressure depends on distance, only in the opposite direction - with an unlimited approach to a point radiator, the sound pressure (as well as the vibrational velocity and displacement of molecules) increases without limit.

The vibrational velocity of molecules in a spherical wave can be determined from the equation of motion of the medium:

Total oscillatory speed v m = ^ sv ^ + to g? phase

/V e star kg

shift relative to sound pressure f= -arctgf ---] (Figure 9.1).

To put it simply, the presence of a phase shift between sound pressure and vibrational velocity is due to the fact that in the near zone, with distance from the center, the sound pressure decreases much faster than it lags.

Rice. 9.1. The dependence of the phase shift φ between the sound pressure R and vibrational speed v from h/c(distance along beam to wavelength)

On fig. 9.1 you can see two characteristic zones:

• 1) near g/H" 1.
• 2) far g/H" 1.

Radiation resistance sphere radius G

This means that not all power is spent on radiation, some is stored in some reactive element and then returned to the emitter. Physically, this element can be associated with the attached mass of the medium oscillating with the emitter:

It is easy to see that the added mass of the medium decreases with increasing frequency.

On fig. 9.2 shows the frequency dependence of the dimensionless coefficients of the real and imaginary components of the radiation resistance. Radiation is efficient if Re(z(r)) > Im(z(r)). For a pulsating sphere, this condition is satisfied for kg > 1.

An oscillatory process propagating in a medium in the form of a wave, the front of which is plane, is called plane sound wave. In practice, a plane wave can be formed by a source whose linear dimensions are large compared to the long waves emitted by it, and if the wave field zone is located at a sufficiently large distance from it. But this is the case in an unbounded environment. If the source fenced some obstacle, then a classic example of a plane wave is oscillations excited by a rigid inflexible piston in a long pipe (waveguide) with rigid walls, if the piston diameter is much less than the length of the radiated waves. The surface of the front in the pipe, due to the rigid walls, does not change as the wave propagates along the waveguide (see Fig. 3.3). We neglect the loss of sound energy due to absorption and scattering in the air.

If the emitter (piston) oscillates according to the harmonic law with a frequency
, and the dimensions of the piston (waveguide diameter) are much smaller than the sound wave length, then the pressure created near its surface is
. Obviously, at a distance X pressure will
, where
is the travel time of the wave from the emitter to the point x. It is more convenient to write this expression as:
, where
- wave number of wave propagation. Work
- determined phase incursion of the oscillatory process at a point remote at a distance X from the emitter.

Substituting the resulting expression into the equation of motion (3.1), we integrate the latter with respect to the vibrational velocity:

(3.8)

In general, for an arbitrary moment of time it turns out that:

. (3.9)

The right side of the expression (3.9) is the characteristic, wave, or specific acoustic resistance of the medium (impedance). Equation (3.) itself is sometimes called the acoustic "Ohm's law". As follows from the solution, the resulting equation is valid in the field of a plane wave. Pressure and vibrational speed in-phase, which is a consequence of the purely active resistance of the medium.

Example: Maximum pressure in a plane wave
Pa. Determine the amplitude of the displacement of air particles in frequency?

Solution: Since , then:

It follows from expression (3.10) that the amplitude of sound waves is very small, at least in comparison with the dimensions of the sound sources themselves.

In addition to the scalar potential, pressure and vibrational velocity, the sound field is also characterized by energy characteristics, the most important of which is intensity - the energy flux density vector carried by the wave per unit time. By definition
is the result of the product of sound pressure and vibrational velocity.

In the absence of losses in the medium, a plane wave, theoretically, can propagate without attenuation over arbitrarily large distances, since the preservation of the shape of a flat front indicates the absence of "divergence" of the wave, and, hence, the absence of attenuation. The situation is different if the wave has a curved front. Such waves include, first of all, spherical and cylindrical waves.

3.1.3. Models of waves with a non-planar front

For a spherical wave, the surface of equal phases is a sphere. The source of such a wave is also a sphere, all points of which oscillate with the same amplitudes and phases, and the center remains motionless (see Fig. 3.4, a).

A spherical wave is described by a function that is a solution of the wave equation in a spherical coordinate system for the potential of a wave propagating from a source:

. (3.11)

Acting by analogy with a plane wave, it can be shown that at distances from the sound source, the wavelengths under study are much greater:
. This means that the acoustic "Ohm's law" is also fulfilled in this case. In practical conditions, spherical waves are excited mainly by compact sources of arbitrary shape, the dimensions of which are much smaller than the length of the excited sound or ultrasonic waves. In other words, a "point" source radiates predominantly spherical waves. At large distances from the source or, as they say, in the “far” zone, a spherical wave behaves like a plane wave in relation to the sections of the wave front that are limited in size, or, as they say: “degenerates into a plane wave”. The requirements for the smallness of the area are determined not only by the frequency, but
- the difference in distances between the compared points. Note that this function
has the feature:
at
. This causes certain difficulties in the rigorous solution of diffraction problems associated with the emission and scattering of sound.

