Let us now see whether the wave equation really describes the basic properties of sound waves in a medium. First of all, we want to deduce that the sound wave, or perturbation, moves at a constant speed. In addition, we need to prove that two different vibrations can freely pass through each other, i.e., the principle of superposition. We also want to prove that sound can propagate both to the right and to the left. All of these properties must be contained in our single equation.

Earlier, we noted that any perturbation that has the form of a plane wave and moves at a constant speed can be written as f(x—vt). Let's now see if f(x— vt) solution of the wave equation. Computing dχ/dx, we get the derivative of the function d χ / dx= f`(x—vt). Differentiating again, we find

Differentiating the same function χ on t, get the value - v, multiplied by the derivative, or d χ / dt = —vf`(x — vt); the second derivative with respect to time gives

It is obvious that f (X — vt) satisfies the wave equation if v equals cs.
Thus, from the laws of mechanics, we obtain that any sound disturbance propagates at a speed cs and besides,

thereby we have related the speed of sound waves to the propertiesenvironment.

It is easy to see that a sound wave can also propagate in the direction of negative X, i.e., a sound disturbance of the form χ(x, t)=g(x+vt) also satisfies the wave equation. The only difference between this wave and the one that propagated from left to right is the sign v, but sign d 2 χ / dt2 does not depend on choice x+vt or X—vt, because this derivative contains only v 2 . It follows that the solution of the equation describes waves traveling in any direction with a speed cs .


Of particular interest is the question of the superposition of solutions. Let's say we found one solution, let's say χ 1 . This means that the second derivative χ 1 . on X equal to the second derivative χ 1 on t, multiplied by 1/c 2 s . And let there be a second solution χ 2 having the same property. Adding these two solutions, we get

Now we want to make sure that χ(x,t) also represents a certain wave, i.e. χ also satisfies the wave equation. This is very easy to prove, since

Hence it follows that d 2 χ/dx 2 = (1/c 2 s)d2χ ldt2, so the validity of the superposition principle has been verified. The very existence of the superposition principle is due to the fact that the wave equation linearly on χ .


Now it would be natural to expect that a plane light wave propagating along the axis X and polarized so that the electric field is directed along the axis at, also satisfies the wave equation

where with is the speed of light. The wave equation for a light wave is one of the consequences of Maxwell's equations. The equations of electrodynamics lead to a wave equation for light, just as the equations of mechanics lead to a wave equation for sound.

Waves. wave equation

In addition to the motions we have already considered, in almost all areas of physics there is another type of motion - waves. A distinctive feature of this movement, which makes it unique, is that it is not the particles of matter that propagate in the wave, but changes in their state (perturbations).

Perturbations that propagate in space over time are called waves . Waves are mechanical and electromagnetic.

elastic wavesare propagating perturbations of the elastic medium.

A perturbation of an elastic medium is any deviation of the particles of this medium from the equilibrium position. Perturbations arise as a result of deformation of the medium in any of its places.

The totality of all points where the wave has reached at a given time forms a surface called wave front .

According to the shape of the front, the waves are divided into spherical and plane. Direction propagation of the wave front is determined perpendicular to the wave front, called beam . For a spherical wave, the rays are a radially divergent beam. For a plane wave, a ray is a beam of parallel lines.

In any mechanical wave, two types of motion simultaneously exist: oscillations of the particles of the medium and the propagation of a disturbance.

A wave in which the oscillations of the particles of the medium and the propagation of the perturbation occur in the same direction is called longitudinal (fig.7.2 a).

A wave in which the particles of the medium oscillate perpendicular to the direction of propagation of perturbations is called transverse (Fig. 7.2 b).

In a longitudinal wave, disturbances represent a compression (or rarefaction) of the medium, and in a transverse wave, they are displacements (shears) of some layers of the medium relative to others. Longitudinal waves can propagate in all media (in liquid, solid, and gaseous), while transverse waves can propagate only in solid ones.

