Modern problems of science and education. Longitudinal and transverse waves Longitudinal vibrations
DEFINITION
Longitudinal wave- this is a wave, during the propagation of which the displacement of the particles of the medium occurs in the direction of the wave propagation (Fig. 1, a).
The cause of the occurrence of a longitudinal wave is compression / extension, i.e. the resistance of a medium to a change in its volume. In liquids or gases, such deformation is accompanied by rarefaction or compaction of the particles of the medium. Longitudinal waves can propagate in any media - solid, liquid and gaseous.
Examples of longitudinal waves are waves in an elastic rod or sound waves in gases.
transverse waves
DEFINITION
transverse wave- this is a wave, during the propagation of which the displacement of the particles of the medium occurs in the direction perpendicular to the propagation of the wave (Fig. 1b).
The cause of a transverse wave is the shear deformation of one layer of the medium relative to another. When a transverse wave propagates in a medium, ridges and troughs are formed. Liquids and gases, unlike solids, do not have elasticity with respect to layer shear, i.e. do not resist shape change. Therefore, transverse waves can propagate only in solids.
Examples of transverse waves are waves traveling along a stretched rope or along a string.
Waves on the surface of a liquid are neither longitudinal nor transverse. If you throw a float on the surface of the water, you can see that it moves, swaying on the waves, in a circular fashion. Thus, a wave on a liquid surface has both transverse and longitudinal components. On the surface of a liquid, waves of a special type can also occur - the so-called surface waves. They arise as a result of the action and force of surface tension.
Examples of problem solving
EXAMPLE 1
| Exercise | Determine the direction of propagation of the transverse wave if the float at some point in time has the direction of velocity indicated in the figure. |
| Decision | Let's make a drawing. Let's draw the surface of the wave near the float after a certain time interval , considering that during this time the float went down, since it was directed down at the moment of time. Continuing the line to the right and to the left, we show the position of the wave at time . Comparing the position of the wave at the initial moment of time (solid line) and at the moment of time (dashed line), we conclude that the wave propagates to the left. |
Under the rod we mean the cylinder П=0х[О, /], when I" diamD. Here D- area on the coordinate plane Ox 2 x 3 (Fig. 62). The rod material is homogeneous and isotropic, and the Ox axis passes through the center of gravity of the section D. The field of external body forces f(r, I)\u003d / (X |, /) e, where e, is the unit vector of the Ox axis. Let the external surface forces on the lateral surface of the cylinder be equal to zero, i.e. Ra= 0 on dd X
Then from (4.8) it follows for 1=0 equality
Own forms X k(j) it is convenient to normalize using the norm of the space /^() to which the function belongs v(s, I), since at each moment of time there exists and is limited by the kinetic energy functional
where S- area of the region D. We have
X*(s) = Jj- sin^-l in the space of velocities R 0 = ji)(s, /): v(s,t)e
As a result, we obtain an orthonormal basis |l r *(^)| ,
where b to „- Kronecker symbol: Functions X k *(s), k= 1,2, are the normal forms of natural oscillations, and u*, k= 1, 2, ..., - natural oscillation frequencies of the system with an infinite number of degrees of freedom.
In conclusion, we note that the function u(s, /) belongs to the configuration space of the system H, = (v(s, t): v(s, t) e e ^(), u(0, 1) = o(1, /) \u003d 0), where U ^ "OO, /]) is the Sobolev space of functions summed together with the squares of the first derivatives on the segment . The space R, is the domain of definition of the potential energy functional of elastic deformations
and contains generalized solutions of the problem under consideration.
Longitudinal waves
Definition 1
A wave in which oscillations occur in the direction of its propagation. An example of a longitudinal wave is a sound wave.
Figure 1. Longitudinal wave
Mechanical longitudinal waves are also called compressional or compressional waves because they produce compression as they move through a medium. Transverse mechanical waves are also called "T-waves" or "shear waves".
Longitudinal waves include acoustic waves (the speed of particles propagating in an elastic medium) and seismic P-waves (created as a result of earthquakes and explosions). In longitudinal waves, the displacement of the medium is parallel to the direction of wave propagation.
sound waves
In the case of longitudinal harmonic sound waves, the frequency and wavelength can be described by the formula:
$y_0-$ oscillation amplitude;\textit()
$\omega -$ wave angular frequency;
$c-$ wave speed.
The usual frequency $\left((\rm f)\right)$ of the wave is given by
The speed of sound propagation depends on the type, temperature, and composition of the medium through which it propagates.
In an elastic medium, a harmonic longitudinal wave travels in a positive direction along the axis.
transverse waves
Definition 2
transverse wave- a wave in which the direction of the molecules of the vibrations of the medium is perpendicular to the direction of propagation. An example of transverse waves is an electromagnetic wave.
Figure 2. Longitudinal and transverse waves
Ripples in a pond and waves on a string are easy to imagine as transverse waves.
Figure 3. Light waves are an example of a transverse wave.
Shear waves are waves that oscillate perpendicular to the direction of propagation. There are two independent directions in which wave motions can occur.
Definition 3
2D shear waves exhibit a phenomenon called polarization.
