Turning to the basic differential equations of oscillations, we will notice that when we multiply them by – = k 2, they will contain terms, some of which have a coefficient of the square of the speed And transverse vibrations, others - square of speed longitudinal hesitation.
The first terms in the case of longitudinal vibrations should disappear from the equations, and we get the first group:
Since the surface p, by our choice, is the surface of a wave, then in the equations of § 7 we must retain one oscillation R and equate the vibrations /?! And R.2, occurring in a plane tangent to the wave. As a result, we find, assuming // =1:
Since A = 0, then equations (1) will take the form:
Multiplying the first of equations (2) by //i // 2, differentiating with respect to p and paying attention to equation (4), we find:
What according to equations (2), B does not depend on either р x or [–]. Therefore, meaning through &F partial derivative of a function F by one of the variables ^, R. 2, we obtain from equation (7):
Substituting into this expression the quantities H 1H 2, found in pp. 3, equating the coefficients at various powers to zero, we find the following conditions that the wave F – i must satisfy
It is known that such relations take place only for sphere, round cylinder and plane.
From here we have, What isothermal wave surfaces can propagate longitudinal vibrations.
So, if the shaking surface or the initial wave does not belong to the surfaces of isothermal waves, then vibrations occur near them mixed , but at considerable distances the wave approaches the form of one of the isothermal waves, and oscillations are detected in the phenomenon longitudinal. STOP!!!
It remains to integrate the given differential equations for the sphere, with using harmonic functions!!!
Tesla's experiments – harmonic oscillator is unacceptable!!!
For spheres in the coordinates we have already used, we have:
Further transformations are insignificant and are not given, since they lead to original equation , which has no physical meaning for soliton-like waves.
The conclusions found are equally applicable to the phenomena of light in homogeneous bodies and, moreover, within the limits of approximation that occur in Boussinesq’s theory!?
From here:"painful moment" identified.
N. Umov mathematical collection, vol. 5, 1870.
Another “terrible” uncertainty
Reasoning similarly, one could easily obtain a similar expression for magnetic energy, and therefore for currents. We see that, even insisting on the simplest of formulas, the problem of energy localization still cannot be solved.
And we have the same thing for energy flow. It is possible to transform the movement of current energy in an arbitrary way by adding another vector (u, v, w) to the Poynting vector, which must satisfy only the equation of incompressible fluids
Being a consequence of general equations, it does not add anything to them.
Therefore, energy localization is logically useless(and sometimes, harmful).
But there is an aspect in which it is important to consider Poynting's theorem.
The main fact from which the law of conservation of energy stems was and remains the experimentally found fact of the impossibility perpetual motion , a fact - independent of our ideas, and can be attributed to portions of energy that the ether should possess in the absence of material bodies.
The law of conservation of energy, in its classical form W = Const, explains this impossibility.
Poynting's theorem, requiring the ability to transform volume integral(somewhat arbitrary) in surface, expresses much less. She easily admits the creation of perpetual motion, without being able to show its impossibility!
In fact, until we introduce the hypothesis delayed potentials, the continuous release of energy from converging waves coming from infinity remains as probable as the loss of energy observed in reality.
If the engine could forever take away only the energy of the ether, regardless of the presence of material bodies, then it could exist perpetual motion . Thus, it becomes clear that before we accept the formula of retarded potentials, we must prove that the accelerated particle loses energy and, as a result, is subjected to a reaction proportional to the derivative of its acceleration.
Just change the sign c to arrive at the converging wave hypothesis.
Then we will discover what a sign radiation vector will also change, and the new hypothesis will lead, say, in the case of a vibrating particle, to a gradual increase in amplitude over time, and in general – to increase the energy of the system?!
In Nature, solitons are:
– on the surface of a liquid, the first solitons discovered in nature are sometimes considered tsunami waves
– various types of water hammer
– sound drums – overcoming “supersonic”
– ionosonic and magnetosonic solitons in plasma
– solitons in the form of short light pulses in the active medium of the laser
– presumably, an example of a soliton is the Giant Hexagon on Saturn
– nerve impulses can be considered in the form of solitons.
Mathematical model, Korteweg-de Vries equation.
One of the simplest and most well-known models that allows the existence of solitons in the solution is the Korteweg-de Vries equation:
u t + uu x + β u xxx = 0.
One possible solution to this equation is solitary soliton:
but also here the oscillator is the harmonic function where r, s,α, U- some are permanent.Uncertainty theorems in harmonic analysis
Harmonic oscillator in quantum mechanics – described by the equation Schrödinger,
(217.5)The equation (217.5) called the Schrödinger equation for stationary states.
The stationary states of a quantum oscillator are determined by the equation Schrödinger kind
(222.2)
Where E – total energy of the oscillator.
In the theory of differential equations it is proven that the equation (222.2) solved only for eigenvalues of energy
(222.3)Formula (222.3) shows that the energy of a quantum oscillator quantized.
The energy is limited from below to be different from zero, as for a rectangular "pits" with infinitely high “walls” (see § 220), a minimum energy value
E 0 = 1/2 ℏ w 0 . The existence of minimum energy is called zero-point energy– is typical for quantum systems and is a direct consequence uncertainty relations.
IN harmonic analysis The uncertainty principle implies that it is impossible to accurately obtain the values of a function and its Fourier map – and therefore make an accurate calculation.
That is, modeling, generation and analogy in compliance with the principles of similarity of processes and forms in Nature, using harmonic oscillator – not possible.
Different types mathematicalsolitons little is known yet and all of them are not suitable for describing objects in three-dimensional space, especially the processes occurring in Nature.
For example, ordinary solitons, which appear in the Korteweg–de Vries equation, are localized in only one dimension if it "run" in the three-dimensional world, then it will look like an endless flat membrane flying forward, to put it mildly, gobbledygook!!!
In nature, such infinite membranes are not observed, which means original equation not suitable for describing three-dimensional objects.
This is where the fallacy of introducing harmonic functions lies – oscillators, connections in the case of mixed oscillations.Connected law of similarity, , but that's another story that will lead to soliton theory from systematic uncertainty, .
Free oscillations of systems with distributed parameters
The main feature of the process of free vibrations of systems with an infinite number of degrees of freedom is expressed in the infinity of the number of natural frequencies and mode shapes. This is also associated with mathematical features: instead of ordinary differential equations that describe the oscillations of systems with a finite number of degrees of freedom, here we have to deal with partial differential equations. In addition to the initial conditions that determine the initial displacements and velocities, it is also necessary to take into account the boundary conditions that characterize the fixation of the system.
6.1. Longitudinal vibrations of rods
When analyzing the longitudinal vibrations of a straight rod (Fig. 67, a), we will assume that the cross sections remain flat and that the particles of the rod do not perform transverse movements, but move only in the longitudinal direction.