Each wave propagates at some speed . Under wave speed υ understand the propagation speed of the disturbance. The speed of a wave is determined by the properties of the medium in which this wave propagates. In solids, the speed of longitudinal waves is greater than the speed of transverse waves.

Wavelengthλ is the distance over which a wave propagates in a time equal to the period of oscillation in its source. Since the speed of the wave is a constant value (for a given medium), the distance traveled by the wave is equal to the product of the speed and the time of its propagation. So the wavelength

It follows from equation (7.1) that particles separated from each other by an interval λ oscillate in the same phase. Then we can give the following definition of the wavelength: the wavelength is the distance between two nearest points oscillating in the same phase.

Let us derive the equation of a plane wave, which allows us to determine the displacement of any point of the wave at any time. Let the wave propagate along the beam from the source with some speed v.

The source excites simple harmonic oscillations, and the displacement of any point of the wave at any moment of time is determined by the equation

S = Asinωt (7. 2)

Then the point of the medium, which is at a distance x from the source of the wave, will also perform harmonic oscillations, but with a time delay of , i.e. the time it takes for the vibrations to propagate from the source to that point. The displacement of the oscillating point relative to the equilibrium position at any moment of time will be described by the relation

(7. 3)

This is the plane wave equation. This wave is characterized by the following parameters:

· S - displacement from the position of the equilibrium point of the elastic medium, to which the oscillation has reached;

· ω - cyclic frequency of oscillations generated by the source, with which the points of the medium also oscillate;

· υ - wave propagation velocity (phase velocity);

x – distance to that point of the medium where the oscillation has reached and the displacement of which is equal to S;

· t – time counted from the beginning of oscillations;

Introducing the wavelength λ into expression (7. 3), the plane wave equation can be written as follows:

(7. 4)

where called the wave number (number of waves per unit length).

wave equation

The plane wave equation (7. 5) is one of the possible solutions of the general differential equation with partial derivatives, which describes the process of propagation of a disturbance in a medium. Such an equation is called wave . Equations (7.5) include variables t and x, i.e. the displacement periodically changes both in time and in space S = f(x, t). The wave equation can be obtained by differentiating (7.5) twice with respect to t:

And twice x

Substituting the first equation into the second, we obtain the equation of a plane traveling wave along the X axis:

(7. 6)

Equation (7.6) is called wave, and for the general case, when the displacement is a function of four variables, it has the form

(7.7)

, where is the Laplace operator

§ 7.3 Wave energy. Vector Umov.

When propagating in a medium of a plane wave

(7.8)

energy transfer takes place. Let us mentally single out the elementary volume ∆V, which is so small that the speed of movement and the deformation at all its points can be considered the same and equal, respectively

The allocated volume has kinetic energy

(7.10)

m=ρ∆V is the mass of matter in the volume ∆V, ρ is the density of the medium].

(7.11)

Substituting into (7.10) the value , we obtain

(7.12)

The maxima of the kinetic energy fall on those points of the medium that pass the equilibrium positions at a given moment of time (S = 0), at these moments of time the oscillatory motion of the points of the medium is characterized by the highest speed.

The considered volume ∆V also has the potential energy of elastic deformation

[E - Young's modulus; - relative elongation or compression].

Taking into account formula (7.8) and the expression for the derivative, we find that the potential energy is equal to

(7.13)

An analysis of expressions (7.12) and (7.13) shows that the maxima of the potential and kinetic energies coincide. It should be noted that this is a characteristic feature of traveling waves. To determine the total volume energy ∆V, you need to take the sum of the potential and kinetic energies:

Dividing this energy by the volume in which it is contained, we obtain the energy density:

(7.15)

It follows from expression (7.15) that the energy density is a function of the x coordinate, i.e., it has different values ​​at different points in space. The energy density reaches its maximum value at those points in space where the displacement is zero (S = 0). The average energy density at each point of the medium is

(7.16)

because the average

Thus, the medium in which the wave propagates has an additional reserve of energy, which is delivered from the source of oscillations to various regions of the medium.