Electromagnetic waves behave in the same way, although it's a little harder to see. Electromagnetic waves are also two-dimensional transverse waves.
Example 1
Prove that the plane undamped wave equation $(\rm y=Acos)\left(\omega t-\frac(2\pi )(\lambda )\right)x+(\varphi )_0$ for the wave shown in the figure , can be written as $(\rm y=Asin)\left(\frac(2\pi )(\lambda )\right)x$. Verify this by substituting the values of the $\ \ x$ coordinate, which are equal to $\frac(\lambda)(4)$; $\frac(\lambda)(2)$; $\frac(0.75)(\lambda)$.
Figure 4
The equation $y\left(x\right)$ for a plane undamped wave does not depend on $t$, which means that the time $t$ can be chosen arbitrarily. We choose the time $t$ such that
\[\omega t=\frac(3)(2)\pi -(\varphi )_0\] \
Substitute this value into the equation:
\ \[=Acos\left(2\pi -\frac(\pi )(2)-\left(\frac(2\pi )(\lambda )\right)x\right)=Acos\left(2\ pi -\left(\left(\frac(2\pi )(\lambda )\right)x+\frac(\pi )(2)\right)\right)=\] \[=Acos\left(\left (\frac(2\pi )(\lambda )\right)x+\frac(\pi )(2)\right)=Asin\left(\frac(2\pi )(\lambda )\right)x\] \ \ \[(\mathbf x)(\mathbf =)\frac((\mathbf 3))((\mathbf 4))(\mathbf \lambda )(\mathbf =)(\mathbf 18),(\mathbf 75)(\mathbf \ cm,\ \ \ )(\mathbf y)(\mathbf =\ )(\mathbf 0),(\mathbf 2)(\cdot)(\mathbf sin)\frac((\mathbf 3 ))((\mathbf 2))(\mathbf \pi )(\mathbf =-)(\mathbf 0),(\mathbf 2)\]
Answer: $Asin\left(\frac(2\pi )(\lambda )\right)x$
Free oscillations of systems with distributed parameters
The main feature of the process of free oscillations of systems with an infinite number of degrees of freedom is expressed in the infinity of the number of natural frequencies and modes of oscillations. This is also related to the features of a mathematical nature: instead of ordinary differential equations describing the oscillations of systems with a finite number of degrees of freedom, here one has to deal with partial differential equations. In addition to the initial conditions that determine the initial displacements and velocities, it is necessary to take into account the boundary conditions that characterize the fixing of the system.
6.1. Longitudinal vibrations of rods
When analyzing the longitudinal vibrations of a rectilinear rod (Fig. 67, a), we will assume that the cross sections remain flat and that the rod particles do not make transverse movements, but move only in the longitudinal direction.
Let be u - longitudinal displacement of the current section of the rod during vibrations; this displacement depends on the location of the section (x coordinates) and on time t. So there is a function of two variables; its definition is the main task. The movement of an infinitely close section is equal, therefore, the absolute elongation of an infinitely small element is (Fig. 67, b), and its relative elongation is .
Accordingly, the longitudinal force in the section with the coordinate X can be written in the form
,(173)
where is the tensile (compressive) stiffness of the rod. The force N is also a function of two arguments - the coordinates X and time t.
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 denoted by the density of the material of the rod, then the mass of the element under consideration is . Therefore, the equation of motion in the projection onto the axis X
,
Considering(173) and assuming A= const , we get
Following the Fourier method, we are looking for a particular solution of the differential equation (175) in the form
,(177)
those. Let's assume that moving 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 obtain
where the primes denote the operation of differentiation with respect to x, and dots on t. Let's rewrite this equation like this:
Here the left side depends only on x, and the right side depends only on t. For the identical fulfillment of this equality (for any x and t ) it is necessary that each of its parts be equal to a constant, which we denote by:
; .(178)
Two equations follow from this:
;
.(179)
The first equation has a solution:
,(180)
indicating an oscillatory character, 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 form of vibrations.
The frequency equation that determines the value of , is compiled by using the 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 give a complete description of the motion. The complete solution is obtained by superimposing all particular solutions:
.
The functions X n (x ) are called own functions tasks and describe their own modes of oscillation. 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 variants of boundary conditions.
Fixed rod end(Fig. 68, a). In the end section, the displacement u must be equal to zero; hence it follows that in this section
X=0(182)
Free rod end(Fig. 68b). In the end section, the longitudinal force
(183)
must be identically equal to zero, which is possible if X"=0 in the end section.
resiliently fixed rod end(Fig. 68, c).
When moving u of the end rod, an elastic reaction of the support occurs 
, where C about - 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.
The force of inertia developed by the mass:
.
Since, according to the first of equations (179), , then the force of inertia can be written as . We get the boundary condition
,
if the mass is at the left end (Fig. 68, e), and
, (184)
if the mass is connected to the right end (Fig. 68, f).
Let us determine the natural frequencies of the cantilever rod (Fig. 68, a").
According to (182) and (183), the boundary conditions
X=0 at x=0;
X"=0 when 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 (when n=2):
Let us determine the natural frequencies of the rod with 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, taking into account (176), has the form
.