Let u - longitudinal movement of the current section of the rod during vibrations; this movement depends on the location of the section (coordinates x) and on time t. So there is a function of two variables; its definition represents the main task. The displacement of an infinitely close section is equal to , therefore, the absolute elongation of an infinitely small element is equal (Fig. 67, b), and its relative elongation is .
Accordingly, the longitudinal force in the section with the coordinate X can be written as
,(173)
where is the rigidity of the rod in tension (compression). The force N is also a function of two arguments - the coordinates X and time t.
Let's consider a rod element located between two infinitely close sections (Fig. 67, c). A force N is applied to the left side of the element, and a force is applied to the right side. If we denote the density of the material of the rod, then the mass of the element in question is . Therefore, the equation of motion in projection onto the axis X
,
Considering(173)andaccepting A= const, we get
Following the Fourier method, we look for a particular solution to the differential equation (175) in the form
,(177)
those. suppose that the movement u can be represented as a product of two functions, one of which depends only on the argument X, and the other only from the argument t. Then, instead of defining a function of two variables u (x, t), it is necessary to define two functions X(x) and T(t), each of which depends on only one variable.
Substituting (177) into (174), we get
where the primes indicate the operation of differentiation with respect to x, and by dots t. Let's rewrite this equation this way:
Here the left side depends only on x, and the right side only on t. For this equality to hold identically (for any x and t) it is necessary that each of its parts be equal to a constant, which we denote by:
; .(178)
This leads to two equations:
;
.(179)
The first equation has a solution:
,(180)
indicating an oscillatory nature, and from (180) it is clear that the unknown quantity has the meaning of the frequency of free oscillations.
The second of equations (179) has a solution:
,(181)
determining the shape of the vibrations.
The frequency equation that determines the value is compiled by using boundary conditions. This equation is always transcendental and has an infinite number of roots. Thus, the number of natural frequencies is infinite, and each frequency value corresponds to its own function T n (t), determined by dependence (180), and its own function Xn (x), determined by dependence (181). Solution (177) is only partial and does not provide a complete description of the motion. The complete solution is obtained by superimposing all partial solutions:
.
The functions X n (x) are called own functions problems and describe their own modes of vibration. They do not depend on the initial conditions and satisfy the orthogonality condition, which for A = const has the form
, If .
Let's consider some options for boundary conditions.
Fixed end of the rod(Fig. 68, a). At the end section, the displacement u must be zero; it follows that in this section
X=0(182)
Free end of the rod(Fig. 68, b). At the end section, the longitudinal force
(183)
must be identically equal to zero, which is possible if at the end section X"=0.
Resilient end of rod(Fig. 68, c).
When moving u end rod, an elastic support reaction occurs 
, where C o is the rigidity of the support. Taking into account (183) for the longitudinal force, we obtain the boundary condition
if the support is located at the left end of the rod (Fig. 68, c), and
if the support is located at the right end of the rod (Fig. 68, d).
Concentrated mass at the end of the rod.
Inertia force developed by the mass:
.
Since, according to the first of equations (179), , the inertia force can be written in the form . We get the boundary condition
,
if the mass is at the left end (Fig. 68, d), and
, (184)
if the mass is connected to the right end (Fig. 68, e).
Let us determine the natural frequencies of the cantilever rod (Fig. 68,a").
According to (182) and (183), the boundary conditions
X=0at x=0;
X"=0 at x= .
Substituting these conditions one by one into solution (181), we obtain
Condition C0 leads to the frequency equation:
The roots of this equation
(n=1,2,…)
determine natural frequencies:
(n=1,2,…).(185)
First (lowest) frequency at n=1:
.
Second frequency (at n=2):
Let us determine the natural frequencies of a rod with a mass at the end (Fig. 68, f).
According to (182) and (184), we have
X=0 at x=0;
at x= .
Substituting these conditions into solution (181), we obtain:
D=0; 
.
Consequently, the frequency equation when taking (176) into account has the form
.
Here the right side represents the ratio of the mass of the rod to the mass of the end load.
To solve the resulting transcendental equation, it is necessary to use some approximate method.
At and the values of the most important lowest root will be 0.32 and 0.65, respectively.
At a small ratio, the load has a decisive influence and an approximate solution gives good results
.
For bars of variable cross-section, i.e. for Аconst, from (173) and (174) the equation of motion is obtained in the form
.
This differential equation cannot be solved in closed form. Therefore, in such cases it is necessary to resort to approximate methods for determining natural frequencies.
6.2. Torsional vibrations of shafts
Torsional vibrations of shafts with a continuously distributed mass (Fig. 69, a) are described by equations that, in structure, completely coincide with the above equations for longitudinal vibrations of rods.
Torque M in section with abscissa X is related to the angle of rotation by a differential dependence similar to (173):
Where Jp-polar moment of inertia of the cross section.
In a section located at a distance dx, the torque is equal to (Fig. 69, b):
Denoting through (where is the density of the shaft material) the intensity of the moment of inertia of the shaft mass relative to its axis (i.e., the moment of inertia per unit length), the equation of motion of an elementary section of the shaft can be written as follows:
,
or similar (174):
.
Substituting expression (186) here, with Jp=const we get, similarly to (175):
, (187)
The general solution to equation (187), like equation (175), has the form
,
(188)
Natural frequencies and eigenfunctions are determined by specific boundary conditions.
In the main cases of fixing the ends, similarly to the case of longitudinal vibrations, we obtain
a) fixed end (=0): X=0;
b) free end (M=0): X"=0;
V) resilient left end: CoХ=GJpX "(Co-stiffness coefficient);
G) resilient right end: -CoX=GJpX ";
e) disk at the left end: 
(Jo is the moment of inertia of the disk relative to the axis of the rod);
e) disk at the right end: 
.
If the shaft is fixed at the left end (x=0), and the right end (x=) is free, then X=0 at x=0 and X"=0 at x=; natural frequencies are determined similarly to (185):
(n=1,2,…).
If the left end is fixed and there is a disk at the right end, we obtain the transcendental equation:
.
If both ends of the shaft are fixed, then the boundary conditions will be X=0 for x=0 and x=. In this case, from (188) we obtain
those.
(n=1,2,…),
from here we find the natural frequencies:
If the left end of the shaft is free, and there is a disk at the right end, then X"=0 for x=0;Jo X=GJpX "for x=.
Using (188) we find
C=0; 
,
or transcendental frequency equation:
.
6.3.Bending vibrations of beams
6.3.1 Basic equation
From the course on the strength of materials, the differential dependencies for bending beams are known:
where EJ is bending rigidity; y=y (x, t) - deflection; M=M(x, t) - bending moment; q is the intensity of the distributed load.