Energy transfer in waves is quantitatively characterized by the energy flux density vector. This vector for elastic waves is called the Umov vector (after the Russian scientist N. A. Umov). The direction of the Umov vector coincides with the direction of energy transfer, and its modulus is equal to the energy transferred by a wave per unit time through a unit area located perpendicular to the direction of wave propagation.

The mechanism of formation of mechanical waves in an elastic medium.

MECHANICAL WAVES

1. The mechanism of formation of mechanical waves in an elastic medium. Longitudinal and transverse waves. Wave equation and its solution. Harmonic waves and their characteristics.

2. Phase velocity and wave dispersion. Wave packet and group velocity.

3. The concept of coherence. Wave interference. standing waves.

4. Doppler effect for sound waves.

If in any place of an elastic medium (solid, liquid or gaseous) oscillations of its particles are excited, then due to the interaction between particles, this oscillation will propagate in the medium from particle to particle with a certain speed. The process of propagation of oscillations in space is called a wave. The locus of points to which oscillations reach by the time t is called the wave front (wave front). Depending on the shape of the front, the wave can be spherical, flat, etc.

The wave is called longitudinal, if the direction of displacement of particles of the medium coincides with the direction of wave propagation.

A longitudinal wave propagates in solid, liquid and gaseous media.

The wave is called transverse, if the displacement of the particles of the medium is perpendicular to the direction of wave propagation. A transverse mechanical wave propagates only in solids (in media with shear resistance, therefore, such a wave cannot propagate in liquids and gases).

Equation to determine offset(x, t) of any point of the medium with the coordinate x at any time t is called the wave equation.

For example, the plane wave equation, i.e. wave propagating in one direction, for example in the direction of the x axis, has the form

Let us introduce the value , which is called the wave number.

If we multiply the wave number by the unit vector of the wave propagation direction , we get a vector called wave vector

Using the Laplace operator (Laplacian) this equation can be written more concisely

(The solution to this equation is the wave equation (28-1), (28-2).)

Definition 1

In the event that a wave propagates in a homogeneous medium, then its motion is generally described by wave equation(by a partial differential equation):

\[\frac((\partial )^2\overrightarrow(s))(\partial t^2)=v^2\left(\frac((\partial )^2\overrightarrow(s))(\partial x ^2)+\frac((\partial )^2\overrightarrow(s))(\partial y^2)+\frac((\partial )^2\overrightarrow(s))(\partial z^2)\ right)\left(1\right)\]

\[\triangle \overrightarrow(s)=\frac(1)(v^2)\frac((\partial )^2\overrightarrow(s))(\partial t^2)\left(2\right), \]

where $v$ is the phase velocity of the wave $\triangle =\frac((\partial )^2)(\partial x^2)+\frac((\partial )^2)(\partial y^2)+\ frac((\partial )^2)(\partial z^2)$ is the Laplace operator. Equation (1.2) is solved by the equation of any wave; these equations satisfy, for example, both plane and spherical waves.

If a plane wave propagates along the $X$ axis, then equation (1) is represented as:

Note 1

If a physical quantity propagates like a wave, then it necessarily satisfies the wave equation. The converse statement is true: if any quantity obeys the wave equation, then it propagates like a wave. The speed of wave propagation will be equal to the square root of the coefficient that stands for the sum of spatial derivatives (in this type of recording).

The wave equation plays a very important role in physics.

Solution of the wave equation for a plane wave

Let us write the general solution of equation (2) for a light wave propagating in vacuum if the scalar function depends only on one of the Cartesian variables, for example $z$, i.e. $s=s(z,t)$, which means , the function $s$ has a constant value at the points of the plane that is perpendicular to the $z-axis. Wave equation (1) in this case will take the form:

where the speed of light propagation in vacuum is equal to $c$.