Here, the right side is 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.
For and the values of the most important lowest root will be 0.32 and 0.65 respectively.
With a small ratio, the load has a decisive influence and good results are obtained by an approximate solution
.
For bars of variable cross section, i.e. at 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, one has to resort to approximate methods for determining natural frequencies.
6.2. Torsional vibrations of shafts
The torsional vibrations of the shaft with a continuously distributed mass (Fig. 69, a) are described by equations that completely coincide in structure with the equations of longitudinal vibrations of the rods given above.
Torque M in section with abscissa X is related to the angle of rotation by a differential dependence similar to (173):
where Jp is the polar moment of inertia of the cross section.
In a section at a distance dx, the torque is (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 of a unit length), the equation of motion of an elementary section of the shaft can be written as follows:
,
or like (174):
.
Substituting expression (186) here, with Jp=const we get, similarly to (175):
, (187)
The general solution of equation (187), as well as equation (175), has the form
,
(188)
Eigenfrequencies 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;
in) elastically fixed left end: СoХ=GJpX "(Co-coefficient of stiffness);
G) elastically fixed 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);
f) 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 (185):
(n=1,2,…).
If the left end is fixed, and there is a disk at the right end, we get the transcendental equation:
.
If both ends of the shaft are fixed, then the boundary conditions will be X=0 at 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 at x=0; Jo X=GJpX" at x=.
Using (188), we find
C=0; 
,
or the transcendental frequency equation:
.
6.3 Flexural vibrations of beams
6.3.1 Basic equation
From the course of resistance of materials, differential dependencies are known for bending beams:
where EJ - bending stiffness; y \u003d 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 oscillations, the load for the elastic skeleton is the distributed inertia forces:
where m is the mass intensity of the beam (mass per unit length), and equation (191) becomes
.
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 identical, 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 motion is oscillatory with frequency .
The second equation defines the shape of the oscillations. The solution of 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)
are 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 eigenfrequencies 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 bar end, you can specify two boundary conditions .
Free rod end(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 end of the rod(Fig. 70b). 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 angle of rotation are equal to zero. Border conditions:
X=0; X"=0 . (204)
At the end of the rod there is a point mass(Fig. 70d). His force of inertia 
can be written using equation (194) as follows: ; it must be equal to the transverse force Q=EJX"""T , so the boundary conditions take the form
; X""=0 .(205)
In the first condition, the plus sign is accepted in the case when the point weight is connected to the left end of the rod, and the 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, e). Here the bending moment is equal to zero, and the transverse force Q=EJX"""T is equal to the reaction of the support 
(C o -coefficient of rigidity of the support).
Border conditions:
X""=0 ; (206)
(the minus sign is taken when the elastic support is left, and the plus sign when it is right).
6.3.3. Frequency equation and eigenforms
An expanded record of the boundary conditions leads to homogeneous equations for the constants C 1 , C 2 , C 3 , C 4 .
For these constants not to be equal to zero, the determinant composed 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 found out, i.e. eigenmodes of oscillations are determined (up to a constant factor).
Let's trace the compilation of frequency equations using examples.
For a beam with hinged ends according to (203) we have the following boundary conditions: X=0; X""=0 when x=0 and x= . With the help of (197)-(200) we obtain from the first two conditions: C 1 =C 3 =0. The remaining two conditions can be written as
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 , then the final frequency equation is written as follows:
. (207)
The roots of this equation are:
,(n=1,2,3,...).
Taking into account (196), we obtain
.(208)
Let's move on to defining our own forms. From the homogeneous equations written above, the following relation follows between the constants C 2 and C 4:
.
Consequently, (197) takes the form
According to (207), we have
,(209)
where is a new constant, the value of which remains undetermined until the initial conditions are introduced into consideration.
6.3.4. Definition of motion by initial conditions
If it is required to determine the movement following the initial perturbation, then it is necessary to specify both initial displacements and initial velocities for all points of the beam:
(210)
and use the orthogonality property of eigenshapes:
.
We write the general solution (201) as follows:
.(211)
The speed is determined by the expression
.(212)
Substituting in the right parts of equations (211) and (212) , and in the left parts - the assumed known initial displacements and velocities, we obtain
.
Multiplying these expressions by and integrating over the entire length, we have
(213)
The infinite sums on the right hand sides have disappeared due to the orthogonality property. From (213) formulas follow for the constants and
(214)
Now these results must be substituted into solution (211).
Again, we emphasize that the choice of the scale of proper shapes is not essential. If, for example, in the expression of its own form (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 cancel each other out. Nevertheless, normalized eigenfunctions are often used, choosing their scale so that the denominators of expressions (214) are equal to one, which simplifies the expressions and .
6.3.5. Influence of constant longitudinal force
Let us consider the case when the oscillating beam experiences the action of a longitudinal force N, the value of which does not change during the oscillation process. In this case, the static bending equation becomes more complicated and takes the form (assuming that the compressive force is considered positive)
.