Combining (189) and (190), we get
.(191)
In the problem of free vibrations, the load for the elastic skeleton is the distributed inertial forces:
where m is the intensity of the mass of the beam (mass per unit length), and equation (191) takes the form
.
In the special case of a constant cross section, when EJ = const, m = const, we have:
.(192)
To solve equation (192), we assume, as above,
y= X ( x)× T ( t ).(193)
Substituting (193) into (192), we arrive at the equation:
.
For this equality to be fulfilled identically, it is necessary that each of the parts of the equality be constant. Denoting this constant by , we obtain two equations:
.(195)
The first equation indicates that the movement is oscillatory with frequency .
The second equation determines the shape of the vibrations. The solution to equation (195) contains four constants and has the form
It is convenient to use the variant of writing the general solution proposed by A.N. Krylov:
(198)
represent the functions of A.N. Krylov.
Let's pay attention to the fact that S=1, T=U=V=0 at x=0. The functions S,T,U,V are interconnected as follows:
Therefore, derivative expressions (197) are written in the form
(200)
In problems of the class under consideration, the number of natural frequencies is infinitely large; each of them has its own time function T n and its own fundamental function X n . The general solution is obtained by imposing partial solutions of the form (193)
.(201)
To determine the natural frequencies and formulas, it is necessary to consider the boundary conditions.
6.3.2. Border conditions
For each end of the bar, you can specify two boundary conditions .
Free end of the rod(Fig. 70, a). The transverse force Q=EJX""T and the bending moment M=EJX""T are equal to zero. Therefore, the boundary conditions have the form
X""=0; X"""=0 .(202)
Hinged supported end of the rod(Fig. 70, b). The deflection y=XT and the bending moment M=EJX""T are equal to zero. Therefore, the boundary conditions are:
X=0 ; X""=0 .(203)
Pinched end(Fig. 70, c). The deflection y=XT and the rotation angle are equal to zero. Border conditions:
X=0; X"=0 . (204)
There is a point mass at the end of the rod(Fig. 70, d). His inertial force 
can be written using equation (194) as follows: ; it must be equal to the shear forceQ=EJX"""T, so the boundary conditions take the form
; X""=0 .(205)
In the first condition, a plus sign is taken when the point load is connected to the left end of the rod, and a minus sign when it is connected to the right end of the rod. The second condition follows from the absence of a bending moment.
Elastically supported end of the rod(Fig. 70, d). Here the bending moment is zero, and the transverse force Q=EJX"""T is equal to the support reaction 
(C o - support rigidity coefficient).
Border conditions:
X""=0 ; (206)
(a minus sign is taken when the elastic support is left, and a plus sign when it is right).
6.3.3. Frequency equation and eigenforms
An expanded recording of the boundary conditions leads to homogeneous equations with respect to the constants C 1, C 2, C 3, C 4.
In order for these constants not to be equal to zero, the determinant made up of the coefficients of the system must be equal to zero; this leads to a frequency equation. During these operations, the relationships between C 1, C 2, C 3, C 4 are clarified, i.e. the natural vibration modes are determined (up to a constant factor).
Let us trace the composition of frequency equations using examples.
For a beam with hinged ends, according to (203), we have the following boundary conditions: X=0; X""=0 for x=0 and x= . Using (197)-(200) we obtain from the first two conditions: C 1 =C 3 =0. The two remaining conditions can be written as
In order for C 2 and C 4 not to be equal to zero, the determinant must be equal to zero:
.
Thus, the frequency equation has the form
.
Substituting the expressions T and U, we get
Since , the final frequency equation is written as follows:
. (207)
The roots of this equation are:
,(n =1,2,3,...).
Taking (196) into account, we obtain
.(208)
Let's move on to defining our own forms. From the homogeneous equations written above, the following relationship between the constants C 2 and C 4 follows:
.
Consequently, (197) takes the form
According to (207), we have
,(209)
where is a new constant, the value of which remains uncertain until the initial conditions are introduced into consideration.
6.3.4. Determination of motion based on initial conditions
If it is necessary to determine the movement following the initial disturbance, then it is necessary to indicate both the initial displacements and the initial velocities for all points of the beam:
(210)
and use the property of orthogonality of eigenforms:
.
We write the general solution (201) as follows:
.(211)
The speed is given by
.(212)
Substituting the initial displacements and velocities assumed to be known into the right-hand sides of equations (211) and (212), and into the left-hand sides, we obtain
.
Multiplying these expressions by and integrating over the entire length, we have
(213)
Infinite sums on the right-hand sides have disappeared due to the orthogonality property. From (213) follow the formulas for constants and
(214)
Now these results need to be substituted into solution (211).
Let us emphasize again that the choice of the scale of eigenforms is unimportant. If, for example, in the expression of the eigenform (209) we take instead a value that is times larger, then (214) will give results that are times smaller; after substitution into solution (211), these differences compensate each other. Nevertheless, they often use normalized eigenfunctions, choosing their scale such that the denominators of expressions (214) are equal to one, which simplifies the expressions and .
6.3.5. Effect of constant longitudinal force
Let us consider the case when an oscillating beam experiences a longitudinal force N, the magnitude of which does not change during the oscillation process. In this case, the static bending equation becomes more complicated and takes the form (provided that the compressive force is considered positive)
.
Assuming and considering the stiffness constant, we obtain the equation of free vibrations
.(215)
We continue to accept a particular solution in the form.
Then equation (215) splits into two equations:
The first equation expresses the oscillatory nature of the solution, the second determines the shape of the oscillations, and also allows you to find the frequencies. Let's rewrite it this way:
(216)
Where K is determined by formula (196), and
The solution to equation (216) has the form
Let us consider the case when both ends of the rod have hinged supports. Conditions on the left end 
give . Satisfying the same conditions on the right end, we get
Equating to zero the determinant composed of coefficients for the quantities and , we arrive at the equation
The roots of this frequency equation are:
Therefore, the natural frequency is determined from the equation
.
From here, taking (217) into account, we find
.(219)
When stretched, the frequency increases, when compressed it decreases. When the compressive force N approaches a critical value, the root tends to zero.
6.3.6. Effect of Chain Forces
Previously, the longitudinal force was considered given and independent of the displacements of the system. In some practical problems, the longitudinal force accompanying the process of transverse vibrations arises due to the bending of the beam and has the character of a support reaction. Consider, for example, a beam on two hinged and fixed supports. When it bends, horizontal reactions of the supports occur, causing the beam to stretch; the corresponding horizontal force is usually called chain force. If the beam oscillates transversely, the chain force will change over time.
If at instant t the deflections of the beam are determined by the function, then the elongation of the axis can be found using the formula
.