The general solution of equation (4) under given conditions will be the expression:

where $s_1\left(z+ct\right)$ is a function describing an arbitrary wave that moves with a speed $c$ in a negative direction with respect to the $z-axis direction, $s_2\left(z-ct\right) $ - a function describing an arbitrary wave that moves with a speed $c$ in a positive direction with respect to the direction of the Z$ axis. It should be noted that during the movement the values ​​of $s_1$ and $s_2$ at any point of the wave and its shape of the wave are unchanged.

It turns out that the wave, which is described by the superposition of two waves (in accordance with formula (5)). Moreover, these component waves move in opposite directions. In this case, it is no longer possible to talk about the speed or direction of the wave. In the simplest case, a standing wave is obtained. In the general case, it is necessary to consider a complex electromagnetic field.

Wave equation and Maxwell's system of equations

The wave equations for oscillations of the vectors of the electric field strength and the vector of magnetic induction of the magnetic field can be easily obtained from the system of Maxwell's equations in differential form. Let us write the system of Maxwell equations for a substance in which there are no free charges and conduction currents:

Let's apply the $rot$ operation to equation (7):

In expression (10), it is possible to change the order of differentiation on the right side of the expression, since spatial coordinates and time are independent variables, therefore, we have:

Let us take into account that equation (6), replace $rot\overrightarrow(B)$ in expression (11) with the right side of formula (6), we have:

Knowing that $rotrot\overrightarrow(E)=graddiv\overrightarrow(E)-(\nabla )^2\overrightarrow(E)$ and using $div\overrightarrow(E)=0$, we get:

Similarly, one can obtain the wave equation for magnetic induction vector. It looks like:

In expressions (13) and (14), the phase velocity of wave propagation $(v)$ is equal to:

Example 1

Exercise: Get the general solution of the wave equation $\frac((\partial )^2s)(\partial z^2)-\frac(1)(c^2)\frac((\partial )^2s)(\partial t^2 )=0(1.1)$ of a plane light wave.

Decision:

Let's introduce independent type variables for the function $s$:

\[\xi =z-ct,\ \eta =z+ct\left(1.2\right).\]

In this case, the partial derivative $\frac(\partial s)(\partial z)$ is equal to:

\[\frac(\partial s)(\partial z)=\frac(\partial s)(\partial \xi)\frac(\partial \xi)(\partial z)+\frac(\partial s)( \partial \eta )\frac(\partial \eta )(\partial z)=\frac(\partial s)(\partial \xi)+\frac(\partial s)(\partial \eta )\left(1.3 \right).\]

The partial derivative $\frac(\partial s)(\partial t)$ is:

\[\frac(\partial s)(\partial t)=\frac(\partial s)(\partial \xi)\frac(\partial \xi)(\partial t)+\frac(\partial s)( \partial \eta)\frac(\partial \eta)(\partial t)=-c\frac(\partial s)(\partial \xi)+c\frac(\partial s)(\partial \eta)\ to \frac(1)(c)\frac(\partial s)(\partial t)=-\frac(\partial s)(\partial \xi)+\frac(\partial s)(\partial \eta) \left(1.4\right).\]

Subtract term by term expression (1.4) from expression (1.3), we have:

\[\frac(\partial s)(\partial z)-\frac(1)(c)\frac(\partial s)(\partial t)=2\frac(\partial s)(\partial \xi) \left(1.5\right).\]

Termwise addition of expressions (1.4) and (1.3) gives:

\[\frac(\partial s)(\partial z)-\frac(1)(c)\frac(\partial s)(\partial t)=2\frac(\partial s)(\partial \eta ) \left(1.6\right).\]

Let us find the product of the left parts of expressions (1.5) and (1.6) and take into account the results written in the right parts of these expressions:

\[\left(\frac(\partial s)(\partial z)-\frac(1)(c)\frac(\partial s)(\partial t)\right)\left(\frac(\partial s )(\partial z)-\frac(1)(c)\frac(\partial s)(\partial t)\right)=\frac((\partial )^2s)(\partial z^2)-\ frac(1)(c^2)\frac((\partial )^2s)(\partial t^2)=4\frac(\partial )(\partial \xi )\frac(\partial s)(\partial \eta )=0\left(1.7\right).\]

