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S.B. Karavashkin, O.N. Karavashkina

4.  Solutions transformation when transforming the model

We can essentially complicate the basic model using the standard procedures. By the superposition, we can take into account the external force acting simultaneously on a few line elements. By the spectral expansion, we can model the response of an elastic line to the external unharmonic action. We can complicate the vibration, extending (2) - (4) to the inclined force action. In this case the solution will describe in the implicit form the inclined waves propagating from the external force action point. The main regularities of vibration process will correspond to those considered in the items 3.1 – 3.4. When transiting to a line with distributed parameters, the inclined pattern of wave process will remain in complete correspondence with results presented in [4].

A heterogeneous elastic infinite line can serve the basic model also for a number of other types of elastic lines having an alike structure. We can obtain for them also the exact analytic solutions by way of corresponding transformation of solutions (2) – (4). For example, using the conventional analogy between the linear and rotary motion of solid body, we can obtain solutions for a system of elastically linked disks. Extending additionally the limiting process to a distributed line, we can investigate the torsion vibrations of heterogeneous shafts, cables and so on. Using the dynamical electromechanical analogy DEMA described in [10] and [12], we can obtain a full complex of solutions for complex electric filters. By way of modelling system (1) simple modification, we can take into account the resistance of an elastic line, remaining the solutions complete, analytical and exact, etc.

As an example, consider a few simple and visual models.

4.1. m1 = m2

The transformation of a heterogeneous elastic line into that homogeneous equalises the parameters characterising both parts of a heterogeneous line. It means that in this case

(24)
If noting (24), the basic system (2) – (4) takes the following form:

for i equless.gif (841 bytes)k 

(25)
for k equless.gif (841 bytes)i equless.gif (841 bytes)n

(26)
and for i equmore.gif (841 bytes)n + 1

(27)

Solutions (25) – (28) completely coincide with the results presented in [1], where the similar solutions were obtained by way of direct use of the presented method to obtain exact analytical solutions. This coincidence of results serves to a definite extent the check of correctness of solutions given here.

4.2.  m2 = 0

In this case we can consider the elastic line as a semi-finite line with unfixed end. With it the equalities

(28)

are true. With regard to (28), the solutions (2) – (4) will take the form:

for i equless.gif (841 bytes)k

(29)
for k equless.gif (841 bytes)i equless.gif (841 bytes)n

(30)
and for i equmore.gif (841 bytes)n + 1

(31)

At i = n the expression (31) will coincide with (30) and will not depend on the index i. Thus we can ignore it in the general system of solutions (29) – (30). The rest two expressions for the first and second sections fully coincide with the corresponding results obtained in [3]. If we go on with this transformation and take additionally k = n , we will obtain the solutions for a homogeneous semi-finite elastic line under external force acting on its free end:

(32)

This result coincides with the solution presented in [1].

 

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