Assuming and assuming the stiffness to be constant, we obtain the equation of free vibrations
.(215)
We still take 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 like this:
(216)
where K is determined by formula (196), and
The solution of equation (216) has the form
Consider the case when both ends of the rod have hinged supports. Conditions at the left end 
give . Satisfying the same conditions at the right end, we get
Equating to zero the determinant, composed of the coefficients at the values and , we arrive at the equation
The roots of this frequency equation are:
Therefore, the natural frequency is determined from the equation
.
Hence, taking into account (217), 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 to be given and independent of the displacements of the system. In some practical problems, the longitudinal force that accompanies the process of transverse vibrations arises due to the bending of the beam and is in the nature of the reaction of the support. Consider, for example, a beam on two hinged-fixed supports. When it is bent, horizontal reactions of the supports occur, causing the beam to stretch; the corresponding horizontal force is called chain force. If the beam makes transverse vibrations, then the chain force will change with time.
If at an instant t the deflections of the beam are determined by the function , then the elongation of the axis can be found by the formula
.
The corresponding chain force can be found using Hooke's law
.
We substitute this result in (215) instead of the longitudinal force N (taking into account the sign)
.(220)
The resulting non-linear integro-differential the equation is simplified by substituting
,(221)
where is a dimensionless function of time, the maximum value of which can be set equal to any number, for example, one; oscillation amplitude.
Substituting (221) into (220), we obtain the ordinary differential equation
,(222)
whose coefficients have the following values:
;.
The differential equation (222) is non-linear, 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 oscillations, 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 comparable, the correction to the frequency becomes significant. If, for example, the oscillation amplitude of a rod of circular cross section is equal to its diameter, then , and the frequency is almost two times greater than in the case of free displacement of the supports.
The case corresponds to the zero value of the radius of inertia, when the bending stiffness of the beam is vanishingly small - a string. In this case, the formula for gives an uncertainty. Revealing this uncertainty, we obtain a formula for the frequency of vibrations of the string
.
This formula refers to the case when, in the equilibrium position, the tension is zero. The problem of string vibrations is often posed under other assumptions: it is assumed that the displacements are small, and the tensile force is given and remains unchanged during the vibrations.
In this case, the formula for the frequency has the form
where N is a constant tensile force.
6.4. Influence of viscous friction
Previously, it was assumed that the material of the rods is ideally elastic and there is no friction. Consider the effect of internal friction, assuming that it is viscous; then the relationship between stresses and strains 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 are looking for a solution to equation (225) in the form
,(226)
where the function is only x coordinates, and the function is only 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 equality be satisfied for any number r, we get
,(227)
where the primes denote differentiation with respect to the coordinate x, and the points are differentiation with respect to time t.
Dividing (227) by the product 
, we arrive at the equality
,(228)
the left side, which can only depend on the coordinate x, and the right one - only from time t. For the identical fulfillment of equality (228), it is necessary that both parts are 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 eigenforms 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 looks like
.(233)
Expression (232) determines the damping rate, and (233) determines the oscillation frequency.
Thus, the complete solution of the problem equation
.(234)
Constant and can always be found by given initial conditions. Let the initial displacements and initial velocities of all sections of the rod be given as follows:
;
,(235)
where and are known functions.
Then for , according to (211) and (212), we have
multiplying both parts 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 vanish. Now it is easy to find from equalities (236) for any number r .
Considering (232) and (234), we note that the higher the number of the mode of vibrations , the faster its damping. In addition, the terms in (234) describe damped oscillations if there is a real number. It can be seen from (233) that this takes place only for a few initial values of r as long as the inequality
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 an aperiodic damped motion. In other words, fluctuations, in the usual sense of the word, express only some 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 bars of variable cross section
In those cases when the distributed mass and cross section of the rod are variable along its length, instead of the equation of longitudinal vibrations (175), one should proceed from the equation
.(238)
The torsional vibration equation (187) should be replaced by the equation
,(239)
and the equation of transverse oscillations (192) - by the equation
.(240)
Equations (238)-(240) with the help of substitutions of the same type ;; can be reduced to ordinary differential equations for the function
ISSN: 2310-7081 (online), 1991-8615 (print) doi: http://dx.doi UDC 517.956.3
PROBLEM OF LONGITUDINAL VIBRATIONS OF ELASTICALLY FIXED LOADED ROD
A. B. Beilin
Samara State Technical University, Russia, 443100, Samara, st. Molodogvardeyskaya, 244.
annotation
One-dimensional longitudinal vibrations of a thick short rod fixed at the ends with the help of concentrated masses and springs are considered. As a mathematical model, an initial-boundary value problem with dynamic boundary conditions for a fourth-order hyperbolic equation is used. The choice of this particular model is due to the need to take into account the effects of deformation of the rod in the transverse direction, the neglect of which, as shown by Rayleigh, leads to an error, which is confirmed by the modern non-local concept of studying the vibrations of solids. The existence of a system of eigenfunctions of the problem under study orthogonal with the load is proved and their representation is obtained. The established properties of the eigenfunctions made it possible to apply the method of separation of variables and prove the existence of a unique solution to the problem.