In turn, cylindrical waves (the surface of the wave front - a cylinder) are emitted by an infinitely long pulsating cylinder (see Fig. 3.4).

In the far zone, the expression for the potential function of such a source tends asymptotically to the expression:


. (3.12)

It can be shown that in this case, too, the relation
. Cylindrical waves, as well as spherical ones, in the far zone degenerate into plane waves.

The weakening of elastic waves during propagation is associated not only with a change in the curvature of the wave front (“divergence” of the wave), but also with the presence of “attenuation”, i.e. sound attenuation. Formally, the presence of damping in a medium can be described by representing the wave number as a complex
. Then, for example, for a plane pressure wave, one can obtain: R(x, t) = P Max
=
.

It can be seen that the real part of the complex wave number describes the spatial traveling wave, and the imaginary part characterizes the attenuation of the wave in amplitude. Therefore, the value of  is called the coefficient of attenuation (attenuation),  is the dimensional value (Neper/m). One "Neper" corresponds to a change in the amplitude of the wave by "e" times when the wave front moves per unit length. In the general case, attenuation is determined by absorption and scattering in the medium:  =  abs +  rass. These effects are determined by different causes and can be considered separately.

In the general case, absorption is associated with the irreversible loss of sound energy when it is converted into heat.

Scattering is associated with the reorientation of part of the energy of the incident wave to other directions that do not coincide with the incident wave.

: such a wave does not exist in nature, since the front of a plane wave begins at $-\backslash mathcal(1)$ and ends at $+\backslash mathcal(1)$ which obviously cannot be. In addition, a plane wave would carry infinite power, and it would take infinite energy to create a plane wave. A wave with a complex (real) front can be represented as a spectrum of plane waves using the Fourier transform in spatial variables.

Quasi-plane wave- a wave whose front is close to flat in a limited area. If the dimensions of the region are large enough for the problem under consideration, then the quasi-plane wave can be approximately considered as a plane wave. A wave with a complex front can be approximated by a set of local quasi-plane waves whose phase velocity vectors are normal to the real front at each of its points. Examples of sources of quasi-plane electromagnetic waves are laser, reflector and lens antennas: phase distribution electromagnetic field in a plane parallel to the aperture (radiating hole), close to uniform. As the distance from the aperture increases, the wave front takes on a complex shape.

Definition

The equation of any wave is the solution of a differential equation called wave. Wave equation for the function $A$ is written in the form

$\backslash Delta\; A(\backslash vec(r),t)\; =\; \backslash frac\; (1)\; (v^2)\; \backslash ,\; \backslash frac\; (\backslash partial^2\; A(\backslash vec(r),t))\; (\backslash partial\; t^2)$ where

• $\backslash Delta$- Laplace operator ;
• $A(\backslash vec(r),t)$- desired function;
• $r$- radius vector of the desired point;
• $v$- wave speed;
• $t$- time.

One-dimensional case

$\backslash Delta\; W\_k\; =\; \backslash cfrac\; (\backslash rho)\; (2)\; \backslash left(\backslash cfrac\; (\backslash partial\; A)\; (\backslash partial\; t)\; \backslash right)^2\; \backslash Delta\; V$ $\backslash Delta\; W\_p\; =\; \backslash cfrac\; (E)\; (2)\; \backslash left(\backslash cfrac\; (\backslash partial\; A)\; (\backslash partial\; x)\; \backslash right)^2\; \backslash Delta\; V\; =\; \backslash cfrac\; (\backslash rho\; v^2)\; (2)\; \backslash left\; (\backslash cfrac\; (\backslash partial\; A)\; (\backslash partial\; x)\; \backslash right)^2\; \backslash Delta\; V\; .$

The total energy is

$W\; =\; \backslash Delta\; W\_k\; +\; \backslash Delta\; W\_p\; =\; \backslash cfrac(\backslash rho)(2)\; \backslash bigg[\; \backslash left(\backslash cfrac\; (\backslash partial\; A)\; (\backslash partial\; t)\; \backslash right)^2\; +\; v^2\; \backslash left(\backslash \; cfrac(\backslash partial\; A)(\backslash partial\; (x))\; \backslash right)^2\; \backslash bigg]\; \backslash Delta\; V\; .$