We find the corresponding chain force using Hooke's law
.
Let's substitute this result into (215) instead of the longitudinal force N (taking into account the sign)
.(220)
The resulting nonlinear integrodifferential the equation is simplified using substitution
,(221)
where is a dimensionless function of time, the maximum value of which can be set equal to any number, for example, unity; amplitude of oscillations.
Substituting (221) into (220), we obtain the ordinary differential equation
,(222)
whose coefficients have the following values:
;.
Differential equation (222) is nonlinear, therefore, the frequency of free oscillations depends on their amplitude.
The exact solution for the frequency of transverse vibrations has the form
where is the frequency of transverse vibrations, calculated without taking into account chain forces; correction factor depending on the ratio of the oscillation amplitude to the radius of gyration of the cross section; the value is given in the reference literature.
When the amplitude and radius of gyration of the cross section are commensurate, the correction to the frequency becomes significant. If, for example, the amplitude of vibration of a round rod is equal to its diameter, then , and the frequency is almost twice as large as in the case of free displacement of the supports.
The case corresponds to a zero value of the radius of inertia, when the bending rigidity of the beam is vanishingly small - a string. At the same time, the formula for gives uncertainty. Revealing this uncertainty, we obtain a formula for the frequency of vibration of the string
.
This formula applies to the case when the tension is zero at the equilibrium position. Often the problem of string oscillations is posed under other assumptions: it is believed that the displacements are small, and the tensile force is given and remains unchanged during the oscillation process.
In this case, the formula for frequency has the form
where N is a constant tensile force.
6.4. Effect of viscous friction
Previously it was assumed that the material of the rods was perfectly elastic and there was no friction. Let us consider the influence of internal friction, assuming that it is viscous; then the relationship between stress and deformation is described by the relations
;
.(223)
Let a rod with distributed parameters perform free longitudinal vibrations. In this case, the longitudinal force will be written in the form
From the equation of motion of the rod element, the relation (174) was obtained
Substituting (224) here, we arrive at the main differential equation
,(225)
which differs from (175) by the second term, which expresses the influence of viscous friction forces.
Following the Fourier method, we look for a solution to equation (225) in the form
,(226)
where the function is only the coordinates x, and the function is only the time t.
In this case, each member of the series must satisfy the boundary conditions of the problem, and the entire sum must also satisfy the initial conditions. Substituting (226) into (225) and requiring that the equality be satisfied for any number r, we get
,(227)
where the primes indicate differentiation with respect to the coordinate x, and the points are differentiation with respect to time t.
Dividing (227) by the product 
, we come to equality
,(228)
left side, which can only depend on the coordinate x, and the right one - only from time t. For equality (228) to be fulfilled identically, it is necessary that both parts be equal to the same constant, which we denote by .
From this follow the equations
(229)
.(230)
Equation (229) does not depend on the viscosity coefficient K and, in particular, remains the same in the case of a perfectly elastic system, when . Therefore, the numbers completely coincide with those found earlier; however, as will be shown below, the value gives only an approximate value of the natural frequency. Note that the eigenshapes are completely independent of the viscous properties of the rod, i.e. the forms of free damped oscillations coincide with the forms of free undamped oscillations.
Now let's move on to equation (230), which describes the process of damped oscillations; its solution has the form
.(233)
Expression (232) determines the rate of decay, and (233) determines the oscillation frequency.
Thus, the complete solution to the problem equation
.(234)
Constant and can always be found based on given initial conditions. Let the initial displacements and initial velocities of all sections of the rod be specified as follows:
;
,(235)
where and are known functions.
Then for , according to (211) and (212), we have
multiplying both sides of these equalities by and integrating over the entire length of the rod, we obtain
(236)
According to the condition of orthogonality of eigenforms, all other terms included in the right-hand sides of these equalities become zero. Now from equalities (236) it is easy to find for any number r.
Considering (232) and (234), we note that the higher the number of the vibration mode, the faster its damping. In addition, the terms included in (234) describe damped oscillations if there is a real number. From (233) it is clear that this occurs only for a few initial values of r as long as the inequality is satisfied
For sufficiently large values r inequality (237) is violated and the quantity becomes imaginary. In this case, the corresponding terms of the general solution (234) will no longer describe damped oscillations, but will represent aperiodic damped motion. In other words, vibrations, in the usual sense of the word, are expressed only by a certain finite part of the sum (234).
All these qualitative conclusions apply not only to the case of longitudinal vibrations, but also to the cases of torsional and bending vibrations.
6.5. Vibrations of variable cross-section bars
In cases where the distributed mass and cross-section of the rod are variable along its length, instead of the longitudinal vibration equation (175), one should proceed from the equation
.(238)
The torsional vibration equation (187) must be replaced by the equation
,(239)
and the equation of transverse vibrations (192) is the equation
.(240)
Equations (238)-(240) with the help of similar substitutions ;;can be reduced to ordinary differential equations for the function
MECHANICS
UDC 531.01/534.112
LONGITUDINAL VIBRATIONS OF A PACK OF RODS
A.M. Pavlov, A.N. Temnov
MSTU im. N.E. Bauman, Moscow, Russian Federation e-mail: [email protected]; [email protected]
In matters of the dynamics of liquid-propellant rockets, an important role is played by the problem of stability of rocket motion when longitudinal elastic oscillations occur. The appearance of such oscillations can lead to the establishment of self-oscillations, which, if the rocket is unstable in the longitudinal direction, can lead to its rapid destruction. The problem of longitudinal oscillations of a package rocket is formulated; a package of rods is used as a calculation model. It is accepted that the liquid in the rocket tanks is “frozen”, i.e. the fluid's own motions are not taken into account. The law of total energy balance for the problem under consideration is formulated and its operator formulation is given. A numerical example is given, for which the frequencies are determined, and the shapes of natural oscillations are constructed and analyzed.
Key words: longitudinal vibrations, frequency and shape of vibrations, package of rods, total energy balance law, self-adjoint operator, vibration spectrum, POGO.
SYSTEM OF RODS LONGITUDINAL VIBRATIONS A.M. Pavlov, AL. Temnov
Bauman Moscow State Technical University, Moscow, Russian Federation e-mail: [email protected]; [email protected]
In questions of dynamics of liquid fuel rockets the problem of motion stability for this rocket has an important role with the appearance of longitudinal elastic vibrations. An occurrence of such kind vibrations can evoke self-vibrations which may cause rapid destruction of the rocket in case of rocket instability within longitudinal direction. The problem on longitudinal vibrations of the liquid fuel rocket based on the packet scheme has been formulated using package rods as computational model. It is assumed that the liquid in the rocket tanks is "frozen", i.e. proper motions of the liquid are not included. For this problem energy conservation principle was formulated and its operator staging is given. There is a numerical example, for which the frequencies have been determined, forms of Eigen vibration were built and analyzed.