Key words: dynamic boundary conditions, longitudinal vibrations, load orthogonality, Rayleigh model.
Introduction. In any working mechanical system, oscillatory processes occur, which can be generated by various reasons. Oscillatory processes can be a consequence of the design features of the system or the redistribution of loads between various elements of a normally operating structure.
The presence of sources of oscillatory processes in the mechanism can make it difficult to diagnose its condition and even lead to a violation of its operation mode, and in some cases to destruction. Various problems associated with a violation of the accuracy and performance of mechanical systems as a result of the vibration of some of their elements are often solved experimentally in practice.
At the same time, oscillatory processes can be very useful, for example, for processing materials, assembling and disassembling joints. Ultrasonic vibrations allow not only to intensify the cutting processes (drilling, milling, grinding, etc.) of materials with high hardness (tungsten-containing, titanium-carbide steels, etc.),
© 2016 Samara State Technical University. Citation Sample
Beilin, A.B., The problem of longitudinal vibrations of an elastically fixed loaded rod, Vestn. Myself. state tech. university Ser. Phys.-Math. Nauki, 2016. V. 20, No. 2. P. 249258. doi: 10.14498/vsgtu1474. About the author
Alexander Borisovich Beilin (Ph.D., Assoc.; [email protected]), Associate Professor, Dept. automated machine and tool systems.
but in some cases become the only possible method for processing brittle materials (germanium, silicon, glass, etc.). The element of the device (waveguide), which transmits ultrasonic vibrations from the source (vibrator) to the tool, is called a concentrator and can have a different shape: cylindrical, conical, stepped, exponential, etc. . Its purpose is to convey fluctuations of the required amplitude to the instrument.
Thus, the consequences of the occurrence of oscillatory processes can be different, as well as the causes that cause them, therefore, the need naturally arises for a theoretical study of the processes of oscillation. The mathematical model of wave propagation in relatively long and thin solid rods, which is based on a second-order wave equation, has been well studied and has long become a classic. However, as shown by Rayleigh, this model is not quite consistent with the study of vibrations of a thick short rod, whereas many details of real mechanisms can be interpreted as short and thick rods. In this case, the deformations of the rod in the transverse direction should also be taken into account. The mathematical model of longitudinal oscillations of a thick short rod, which takes into account the effects of the transverse motion of the rod, is called the Rayleigh rod and is based on a fourth-order hyperbolic equation
^ ^ - IX (a(x) e) - dx (b(x)) =; (xL (1)
whose coefficients have a physical meaning:
g(x) = p(x)A(x), a(x) = A(x)E(x), b(x) = p(x)u2(x)1p(x),
where A(x) is the cross-sectional area, p(x) is the mass density of the rod, E(x) is Young's modulus, V(x) is Poisson's ratio, 1P(x) is the polar moment of inertia, u(x, b) - longitudinal displacements to be determined.
Rayleigh's ideas have found their confirmation and development in modern works devoted to the processes of vibrations, as well as the theory of plasticity. The review article substantiates the shortcomings of classical models describing the state and behavior of solids under load, in which a priori the body is considered an ideal continuum. The modern level of development of natural science requires the construction of new models that adequately describe the processes under study, and the mathematical methods developed in the last few decades provide this opportunity. On this path, in the last quarter of the last century, a new approach was proposed to the study of many physical processes, including those mentioned above, based on the concept of nonlocality (see the article and the list of references in it). One of the classes of non-local models identified by the authors is called “weakly non-local”. Mathematical models belonging to this class can be implemented by introducing high-order derivatives into the equation describing a certain process, which make it possible to take into account, in some approximation, the interaction of the internal elements of the object of study. Thus, the Rayleigh model is relevant in our time.
1. Statement of the problem. Let the ends of the rod x = 0, x = I be attached to a fixed base with the help of concentrated masses N1, M2 and springs, the stiffnesses of which are K1 and K2. We will assume that the rod is a body of revolution about the 0x axis and the initial moment of time is at rest in the equilibrium position. Then we come to the following initial-boundary value problem.
Task. Find in the area Qt \u003d ((0,1) x (0, T) : 1, T< те} "решение уравнения (1), удовлетворяющее начальным данным
u(x, 0) = (p(x), u(x, 0) = φ(x) and boundary conditions
a(0)ux(0, r) + b(0)uu(0, r) - k^(0, r) - M1u(0, r) = 0, a(1)ux(1, r) + b(1)uu(1, r) + K2u(1, r) + M2uu(1, r) = 0. ()
The article considers some special cases of problem (1)-(2) and gives examples in which the coefficients of the equation have an explicit form and M\ = M2 = 0. The article proves the unambiguous weak solvability of the problem in the general case.
Conditions (2) are determined by the method of fixing the rod: its ends are attached to fixed bases with the help of some devices having masses M1, M2, and springs with stiffnesses K1, K2, respectively. The presence of masses and allowance for transverse displacements leads to conditions of the form (2) containing time derivatives. Boundary conditions that include time derivatives are called dynamic. They can arise in various situations, the simplest of which are described in a textbook, and much more complex ones in a monograph.