The energy density, respectively, is equal to

$\backslash omega\; =\; \backslash cfrac\; (W)\; (\backslash Delta\; V)\; =\; \backslash cfrac(\backslash rho)(2)\; \backslash bigg[\; \backslash left(\backslash cfrac\; (\backslash partial\; A)\; (\backslash partial\; t)\; \backslash right)^2\; +\; v^2\; \backslash left(\backslash cfrac\; (\backslash partial\; A)\; (\backslash partial\; (x))\; \backslash right)^2\; \backslash bigg]\; =\; \backslash rho\; A^2\; \backslash omega^2\; \backslash sin^2\; \backslash left(\backslash omega\; t\; -\; k\; x\; +\; \backslash varphi\_0\; \backslash right)\; .$

Polarization

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Literature

• Saveliev I.V.[Part 2. Waves. Elastic waves.] // Course of general physics / Edited by L.I. Gladnev, N.A. Mikhalin, D.A. Mirtov. - 3rd ed. - M .: Nauka, 1988. - T. 2. - S. 274-315. - 496 p. - 220,000 copies.

Notes

see also

An excerpt characterizing the plane wave

- It's a pity, a pity for the young man; give me a letter.
As soon as Rostov had time to hand over the letter and tell the whole story of Denisov, quick steps with spurs pounded from the stairs and the general, moving away from him, moved to the porch. The gentlemen of the sovereign's retinue ran down the stairs and went to the horses. The landlord Ene, the same one who was in Austerlitz, brought the sovereign's horse, and on the stairs there was a slight creak of steps, which Rostov now recognized. Forgetting the danger of being recognized, Rostov moved with several curious residents to the very porch and again, after two years, he saw the same features he adored, the same face, the same look, the same gait, the same combination of greatness and meekness ... And a feeling of delight and love for the sovereign with the same strength resurrected in the soul of Rostov. The sovereign in the Preobrazhensky uniform, in white leggings and high boots, with a star that Rostov did not know (it was legion d "honneur) [star of the Legion of Honor] went out onto the porch, holding his hat under his arm and putting on a glove. He stopped, looking around and that's all illuminating his surroundings with his gaze. He said a few words to some of the generals. He also recognized the former head of the division Rostov, smiled at him and called him to him.
The whole retinue retreated, and Rostov saw how this general said something to the sovereign for quite some time.
The emperor said a few words to him and took a step to approach the horse. Again a crowd of retinues and a crowd of the street, in which Rostov was, moved closer to the sovereign. Stopping by the horse and holding the saddle with his hand, the emperor turned to the cavalry general and spoke loudly, obviously with a desire that everyone could hear him.
“I can’t, General, and therefore I can’t, because the law is stronger than me,” said the emperor and put his foot in the stirrup. The general bowed his head respectfully, the sovereign sat down and galloped down the street. Rostov, beside himself with delight, ran after him with the crowd.