If we integrate the expression (1.7) over $\xi $, then we get a function that does not depend on this variable and can depend only on $\eta $, which means that it is an arbitrary function of $\Psi(\eta)$. In this case, equation (1.7) will take the form:

\[\frac(\partial s)(\partial \eta )=\Psi \left(\eta \right)\left(1.8\right).\]

Let us integrate (1.8) over $\eta $ we have:

where $s_1\left(3\right)$ is the antiderivative, $s_2\left(\xi \right)$ is the integration constant. Moreover, the functions $s_1$ and $s_2$ are arbitrary. Taking into account expressions (1.2), the general solution of equation (1.1) can be written as:

Answer:$s\left(z,t\right)=s_1\left(z+ct\right)+s_2\left(z-ct\right).$

Example 2

Exercise: Determine from the wave equation what is the phase velocity of propagation of a plane light wave.

Decision:

Comparing the wave equation, for example, for the field vector, obtained from Maxwell's equations:

\[(\nabla )^2\overrightarrow(E)-\varepsilon (\varepsilon )_0\mu (\mu )_0\frac((\partial )^2\overrightarrow(E))(\partial t^2) =0(2.1)\]

with the wave equation:

\[\triangle \overrightarrow(s)=\frac(1)(v^2)\frac((\partial )^2\overrightarrow(s))(\partial t^2)(2.2)\]

allows us to conclude that the wave propagation velocity $(v)$ is equal to:

But here it is necessary to note that the concept of the speed of an electromagnetic wave has a certain meaning only with waves of a simple configuration; for example, the category of plane waves is suitable for such waves. So $v$ will not be the speed of wave propagation in the case of a derivative solution of the wave equation, which include, for example, standing waves.

Answer:$v=\frac(c)(\sqrt(\mu \varepsilon )).$

One of the most common second-order partial differential equations in engineering practice is the wave equation, which describes various types of oscillations. Since oscillations are a non-stationary process, one of the independent variables is time t. In addition, independent variables in the equation are also spatial coordinates x, y,z. Depending on their number, one-dimensional, two-dimensional and three-dimensional wave equations are distinguished.

One-dimensional wave equation- an equation that describes the longitudinal vibrations of a rod, the sections of which perform plane-parallel oscillatory movements, as well as transverse vibrations of a thin rod (string) and other problems. 2D wave equation used to study the vibrations of a thin plate (membrane). 3D wave equation describes the propagation of waves in space (for example, sound waves in a liquid, elastic waves in a continuous medium, etc.).

Consider a one-dimensional wave equation, which can be written as

For transverse vibrations of the string, the desired function U(x, t) describes the position of the string at the moment t. In this case a 2 = Т/ρ, where T - string tension, ρ - its linear (linear) density. The fluctuations are assumed to be small, i.e. the amplitude is small compared to the length of the string. In addition, equation (2.63) is written for the case of free oscillations. In the case of forced oscillations, a certain function is added to the right side of the equation f(x, t), characterizing external influences, while the resistance of the medium to the oscillatory process is not taken into account.

The simplest problem for equation (2.63) is the Cauchy problem: at the initial moment of time, two conditions are set (the number of conditions is equal to the order of the derivative with respect to t):

These conditions describe the initial shape of the string and the speed of its points.

In practice, it is more often necessary to solve not the Cauchy problem for an infinite string, but a mixed problem for a bounded string of some length l. In this case, boundary conditions are set at its ends. In particular, when the ends are fixed, their displacements are equal to zero, and the boundary conditions have the form

Let us consider some difference schemes for solving problem (2.63)-(2.65). The simplest is the explicit three-layer cross scheme (the template is shown in Figure 2.21). Let us replace in equation (2.63) the second derivatives of the desired function U on t and X their finite-difference relations using the values ​​of the grid function at the grid nodes:

Rice. 2.21. Explicit Schema Pattern

From here, one can find an explicit expression for the value of the grid function on ( j + 1)th layer:

Here, as usual in three-layer schemes, to determine the unknown values ​​on ( j + 1)th layer needs to know the solutions on j-om and ( j- 1)th layers. Therefore, it is possible to start counting using formulas (2.66) only for the second layer, and the solutions on the zero and first layers must be known. They are found using the initial conditions (2.64). On the zero layer we have

To obtain a solution on the first layer, we use the second initial condition (2.64). We replace the derivative with a finite-difference approximation. In the simplest case, one assumes

(2.68)

From this relation, one can find the values ​​of the grid function at the first time layer:

Note that the approximation of the initial condition in the form (2.68) worsens the approximation of the original differential problem: the approximation error becomes of the order of , i.e. first order in τ, although the scheme (2.66) itself has the second order of approximation in h and τ. The situation can be corrected if, instead of (2.69), we take a more accurate representation:

(2.70)

Instead, take . And the expression for the second derivative can be found using the original equation (2.63) and the first initial condition (2.64). Get

Then (2.70) takes the form:

Difference scheme (2.66), taking into account (2.71), has an approximation error of the order

When solving a mixed problem with boundary conditions of the form (2.65), i.e. when the values ​​of the function itself are given at the ends of the segment under consideration, the second order of approximation is preserved. In this case, for convenience, the extreme nodes of the grid are located at the boundary points ( x0=0, xI = l). However, boundary conditions can also be specified for the derivative.

For example, in the case of free longitudinal vibrations of a rod, the condition

If this condition is written in a difference form with the first order of approximation, then the approximation error of the scheme becomes of the order of . Therefore, to preserve the second order of this scheme in terms of h it is necessary to approximate the boundary condition (2.72) with the second order.

The considered difference scheme (2.66) for solving problem (2.63) - (2.65) is conditionally stable. Necessary and sufficient condition for stability:

Consequently, under this condition and taking into account the approximation, scheme (2.66) converges to the original problem at a rate O(h2 + τ 2 ). This scheme is often used in practical calculations. It provides acceptable solution accuracy. U(x, t), which has continuous fourth-order derivatives.

Rice. 2.22. Algorithm for solving the wave equation

The algorithm for solving problem (2.63)-(2.65) with the help of this explicit difference scheme is shown in fig. 2.22. Here the simplest version is presented, when all the values ​​of the grid function, which form a two-dimensional array, are stored in the computer memory during the calculation, and after solving the problem, the results are displayed. It would be possible to provide for storing the solution on only three layers, which would save memory. The results in this case can be displayed during the calculation process (see Fig. 2.13).

There are other difference schemes for solving the wave equation. In particular, it is sometimes more convenient to use implicit schemes to get rid of the restrictions on the step size imposed by condition (2.73). These schemes are usually absolutely stable, but the algorithm for solving the problem and the computer program become more complicated.

Let us construct the simplest implicit scheme. The second derivative with respect to t in equation (2.63) we approximate, as before, by the three-point pattern using the values ​​of the grid function on the layers j- 1, j, j + 1. Derivative to X we replace the half-sum of its approximation to ( j + 1)-ohm and ( j- 1)th layers (Fig. 2.23):

Rice. 2.23. Implicit schema pattern

From this relation, one can obtain a system of equations for unknown values ​​of the grid function on ( j+ 1)th layer:

The resulting implicit scheme is stable and converges at a rate of . The system of linear algebraic equations (2.74) can, in particular, be solved by the sweep method. This system should be supplemented with difference initial and boundary conditions. Thus, expressions (2.67), (2.69) or (2.71) can be used to calculate the values ​​of the grid function on the zero and first time layers.

For two or three independent spatial variables, the wave equations take the form

Difference schemes can also be constructed for them by analogy with the one-dimensional wave equation. The difference is that it is necessary to approximate the derivatives with respect to two or three spatial variables, which naturally complicates the algorithm and requires much more memory and computation time. Two-dimensional problems will be considered in more detail below for the heat equation.