2. Study of natural oscillations of the rod. Consider a homogeneous equation corresponding to equation (1). Since the coefficients depend only on x, we can separate the variables by representing u(x, z) = X(x)T(z). We get two equations:
m""(r) + \2m(r) = 0,
((a(x) - A2b(x))X"(x))" + A2dX(x) = 0. (3)
Equation (3) is accompanied by boundary conditions
(a(0) - \2b(0))X"(0) - (K1 - \2M1)X(0) = 0,
(a(1) - \2b(1))X"(1) + (K2 - \2M2)X(I) = 0. (4)
Thus, we have arrived at the Sturm-Liouville problem, which differs from the classical one in that the spectral parameter Λ is included in the coefficient of the highest derivative of the equation, as well as in the boundary conditions. This circumstance does not allow us to refer to results known from the literature, so our immediate goal is to study problem (3), (4). For the successful implementation of the method of separation of variables, we need information about the existence and location of eigenvalues, about qualitative
properties of eigenfunctions: do they have the property of orthogonality?
Let us show that A2 > 0. Let us assume that this is not the case. Let X(x) be an eigenfunction of problem (3), (4) corresponding to the value A = 0. We multiply (3) by X(x) and integrate the resulting equality over the interval (0,1). Integrating by parts and applying boundary conditions (4), after elementary transformations we obtain
1(0) - A2b(0))(a(1) - A2b(1)) I (dX2 + bX"2)dx+
N\X 2(0) + M2X 2(1)
I aX "2<1х + К\Х2(0) + К2Х2(1). Jo
We note that from the physical meaning of the functions a(x), b(x), g(x) are positive, Kr, Mr are non-negative. But then it follows from the resulting equality that X "(x) \u003d 0, X (0) \u003d X (1) \u003d 0, therefore, X (x) \u003d 0, which contradicts the assumption made. Therefore, the assumption that that zero is an eigenvalue of problem (3), (4) is false.
The representation of the solution of equation (3) depends on the sign of the expression a(x) - - A2b(x). Let us show that a(x)-A2b(x) > 0 Vx e (0,1). We fix arbitrarily x e (0, 1) and find the values at this point of the functions a(x), b(x), g(x). We write equation (3) in the form
X "(x) + VX (x) \u003d 0, (5)
where we marked
at the chosen fixed point, and conditions (4) can be written in the form
X "(0) - aX (0) \u003d 0, X" (1) + bX (I) \u003d 0, (6)
where a, b are easy to calculate.
As is known, the classical Sturm-Liouville problem (5), (6) has a countable set of eigenfunctions for V > 0, whence, due to the arbitrariness of x, the desired inequality follows.
The eigenfunctions of problem (3), (4) have the property of orthogonality with the load , expressed by the relation
I (dXm (x) Xn (x) + bX "m (x) X" p (x))<х+ ■)о
M1Xm(0)Xn(0) + M2Xm(1)Xn (I) = 0, (7)
which can be obtained in a standard way (see, for example, ), the implementation of which in the case of the problem under consideration is associated with elementary but painstaking calculations. Let us briefly present its derivation, omitting the argument of the functions Xr(x) in order to avoid cumbersomeness.
Let λm, λn be different eigenvalues, λm, λn be the eigenfunctions of problem (3), (4) corresponding to them. Then
((a - L2mb)X"t)" + L2tdXm = 0, ((a - L2nb)X"n)" + L2pdXp = 0.
We multiply the first of these equations by Xn, and the second by Xm, and subtract the second from the first. After elementary transformations, we obtain the equality
(Lt - Lp) YHtXp \u003d (aXtXP) "- LP (bXtX" p) "- (aX "tXp)" + Rt (bXtXp)",
which we integrate over the interval (0,1). As a result, taking into account (4) and reducing by (Лт - Лп), we obtain relation (7).
Proved statements about the properties of eigenvalues and eigenfunctions of the Sturm-Liouville problem (3), (4) allow us to apply the method of separation of variables to find a solution to the problem.
3. Solvability of the problem. Denote
C(CT) = (u: u e C(St) P C2(St), uixx e C^m)).
Theorem 1. Let a, b e C1 , e C. Then there exists at most one solution u e C(m) of problem (1), (2).
Proof. Let us assume that there are two different solutions to problem (1), (2), u1(x, z) and u2(x, z). Then, due to the linearity of the problem, their difference u = u1 - u2 is a solution to the homogeneous problem corresponding to (1), (2). Let us show that its solution is trivial. We note in advance that, from the physical meaning of the coefficients of the equation and the boundary conditions, the functions a, b, q are positive everywhere in Qm, while M^, K^ are non-negative.
Multiplying equality (1) by u and integrating over the domain Qt, where t e and arbitrarily, after simple transformations, we obtain
/ (di2(x, m) + au2x(x, m) + buXl(x, m)) ux + ./o
K1u2(0, m) + M1u2(0, m) + K2u2(1, m) + M2u2(1, m) = 0,
whence, by virtue of the arbitrariness of m, the assertion of the theorem immediately follows. □
Let us prove the existence of a solution for the case of constant coefficients.