Keywords: longitudinal vibrations, eigen modes and frequencies, rods model, energy conservation principle, selfadjoint operator, vibration spectrum, POGO.
Introduction. Currently, in Russia and abroad, launch vehicles of a package layout with identical side blocks evenly distributed around the central block are often used to launch a payload into the required orbit.
Studies of vibrations of package structures encounter certain difficulties associated with the dynamic effect of the side and central blocks. In the case of symmetry of the launch vehicle layout, the complex, spatial interaction of the blocks of the package design can be divided into a finite number of types of vibrations, one of which is the longitudinal vibrations of the central and side blocks. A mathematical model of longitudinal vibrations of such a structure in the form of a package of thin-walled rods is discussed in detail in the work. Rice. 1. Scheme of the central- This article presents the theoretical rod and computational results of the longitudinal
vibrations of a package of rods, complementing the study carried out by A.A. Pity.
Formulation of the problem. Let us consider other longitudinal vibrations of a package of rods consisting of a central rod of length l0 and N side rods of the same length j = l, (l0 > lj), j = 1, 2,..., N, fastened at point A (xA = l) (Fig. 1) with central spring elements with stiffness k.
Let us introduce a fixed reference frame OX and assume that the rigidity of the rods EFj (x), the distributed mass mj (x) and the disturbance q (x,t) are bounded functions of the coordinate x:
0 0 < mj < mj (x) < Mj; (1) 0 Let, during longitudinal vibrations, displacements Uj (x, t) arise in sections of rods with coordinate x, determined by the equations mj (x) ^ - ¿(eFj (x) ^ = qj (x,t), j = 0,1, 2,..., N, (2) boundary conditions for the absence of normal forces at the ends of the rods 3 =0, x = 0, ^ = 1, 2, 0, x = 0, x = l0; conditions of equality of normal forces arising in the rods, EF-3 = F x = l elastic forces of spring elements FпPJ = к (ш (ха) - у (¡,)); (4) EUodX (xa - 0) - EFodX (xa + 0) = , x = xa; condition of equality of displacements at point xa of the central rod Shch (ha-o) = Shch (xa+o) and initial conditions Shch y (x, 0) - Shch (x); , _ u(x, 0) = u(x), where u(x, 0) = "d^1(x, 0). Law of total energy balance. Let's multiply equation (2) by u(x,ξ), integrate over the length of each rod and add up the results using boundary conditions (3) and matching condition (4). As a result we get (( 1 ^ [ (diL 2 TZ (x) "BT" (x+ dt | 2 ^ J 3 w V dt N x „ h 2 .. N „ i. 1 ^ Г „„ , f dп3\ , 1 ^ Гj 1 N /* i dpl 2 1 N fl j EF3 dx +2^Уо И (x - -)(no - Uj)2 dx = / ^ (x, £) them y (x, £) (x, (6) where 8 (x - ¡y) is the Dirac delta function. In equation (6), the first term in curly brackets represents the kinetic energy T (¿) of the system, the second is the potential energy Pr (£), caused by the deformation of the rods, and the third is the potential energy Pk (£) of the spring elements, which in the presence of elastic deformations rods can be written in the form Pk (*) = 2 £ / Cy (¡y) 8 (x - ¡1) E^ (¡y) (ddit (¡1)) 2 (x, Cy = Eu. Equation (6) shows that the change in total energy per unit time of the mechanical system under consideration is equal to the power external influence. In the absence of external disturbance q (x,t), we obtain the law of conservation of total energy: T (t) + Pr (t) + Pk (t) = T (0) + Pr (0) + Pk (0). Cinematography. The energy balance law shows that for any time t the functions Uj (x, t) can be considered as elements of the Hilbert space L2j(; m3 (x)), defined on the length ¡i by the scalar product (us,Vk)j = J mj (x) usVkdx 0 and the corresponding norm. Let us introduce the Hilbert space H equal to the orthogonal sum L2j, H = L20 Ф L21 Ф... Ф L2N, the vector function U = (uo, Ui,..., uN)т and the operator A acting in the space H according to the relation AU = diag (A00U0, A11U1,..., Annun). mj(x)dx\jdx" operators defined on set B (A33) С Н of functions satisfying conditions (3) and (4). The original problem (1)-(5) together with the initial conditions will be written in the form Au = f (*), u (0) = u0, 17(0) = u1, (7) where f (*) = (to (*),51 (*),..., Yam (¿))t. Lemma. 1. If the first two conditions (1) are satisfied, then the operator A in the evolution problem (7) is an unbounded, self-adjoint, positive definite operator in the space H (Au,K)n = (u,AK)n, (Au, u)i > c2 (i, u)i. 2. Operator A generates an energy space NA with a norm equal to twice the potential energy of oscillations of a package of rods 3\^I h)2 = 2П > 0. (8) IIUIIA = £/ EF^^J dx + k £ (uo - U)2 = 2П > 0. < Оператор А неограничен в пространстве Н, поскольку неограничен каждый диагональный элемент А33. Самосопряженность и положительная определенность оператора А проверяются непосредственно: (AU, v)h =/m (x) (-^| (EFo (x) ^j) Vo (x) dx+ +£ jm(x) (- jx) | (ef- (x) dndxa))v-(x) dx=... = EFo (x) uo (x) vo (x) dx - EFo (x) U) (x) vo (x) J EFo (x) uo (x) vo (x) dx - EFo (x) uo (x) ?o (x) + ^^ / EF- (x) u- (x) vo (x) dx - ^^ EF- (x) u- (x) v- (x) J EFo (x) uo (x) v" (x) dx - EFo (xa - 0) uo (xa - 0) vo (xa) + 0 EFo (xa + 0) uo (xa + 0) vo (xa) - £ EF- (/-) u- (/-) v- (/-) + J EF- (x) u- (x) v- (x) dx = J EFo (x) uo (x) vo (x) dx+ -=100 + £ / EF.,- (x) u- (x) g?- (x) dx+ o O(xa)- £ EF- (/-) u- (/-) v?"- (/-) = EFo (x) uo (x) v?"o (x) dx+ -=10 + £ / EF- (x) u- (x) v- (x) dx+ -=1 0 - + £ k (uo (xa) - u- (/-)) (vo (xa) - v- (/-)) = (U, A?)H (AU, U)H = ... = I EF0 (x) u"2 (x) dx - EF0 (x) u0 (x) u0 (x) J EF0 (x) u"0 (x) dx - EF0 (x) u0 (x) u0 (x) + ^^ / EFj (x) u"2 (x) dx - ^^ EFj (x) uj (x) u3 (x) "J EF°(x) u"2 (x) dx 4EF0 (x) u"2 (x) dx+£ JEFj (x) u"2 (x) dx У^ k (u0 (l) uj (l) - u2 (/)) + u0 (l) ^ k (u0 (l) - uj (l)) = EF0 (x) u"2 (x) dx + / EF0 (x) u"0 (x) dx + S / EFj (x) u"2 (x) dx + k ^ (u0 (l) - uj (l))2 > c2 (U, U)H From the above results it follows that the energy norm of the operator A is expressed by formula (8). Solvability of the evolutionary problem. Let us formulate the following theorem. Theorem 1. Let the conditions be satisfied U0 £ D (A1/2) , U0 £ H, f (t) £ C (; H), then problem (7) has a unique weak solution U (t) on the interval, defined by the formula U (t) = U0 cos (tA1/2) +U1 sin (tA1/2) +/sin ((t - s) A1/2) A-1/2f (s) ds. 5 in the absence of external disturbance f (£), the law of conservation of energy is satisfied 1 II A 1/2UИ2 = 1 1 II A1/2U 0|H. < Эволюционная задача (7) - это стандартная задача Коши для дифференциального операторного уравнения гиперболического типа, для которого выполнены все условия теоремы о разрешимости . Natural vibrations of a package of rods. Let us assume that the rod system