On the square where the sovereign went, the battalion of the Preobrazhenians stood face to face on the right, the battalion of the French guards in bear hats on the left.
While the sovereign was approaching one flank of the battalions, which had made guard duty, another crowd of horsemen jumped to the opposite flank, and ahead of them Rostov recognized Napoleon. It couldn't be anyone else. He rode at a gallop in a small hat, with St. Andrew's ribbon over his shoulder, in a blue uniform open over a white camisole, on an unusually thoroughbred Arabian gray horse, on a crimson, gold embroidered saddle. Riding up to Alexander, he raised his hat, and with this movement, the cavalry eye of Rostov could not fail to notice that Napoleon was badly and not firmly sitting on his horse. The battalions shouted: Hooray and Vive l "Empereur! [Long live the Emperor!] Napoleon said something to Alexander. Both emperors got off their horses and took each other's hands. Napoleon had an unpleasantly fake smile on his face. Alexander with an affectionate expression said something to him .
Rostov did not take his eyes off, despite the trampling by the horses of the French gendarmes, besieging the crowd, followed every movement of Emperor Alexander and Bonaparte. As a surprise, he was struck by the fact that Alexander behaved as an equal with Bonaparte, and that Bonaparte was completely free, as if this closeness with the sovereign was natural and familiar to him, as an equal, he treated the Russian Tsar.
Alexander and Napoleon with a long tail of retinue approached the right flank of the Preobrazhensky battalion, right on the crowd that was standing there. The crowd unexpectedly found itself so close to the emperors that Rostov, who was standing in the front ranks of it, became afraid that they would not recognize him.
- Sire, je vous demande la permission de donner la legion d "honneur au plus brave de vos soldats, [Sir, I ask you for permission to give the Order of the Legion of Honor to the bravest of your soldiers,] - said a sharp, precise voice, finishing each letter This was said by Bonaparte, small in stature, looking directly into Alexander's eyes from below.
- A celui qui s "est le plus vaillament conduit dans cette derieniere guerre, [To the one who showed himself the most bravely during the war,]" Napoleon added, rapping out each syllable, with outrageous calmness and confidence for Rostov, looking around the ranks of the Russians stretched out in front of him soldiers, keeping everything on guard and looking motionlessly into the face of their emperor.
- Votre majeste me permettra t elle de demander l "avis du colonel? [Your Majesty will allow me to ask the colonel's opinion?] - Alexander said and took a few hasty steps towards Prince Kozlovsky, the battalion commander. Meanwhile, Bonaparte began to take off his white glove, small hand and tearing it, he threw it in. The adjutant, hastily rushing forward from behind, picked it up.
- To whom to give? - not loudly, in Russian, Emperor Alexander asked Kozlovsky.
- Whom do you order, Your Majesty? The sovereign grimaced with displeasure and, looking around, said:
“Yes, you have to answer him.
Kozlovsky looked back at the ranks with a resolute look, and in this look captured Rostov as well.
“Is it not me?” thought Rostov.
- Lazarev! the colonel commanded, frowning; and the first-ranking soldier, Lazarev, briskly stepped forward.
– Where are you? Stop here! - voices whispered to Lazarev, who did not know where to go. Lazarev stopped, glancing fearfully at the colonel, and his face twitched, as happens with soldiers called to the front.
Napoleon slightly turned his head back and pulled back his small plump hand, as if wanting to take something. The faces of his retinue, guessing at the same moment what was the matter, fussed, whispered, passing something to one another, and the page, the same one whom Rostov had seen yesterday at Boris, ran forward and respectfully leaned over the outstretched hand and did not make her wait for a single moment. one second, put an order on a red ribbon into it. Napoleon, without looking, squeezed two fingers. The Order found itself between them. Napoleon approached Lazarev, who, rolling his eyes, stubbornly continued to look only at his sovereign, and looked back at Emperor Alexander, showing by this that what he was doing now, he was doing for his ally. Small white hand with the order touched the button of the soldier Lazarev. It was as if Napoleon knew that in order for this soldier to be happy, rewarded and distinguished from everyone else in the world forever, it was only necessary that Napoleon’s hand deign to touch the soldier’s chest. Napoleon only put the cross on Lazarev's chest and, letting go of his hand, turned to Alexander, as if he knew that the cross should stick to Lazarev's chest. The cross really stuck.

plane wave is a wave whose front is a plane. Recall that the front is an equiphase surface, i.e. surface of equal phases.

We accept that at point O (Fig. 5.1) there is a point source, a plane R perpendicular to the Z axis, points M j and M 2 lie in a plane R. We also accept that the source O is so far from the plane R, what omj | | OM 2 . This means that all points in the plane R, which is the wave front, are equal, i.e. when moving in a plane R there is no process state change:

Rice. 5.1.

Let's solve the Helmholtz equations

with respect to the field vectors and study the resulting solutions.

In this case, out of six equations, only two equations remain:

Plane waves in a vacuum

Solution differential equations(5.1) has the form

where are the roots of the characteristic equation

Passing from complex vectors to their instantaneous values, we obtain

The first term is the forward wave, and the second is the backward wave. Consider the first term in equation (5.2). On fig. 5.2 in accordance with this equation shows the distribution of tension electric field at time t and At. Points 1 and 2 correspond to the maxima of the electric field strength. The position of the maximum has shifted over time At at a distance Az:

The equality of function values ​​is ensured by the equality of arguments: ooAt = kAz. In this case, we obtain the equation for the phase velocity

Pic. 5.2. Graph of changes in electric field strength

For vacuum UV =- , C ° = -j2== 3 10 8 m/s.

W 8 oMo-o V E oMo

This means that in a vacuum the propagation velocity electromagnetic wave equal to the speed of light. Consider the second term in equation (5.2):

It gives UV =-. This corresponds to a wave propagating towards the source.

Let's define the distance X between field points with phases differing by 360°. This distance is called the wavelength. Because the

where to is the wavenumber (propagation constant), then

Vacuum wavelength X 0= c / /, where c is the speed of light.

Phase velocity and wavelength in other media, respectively

As follows from the formula for the phase velocity, it does not depend on the frequency of the electromagnetic field, which means that a lossless medium is non-dispersive.