Theorem 2. Let<р е С2, <р(0) = <р(1) = (0) = ц>"(\) = 0, has a third-order piecewise continuous derivative in (0,1), φ e C 1, φ(0) = φ(1) = 0, and has a second-order piecewise continuous derivative in (0,1), f e C(C^m), then the solution to problem (1), (2) exists and can be obtained as the sum of a series of eigenfunctions.
Proof. We will, as usual, look for a solution to the problem in the form of a sum
where the first term is the solution of the formulated problem for the homogeneous equation corresponding to (1), the second is the solution of equation (1) that satisfies zero initial and boundary conditions. Let us use the results of the studies carried out in the previous paragraph and write down the general solution of equation (3):
X(x) = Cr cos A J-+ C2 sin Aw-^rrx.
\¡ a - A2b \¡ a - A2b
Applying boundary conditions (4), we arrive at a system of equations for Cj!
(a - A2b)c2 - (Ki - A2Mi)ci = 0,
(-A(a - A2b) sin Ayja-A¡bl + (K - A2M2) cos A^O-A^l) ci+
Equating its determinant to zero, we obtain the spectral equation ctg \u003d (a - A4) A2 "- (K - A? Mí) (K2 - A "M). (eight) b Va - A2b A^q(a - A2b)(Ki + K2 - A2(Mi + M2)) Let us find out whether this transcendental equation has a solution. To do this, consider the functions that are in its left and right parts, and examine their behavior. Without limiting the generality too much, we set Mi = M2 = M, Kg = K2 = K, which will slightly simplify the necessary calculations. Equation (8) takes the form x I q , Aja - A2b Jq K - A2M ctg A\Z-^l = a - A2b 2(K - A2M) 2A^^0-A2b" and write the spectral equation in new notation! aqlß Kql2 + ß2 (Kb - aM) 2Kql2 + 2^2(Kb - aM) 2/j.aql An analysis of the functions of the left and right parts of the last equation allows us to assert that there is a countable set of its roots and, therefore, a countable set of eigenfunctions of the Sturm-Liouville problem (3), (4), which, taking into account the relation obtained from the system with respect to c¿, can be written v / l l I q K - x2pm. l i q Xn(x) = COS XnJ-myx + ----sin XnJ-myx. V a - A2b AnVa - ftb^q V a - A2b Now we turn to finding a solution that also satisfies the initial conditions. We can now easily find the solution of the problem for the homogeneous equation in the form of a series u(x,t) = ^Tn(t)Xn(x), whose coefficients can be found from the initial data using the orthogonality property of the functions Xn(x), whose norm can be obtained from relation (7): ||X||2 = f (qX2 + bX%)dx + MiX2(0) + M2x2(l). ■Jo The process of finding the function v(x,t) is also essentially standard, but we still notice that, looking for a solution in the traditional form v(x,t) = ^ Tn(t)Xn(x), we get two equations. Indeed, taking into account the form of the eigenfunctions, let us specify the structure of the series in which we are looking for a solution: j(x,t) = ^ (Vn(t)cos Xn^J a b x+ Wn(t) K-XnM~sin X^ GAirx). (nine) v JXnVa - xnb^q V a - xn" To satisfy the zero initial conditions y(x, 0) = y^x, 0) = 0, we require that Yn(0) = Yn(0) = 0, Wn(0) = W(0) = 0. Expanding f( x, d) into a Fourier series with respect to the eigenfunctions Xn(x), we find the coefficients ¡n(b) and dn(b). Substituting (9) into equation (1), written with respect to y(x, b), after a series of transformations, we obtain equations for finding Yn(b) and Shn(b): uc® + >&pYu = ™ + xn Wn (<) = Xn (-a-iKrW g Taking into account the initial conditions Yn(0) = Y,(0) = 0, Shn(0) = W,(0) = 0, we arrive at the Cauchy problems for each of the functions Yn(b) and Shn(b), whose unique solvability guaranteed by the conditions of the theorem. The properties of the initial data formulated in the theorem leave no doubt about the convergence of all the series that have arisen in the course of our research and, therefore, about the existence of a solution to the problem. □ Conclusion. The existence of a system of eigenfunctions of the problem under study orthogonal with the load is proved and their representation is obtained. The established properties of the eigenfunctions made it possible to prove the existence of a unique solution to the problem. Note that the results obtained in the article can be used both for further theoretical studies of problems with dynamic boundary conditions, and for practical purposes, namely, for calculating the longitudinal vibrations of a wide range of technical objects. Alexander Borisovich Beilin: http://orcid.org/0000-0002-4042-2860 REFERENCES 1. Nerubay M. S., Shtrikov B. L., Kalashnikov V. V. Ultrasonic mechanical processing and assembly. Samara: Samara book publishing house, 1995. 