is not affected by the field of external forces: f (t) = 0. In this case, the movements of the rods will be called free. Free movements of the rods, depending on time t according to the law exp (iwt), will be called natural vibrations. Taking U (x, t) = U (x) eiWÍ in equation (7), we obtain the spectral problem for the operator A: AU - AEU = 0, L = w2. (9) The properties of the operator A allow us to formulate a theorem about the spectrum and properties of eigenfunctions. Theorem 2. Spectral problem (9) about natural vibrations of a package of rods has a discrete positive spectrum 0 < Ai < Л2 < ... < Ak < ..., Ak ^ то and a system of eigenfunctions (Uk (x))^=0, complete and orthogonal in the spaces H and HA, and the following orthogonality formulas are satisfied: (Ufe, Us)H = £ m (xj UfejMSjdx = j=0 0 (Uk= £/T^) d*+ K (“feo - Mfej) (uso -) = Afeífes. j=i Study of the spectral problem in the case of a homogeneous package of rods. Presenting the displacement function m- (x, £) in the form m- (x, £) = m- (x), after separating the variables we obtain spectral problems for each rod: ^Ou + Lm = 0, ^ = 0,1,2,..., N (10) which we write in matrix form 4 £ + Li = 0, A = -,-,-,...,- \ t0 t1 t2 t « u = (u0, u1, u2,..., u«)t. Solution and analysis of the results obtained. Let us denote the displacement functions for the central rod in the section as u01 and in the section as u02 (g). In this case, for the function u02 we move the origin of coordinates to the point with coordinate /. For each rod we present the solution to equation (10) in the form To find the unknown constants in (11), we use the boundary conditions formulated above. From homogeneous boundary conditions it is possible to determine some constants, namely: C02 = C12 = C22 = C32 = C42 = ... = CN 2 = 0. As a result, it remains to find N + 3 constants: C01, C03, C04, C11, C21, C31, C41,..., CN1. To do this, we solve N + 3 equations for N + 3 unknowns. Let us write the resulting system in matrix form: (A) (C) = (0) . Here (C) = (C01, C03, C04, C11, C21, C31, C41,..., Cn 1)t is the vector of unknowns; (A) - characteristic matrix, cos (A1) EF0 A sin (A1) + L sin (L (Zo - 1)) L cos (L (Zo - 1)) 0 00 0 \ -1 0 0000 0 y 00 00 0 000Y a = k soe ^ ^A-L^ ; in = -k co8((.40-01L)1/2 ^ ; 7 = (A4"-1 l) 1/2 ap ((A"1l) 1/2 + k sov ((A"1l) 1/2; (~ \ 1/2 ~ Л= ^Л] ; A--: 3 = 0. To find a non-trivial solution, we take the constant C01 € M as a variable. We have two options: C01 = 0; C01 = 0. Let C01 = 0, then C03 = C04 = 0. In this case, a nontrivial solution can be obtained if 7 = 0 from (12) when the additional condition is met £ s-1 = 0, (13) which can be obtained from the third equation of system (12). As a result, we obtain a simple frequency equation EP (A"1 L)1/2 W ((A"1^1/2 P + zz \V zz K cos ^ (A-/a) 1/2 ^ = 0, j G , coinciding with the frequency equation for a rod elastically fixed at one end, which can be considered as the first partial system. In this case, all possible combinations of movements of the side rods that satisfy condition (13) can be conditionally divided into groups corresponding to different combinations of phases (in the case under consideration, the phase is determined by the sign C.d). If we assume the side rods are identical, then we have two options: 1) Сд = 0, then the number of such combinations n for different N can be calculated using the formula n = N 2, where is the division function without a remainder; 2) any (or any) of the constants C- are equal to 0, then the number of possible combinations increases and can be determined by the formula £ [(N - m) div 2]. Let Coi = 0, then Cn = C21 = C31 = C41 = ... = CN1 = = C01 (-v/t), where in and y are the complexes included in (12). From system (12) we also have: C03 = C01 cos (А/); C04=C03 tg (L (/0 - /)) = C01 cos (A/) x x tg (L (/0 - /)), i.e. all constants are expressed through C01. The frequency equation takes the form EFo U-o1 L tg A-1 L) " (lo - l)) - K2 cos | í a!-,1 L As an example, consider a system with four side bars. In addition to the method described above, for this example, you can write down the frequency equation for the entire system by calculating the determinant of matrix A and equating it to zero. Let's look at it Y4 (L sin (L (/o - /)) cos (L/) EFoЛ+ L cos (L (/o - /)) (EFoЛ sin (L/) + 4v)) - 4av3L cos (L(/0 - /)) = 0. Graphs of transcendental frequency equations for the cases considered above are presented in Fig. 2. The following were taken as initial data: EF = 2,109 N; EF0 = 2.2 109 N; k = 7 107 N/m; m = 5900 kg/m; mo = 6000 kg/m; / = 23; /о = 33 m. The values of the first three oscillation frequencies of the considered circuit are given below: n...................................... and, glad/s................................... 1 2 3 20,08 31,53 63,50 Rice. 2. Graphs of transcendental frequency equations for Coi = 0 (i) and Coi = 0 (2) Let us present the vibration modes corresponding to the obtained solutions (in the general case, the vibration modes are not normalized). The vibration forms corresponding to the first, second, third, fourth, 13 and 14 frequencies are shown in Fig. 3. At the first vibration frequency, the side rods vibrate with the same shape, but in pairs in antiphase Fig.3. Forms of vibration of the side (1) and central (2) rods, corresponding to the first V = 3.20 Hz (a), the second V = 5.02 Hz (b), the third V = 10.11 Hz (c), the fourth V = 13.60 Hz (d), 13th V = 45.90 Hz (d) and 14th V = 50.88 Hz (f) frequencies (Fig. 3, a), with the second, the central rod oscillates, and the side ones oscillate in the same shape in phase (Fig. 3, b). It should be noted that the first and second vibration frequencies of the rod system under consideration correspond to vibrations of a system consisting of solid bodies. When the system oscillates with the third natural frequency, nodes appear for the first time (Fig. 3c). The third and subsequent frequencies (Fig. 3d) correspond to elastic vibrations of the system. With an increase in the frequency of vibrations, associated with a decrease in the influence of elastic elements, the frequencies and shapes of vibrations tend to be partial (Fig. 3, e, f). The curves of functions, the points of intersection of which with the abscissa axis are solutions of transcendental equations, are presented in Fig. 4. According to the figure, the natural frequencies of oscillations of the system are located near the partial frequencies. As noted above, with increasing frequency, the convergence of natural frequencies with partial ones increases. As a result, the frequencies at which the entire system oscillates are conventionally divided into two groups: those close to the partial frequencies of the side rod and frequencies close to the partial frequencies of the central rod. Conclusions. The problem of longitudinal vibrations of a package of rods is considered. The properties of the posed boundary value problem and the spectrum of its eigenvalues are described. A solution to the spectral problem for an arbitrary number of homogeneous side rods is proposed. For a numerical example, the values of the first oscillation frequencies are found and the corresponding shapes are constructed. Some characteristic properties of the constructed vibration modes were also identified. Rice. 4. The curves of functions, the points of intersection of which with the abscissa axis are solutions to transcendental equations, for CoX = 0 (1), Cox = 0 (2) coincide with the first partial system (side rod fixed to the elastic element at point x = I) and second partial system (5) (central rod fixed to four elastic elements at point A) LITERATURE 1. Kolesnikov K.S. Dynamics of rockets. M.: Mechanical Engineering, 2003. 