Let us establish a connection between the directions of the vectors of the electric and magnetic fields. Let's start with Maxwell's equations:

We replace vector equations with scalar ones, i.e. equate the projections of the vectors in the last equations:

We take into account that in system (5.3)

then we get

It is obvious from condition (5.4) that plane waves have no longitudinal components, since Ez= Oh, H 2= 0. Compose the scalar product (E, R), expressing E x and E y from expressions (5.4):

Since the dot product of vectors is zero, the vectors Yo and I in a plane wave are perpendicular to each other. Due to the fact that they do not have longitudinal components, ? and I are perpendicular to the direction of propagation. Let us determine the ratio of the amplitudes of the vectors of the electric and magnetic fields.

Accept that a vector? directed along the axis X, respectively E y - 0, H X - 0.

From equation (5.4) E x=-I am at ~-E x. Hence =-=,/- -Z, soe litter Well soy v e

where Z is the wave resistance of the medium with macroscopic parameters e and p;

Z 0 - vacuum impedance. With a high degree of accuracy, this value can be considered as the wave resistance of dry air.

Let's write expressions for instantaneous values Me and? incident wave using equation (5.2). As a result, we get

likewise

As the incident wave moves along the axis z amplitude? and I remain unchanged, i.e. there is no damping of the wave, since there are no conduction currents and no energy release in the form of heat in the dielectric.

On fig. 5.3, a spatial curves are shown, which are graphs of the instantaneous values ​​of R and?. These graphs are built according to the obtained equations for the moment of time cot= 0. For a later point in time, for example for cot + |/ n = p/2, similar curves are shown in Figs. 5.3, b.

Rice. 5.3.

a- at a )t= 0; b - at u>t= n/2

As seen in fig. 5.3, a and b, vector E when the wave moves, it remains directed along the axis X, and the vector I - along the axis y, phase shift between I and? no.

The Poynting vector of the incident wave is directed along the axis z. Its modulus changes according to the law П = C 2 Z sin 2 ^cot + --zj. Because the

sin2a = (1 - cos2a)/2, to 1-cosf 2cot+-- z] , i.e. vector

2 L V v)_

Pointing has a constant component C 2 Z /2 and a time-varying variable with double the angular frequency.

Based on the analysis of the solution of wave equations, the following conclusions can be drawn.

• 1. In vacuum, plane waves propagate at the speed of light; in other media, the speed is ^/e,.p r times less.
• 2. Vectors of electric and magnetic fields have no longitudinal components and are perpendicular to each other.
• 3. The ratio of the amplitudes of the electric and magnetic fields is equal to the wave resistance of the medium in which electromagnetic waves propagate.

> Spherical and plane waves

Learn to differentiate spherical and plane waves. Read what wave is called flat or spherical, the source, the role of the wave front, the characteristic.

spherical waves arise from a point source in a spherical pattern, and flat are infinite parallel planes normal to the phase velocity vector.

Learning task

• Calculate sources of spherical and plane wave patterns.

Key Points

• Waves create constructive and destructive interference.
• Spherical ones arise from a single point source in a spherical shape.
• Flat water is frequency, the wave fronts of which act as infinite parallel planes with a stable amplitude.
• In reality, it will not work to get an ideal plane wave, but many are approaching such a state.

Terms

• Destructive interference - the waves interfere with each other, and the points do not match.
• Constructive - the waves interfere and the points are located in identical phases.
• A wave front is an imaginary surface extending through oscillating points in the medium phase.

spherical waves

What is a spherical wave? Christian Huygens succeeded in developing a method for determining the method and place of wave propagation. In 1678, he suggested that every point that a light hindrance encounters turns into a source of a spherical wave. The summation of the secondary waves calculates the view at any time. This principle showed that upon contact, waves create destructive or constructive interference.

Constructive ones are formed if the waves are completely in phase with each other, and the final one is amplified. In destructive waves, they do not match in phase and the final one is simply reduced. Waves originate from a single point source, so they form in a spherical pattern.

If the waves are generated from a point source, then they act as spherical

This principle applies the law of refraction. Each point on a wave creates waves that interfere with each other constructively or destructively.

plane waves

Now let's understand what kind of wave is called a plane wave. The plane represents a frequency wave, the fronts of which are infinite parallel planes with a stable amplitude, located perpendicular to the phase velocity vector. In reality, it is impossible to get a true plane wave. Only a flat one with an infinite length can match it. True, many waves approach this state. For example, an antenna generates a field that is approximately flat.

Flat ones display an infinite number of wavefronts normal to the side of propagation