191 p. 2. Khmelev V. N., Barsukov R. V., Tsyganok S. N. Ultrasonic dimensional processing of materials. Barnaul: Altai Technical University im. I.I. Polzunova, 1997. 120 p. 3. Kumabe D. Vibration cutting. M.: Mashinostroenie, 1985. 424 p. 4. A. N. Tikhonov and A. A. Samarskii, Equations of Mathematical Physics. M.: Nauka, 2004. 798 p. 5. Strett J. V. Theory of sound. T. 1. M.: GITTL, 1955. 504 p. 6. Rao J. S. Advanced Theory of Vibration: Nonlinear Vibration and One Dimensional Structures. New York: John Wiley & Sons, Inc., 1992. 431 pp. 7. Fedotov I. A., Polyanin A. D., Shatalov M. Yu. The theory of free and forced vibrations of a solid rod based on the Rayleigh model// DAN, 2007. V. 417, no. pp. 56-61. 8. Bazant Z., Jirasek M. Nonlocal Integral Formulations of Plasticity and Damage: Survey of Progress// J. Eng. Mech., 2002. vol. 128, no. 11.pp. 1119-1149. doi: 10.1061/(ASCE) 0733-9399(2002)128:11(1119). 9. A. B. Beilin and L. S. Pulkina, “Problem of Longitudinal Vibrations of a Rod with Dynamic Boundary Conditions,” Vestn. SamGU. Natural Science Ser., 2014. No. 3 (114). pp. 9-19. 10. M. O. Korpusov, Fracture in nonclassical wave equations. M.: URSS, 2010. 237 p. Received 10/II/2016; in the final version - 18/V/2016; accepted for publication - 27/V/2016. Vestn. Samar. gos. Techn. Unta. Ser. Phys.-mat. science 2016, vol. 20, no. 2, pp. 249-258 ISSN: 2310-7081 (online), 1991-8615 (print) doi: http://dx.doi.org/10.14498/vsgtu1474 MSC: 35L35, 35Q74 A PROBLEM ON LONGITUDINAL VIBRATION OF A BAR WITH ELASTIC FIXING Samara State Technical University, 244, Molodogvardeyskaya st., Samara, 443100, Russian Federation. In this paper, we study longitudinal vibration in a thick short bar fixed by point forces and springs. For mathematical model we consider a boundary value problem with dynamical boundary conditions for a forth order partial differential equation. The choice of this model depends on a necessity to take into account the result of a transverse strain. It was shown by Rayleigh that neglect of a transverse strain leads to an error. This is confirmed by modern nonlocal theory of vibration. We prove the existence of orthogonal with load eigenfunctions and derive representation of them. Established properties of eigenfunctions make possible using the separation of variables method and finding a unique solution of the problem. Keywords: dynamic boundary conditions, longitudinal vibration, loaded orthogonality, Rayleigh's model. Alexander B. Beylin: http://orcid.org/0000-0002-4042-2860 1. Nerubai M. S., Shtrikov B. L., Kalashnikov V. V. Ul "trazvukovaia mekhanicheskaia obrabotka i sborka. Samara, Samara Book Publ., 1995, 191 pp. (In Russian) 2. Khmelev V. N., Barsukov R. V., Tsyganok S. N. Ul "trazvukovaia razmernaia obrabotka materialov. Barnaul, 1997, 120 pp. (In Russian) 3. Kumabe J. Vibration Cutting. Tokyo, Jikkyou Publishing Co., Ltd., 1979 (In Japanese). 4. Tikhonov A. N., Samarsky A. A. Uravneniia matematicheskoi fiziki. Moscow, Nauka, 2004, 798 pp. (In Russian) 5. Strutt J. W. The theory of sound, vol. 1. London, Macmillan and Co., 1945, xi+326 pp. 6. Rao J. S. Advanced Theory of Vibration: Nonlinear Vibration and One Dimensional Structures. New York, John Wiley & Sons, Inc., 1992, 431 pp. Beylin A.B. A problem on longitudinal vibration of a bar with elastic fixing, Vestn. Samar. gos. Technology. Univ., Ser. Phys.-Mat. Science, 2016, vol. 20, no. 2, pp. 249-258. doi: 10.14498/vsgtu1474. (In English) Author Details: Alexander B. Beylin (Cand. Techn. Sci.; [email protected]), Associate Professor, Dept. of Automation Machine Tools and Tooling Systems. 7. Fedotov I. A., Polyanin A. D., Shatalov M. Yu. Theory of free and forced vibrations of a rigid rod based on the Rayleigh model, Dokl. Phys., 2007, vol.52, no. 11, pp. 607-612. doi: 10.1134/S1028335807110080. 8. Bazant Z., Jirasek M. Nonlocal Integral Formulations of Plasticity and Damage: Survey of Progress, J. Eng. Mech., 2002, vol.128, no. 11, pp. 1119-1149. doi: 10.1061/(ASCE) 0733-9399(2002)128:11(1119). 9. Beylin A. B., Pulkina L. S. A promlem on longitudinal vibrations of a rod with dynamic boundary conditions, Vestnik SamGU. Estestvenno-Nauchnaya Ser., 2014, no. 3(114), pp. 919 (In Russian). 10. Korpusov M. O. Razrushenie v neklassicheskikh volnovykh uravneniakh. Moscow, URSS, 2010, 237 pp. (In Russian) Received 10/II/2016; received in revised form 18/V/2016;
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