520 p. 2. Ballistic missiles and launch vehicles / O.M. Alifanov, A.N. Andreev, V.N. Gushchin et al. M.: Bustard, 2004. 511 p. 3. Rabinovich B.I. Introduction to the dynamics of spacecraft launch vehicles. M.: Mechanical Engineering, 1974. 396 p. 4. Parameter study on POGO stability of liquid rockets / Z. Zhao, G. Ren, Z. Yu, B. Tang, Q. Zhang // J. of Spacecraft and Rockets. 2011. Vol. 48. Is. 3. P. 537-541. 5. Balakirev Yu.G. Methods for analyzing longitudinal vibrations of liquid-propelled launch vehicles // Cosmonautics and Rocket Science. 1995. No. 5. P. 50-58. 6. Balakirev Yu.G. Features of the mathematical model of a liquid rocket of a batch arrangement as a control object // Selected problems of strength of modern mechanical engineering. 2008. pp. 43-55. 7. Dokuchaev L.V. Improving methods for studying the dynamics of a package launch vehicle, taking into account their symmetry // Cosmonautics and Rocket Science. 2005. No. 2. P. 112-121. 8. Pozhalostin A.A. Development of approximate analytical methods for calculating natural and forced vibrations of elastic shells with liquid: dis. ... Dr. Tech. Sci. M., 2005. 220 p. 9. Crane S.G. Linear differential equations in Banach spaces. M.: Nauka, 1967. 464 p. 10. Kopachevsky I.D. Operator methods of mathematical physics. Simferopol: LLC "Forma", 2008. 140 p. Kolesnikov K.S. Dinamika raket. Moscow, Mashinostroenie Publ., 2003. 520 p. Alifanov O.N., Andreev A.N., Gushchin V.N., eds. Ballisticheskie rakety i rakety-nositeli. Moscow, Drofa Publ., 2003. 511 p. Rabinovich B.I. Vvedenie v dinamiku raket-nositeley kosmicheskikh apparatov. Moscow, Mashinostroenie Publ., 1974. 396 p. Zhao Z., Ren G., Yu Z., Tang B., Zhang Q. Parameter study on POGO stability of liquid fuel rocket. J. Spacecraft and Rockets, 2011, vol. 48, iss. 3, pp. 537-541. Balakirev Yu.G. Methods of analysis of longitudinal vibrations of launch vehicles with liquid propellant engine. Kosm. i raketostr. , 1995, no. 5, pp. 50-58 (in Russ.). Balakirev Yu.G. Osobennosti matematicheskoy modeli zhidkostnoy rakety paketnoy komponovki kak ob"ekta upravlenii. Sb. "Izbrannye problemy prochnosti sovremennogo mashinostroeniya". Moscow, Fizmatlit Publ., 2008. 204 p. (cited pp. 4355). Dokuchaev L.V. Improvement of methods for studying the dynamics of clustered launch vehicles considering their symmetry. Kosm. i raketostr. , 2005, no. 2, pp. 112-121 (in Russ.). Pozhalostin A.A. Razrabotka priblizhennykh analiticheskikh metodov rascheta sobstvennykh i vynuzhdennykh kolebaniy uprugikh obolochek s zhidkost"yu. Diss. doct. tekhn. nauk . Kreyn S.G. Lineynye differentsial"nye uravneniya v Banakhovykh prostranstvakh. Moscow, Nauka Publ., 1967. 464 p. Kopachevskiy I.D. Operatornye metody matematicheskoy fiziki. Simferopol", Forma Publ., 2008. 140 p. The article was received by the editor on April 28, 2014 Pavlov Arseniy Mikhailovich - student of the Department of Spacecraft and Launch Vehicles at Moscow State Technical University. N.E. Bauman. Specializes in the field of rocket and space technology. MSTU im. N.E. Baumash, Russian Federation, 105005, Moscow, 2nd Baumanskaya st., 5. Pavlov A.M. - student of "Spacecrafts and Launch Vehicles" department of the Bauman Moscow State Technical University. Specialist in the field of rocket-and-space technology. Bauman Moscow State Technical University, 2-ya Baumanskaya st. 5, Moscow, 105005 Russian Federation. Temnov Alexander Nikolaevich - Ph.D. physics and mathematics Sciences, Associate Professor of the Department of Spacecraft and Launch Vehicles, Moscow State Technical University. N.E. Bauman. Author of more than 20 scientific papers in the field of fluid and gas mechanics and rocket and space technology. MSTU im. N.E. Baumash, Russian Federation, 105005, Moscow, 2nd Baumanskaya st., 5. Temnov A.N. - Cand. Sci. (Phys.-Math.), assoc. professor of "Spacecrafts and Launch Vehicles" department of the Bauman Moscow State Technical University. Author of more than 20 publications in the field of fluid and gas mechanics and rocket-and-space technology. Bauman Moscow State Technical University, 2-ya Baumanskaya st. 5, Moscow, 105005 Russian Federation. Let us consider a uniform rod of length, i.e., a body of cylindrical or some other shape, to stretch or bend which a certain force must be applied. The latter circumstance distinguishes even the thinnest rod from a string, which, as we know, bends freely. In this chapter, we will apply the method of characteristics to the study of longitudinal vibrations of a rod, and we will limit ourselves to studying only such vibrations in which the cross sections, moving along the axis of the rod, remain flat and parallel to each other (Fig. 6). Such an assumption is justified if the transverse dimensions of the rod are small compared to its length. If the rod is slightly stretched or compressed along the longitudinal axis and then left to itself, then longitudinal vibrations will arise in it. Let us direct the axis along the axis of the rod and assume that in a state of rest the ends of the rod are at the points Let the abscissa of a certain section of the rod when the latter is at rest. Let us denote by the displacement of this section at the moment of time, then the displacement of the section with the abscissa will be equal to From here it is clear that the relative elongation of the rod in the section with the abscissa x is expressed by the derivative Now assuming that the rod undergoes small oscillations, we can calculate the tension in this section. Indeed, applying Hooke’s law, we find that where is the elastic modulus of the rod material, its cross-sectional area. Let us take a rod element enclosed between two sections, the abscissas of which at rest are respectively equal. This element is acted upon by tension forces applied in these sections and directed along the axis. The resultant of these forces has the magnitude and is also directed along . On the other hand, the acceleration of the element is equal, as a result of which we can write the equality where is the volumetric density of the rod. Putting and reducing by we obtain the differential equation of longitudinal vibrations of a homogeneous rod The form of this equation shows that the longitudinal vibrations of the rod are of a wave nature, and the speed a of propagation of longitudinal waves is determined by formula (4). If the rod is also acted upon by an external force calculated per unit of its volume, then instead of (3) we obtain This is the equation of forced longitudinal vibrations of the rod. As in dynamics in general, the equation of motion (6) alone is not enough to completely determine the motion of the rod. It is necessary to set the initial conditions, i.e. set the displacements of the sections of the rod and their velocities at the initial moment of time where and are given functions in the interval ( In addition, boundary conditions at the ends of the rod must be specified. For example. In this section we will consider the problem of longitudinal vibrations of a homogeneous rod. A rod is a cylindrical (in particular, prismatic) body, to stretch or compress which a certain force must be applied. We will assume that all forces act along the axis of the rod and each of the cross sections of the rod (Fig. 23) moves translationally only along the axis of the rod. Usually this assumption is justified if the transverse dimensions of the rod are small compared to its length, and the forces acting along the axis of the rod are relatively small. In practice, longitudinal vibrations most often occur when the rod is first slightly stretched or, conversely, compressed and then left to its own devices. In this case, free longitudinal vibrations arise in it. Let us derive the equations for these oscillations. Let's direct the abscissa axis along the axis of the rod (Fig. 23); in a state of rest, the ends of the rod have abscissas, respectively. Consider the cross section; - its abscissa is at rest. The displacement of this section at any time t will be characterized by a function to find which we must create a differential equation. Let us first find the relative elongation of the section of the rod limited by the sections. If the abscissa of the section is at rest, then the displacement of this section at time t, accurate to infinitesimals of higher order, is equal to Therefore, the relative elongation of the rod in the section with the abscissa at time t is equal to Assuming that the forces causing this elongation obey Hooke’s law, we will find the magnitude of the tension force T acting on the section: where is the cross-sectional area of the rod, and is the elastic modulus (Young’s modulus) of the rod material. Formula (5.2) should be well known to the reader from the course on strength of materials. Accordingly, the force acting on the section is equal to Since the forces replace the action of the discarded parts of the rod, their resulting force is equal to the difference Considering the selected section of the rod to be a material point with mass , where is the volumetric density of the rod, and applying Newton’s second law to it, we create the equation By abbreviating by and introducing the notation we obtain the differential equation of free longitudinal vibrations of the rod If we additionally assume that an external force calculated per unit volume and acting along the axis of the rod is applied to the rod, then a term will be added to the right side of relation (5 3) and equation (5.4) will take the form which exactly coincides with the equation of forced oscillations of the string. Let us now move on to establishing the initial and boundary conditions of the problem and consider the practically most interesting case, when one end of the rod is fixed and the other is free. At the free end, the boundary condition will have a different form. Since there are no external forces at this end, the force T acting in the section must also be equal to zero, i.e. Oscillations occur because at the initial moment the rod was deformed (stretched or compressed) and certain initial velocities were imparted to the points of the rod. Therefore, we must know the displacement of the cross sections of the rod at the moment as well as the initial velocities of the points of the rod So, the problem of free longitudinal vibrations of a rod fixed at one end, arising due to initial compression or tension, led us to the equation with initial conditions and boundary conditions It is the last condition that, from a mathematical point of view, distinguishes the problem under consideration from the problem of oscillations of a string fixed at both ends. We will solve the problem posed by the Fourier method, i.e., find partial solutions of the equation that satisfy the boundary conditions (5.8) in the form Since the further course of the solution is similar to that already outlined in § 3, we will limit ourselves to only brief instructions. Differentiating the function , substituting the resulting expressions into (5.6) and separating the variables, we obtain (We leave it to the reader to independently establish that, due to boundary conditions, the constant on the right side cannot be a positive number or zero.) The general solution of the equation has the form Due to the conditions imposed on the function we will have Solutions that are not identically equal to zero will be obtained only if the condition is met, i.e., for , where k can take values So, the eigenvalues of the problem are the numbers Each has its own function As we already know, by multiplying any of the eigenfunctions by an arbitrary constant, we will obtain a solution to the equation with the set boundary conditions. It is easy to check that by giving the number k negative values, we will not obtain new eigenfunctions (for example, at will result in a function that differs from the eigenfunction ) only in sign), Let us first prove that the eigenfunctions (5.11) are orthogonal in the interval . Indeed, when If then It is possible to prove the orthogonality of eigenfunctions in another way, not relying on their explicit expressions, but using only the differential equation and boundary conditions. Let and be two distinct eigenvalues, and be the corresponding eigenfunctions. By definition, these functions satisfy the equations and boundary conditions. Let's multiply the first equation by the second by and subtract one from the other.
(5.2)
