lec18.ppt
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This document discusses solving a mass-spring system as an eigenvalue problem. It: 1) Sets up differential equations to model the displacements of two masses connected by springs. 2) Transforms the coupled differential equations into a matrix eigenvalue equation. 3) Solves the eigenvalue equation to obtain the frequencies of oscillation for the two masses. 4) Combines the eigenvectors with complex exponential functions to obtain general solutions for the displacements of each mass over time.
![NPTEL Physics Mathematical Physics - 1
2 1
2 = μ = (μ2
) = 基 = [
μ1 5 2
2 2 2
] ( ) (5)
One may try a solution of the form,
= 基
(6)
Putting this in (5) leads to the eigenvalue equation,
基 = y where = 2
and = [ 5 2 ]
2 2
A has the
eigenvalues
-
[5 2
2 2
] = 0
1 = 1 and 2 = 6
So, = 1 and 6
= and 6 respectively. The
corresponding e-vectors are-
1 = ( ) corresponding to
1
2
and 2 = ( 2 ) corresponding to 6
1
Thus we obtain four complex solutions
1 = 1( 賊 )
26 = 2(6 賊 6)
Thus the real solutions are
1, 1, 26, 26
A general solution is obtained by taking a linear combination of all the above
solutions.
= 1(1 + 1) + (226 + 26)
with arbitrary constants 1, 2, 1, 2 which can be defined from the initial
conditions. From the structure of the vectors 1 and 2, one can write the final
solutions as-
Joint initiative of IITs and IISc Funded by MHRD Page 10 of 17](https://image.slidesharecdn.com/lec18-231023100908-fc5616e1/85/lec18-ppt-2-320.jpg)
lec18.ppt
- 1. NPTEL Physics Mathematical Physics - 1 Lecture 18 Eigenvalue Problems Example 1. Consider a mass-spring system given by, We shall show that this corresponds to an eigenvalue problem, solution of which will yield the displacements of the two masses and their frequencies of oscillation. Let 1,2 be the displacement variables for the two masses 1,2. One can write the differential equations as, 2 21 = ( + ) + 1 2 1 2 2 for mass 1 (1) 2 2 2 = 2 1 2 2 : for mass (2) 2 Putting values for 1 and 2, 2 1 2 = 51 + 22 (3) 2 2 2 = 21 22 (4) The coupled equations (3) and (4) can be solved using the following eigenvalue equation. Joint initiative of IITs and IISc Funded by MHRD Page 9 of 17
- 2. NPTEL Physics Mathematical Physics - 1 2 1 2 = μ = (μ2 ) = 基 = [ μ1 5 2 2 2 2 ] ( ) (5) One may try a solution of the form, = 基 (6) Putting this in (5) leads to the eigenvalue equation, 基 = y where = 2 and = [ 5 2 ] 2 2 A has the eigenvalues - [5 2 2 2 ] = 0 1 = 1 and 2 = 6 So, = 1 and 6 = and 6 respectively. The corresponding e-vectors are- 1 = ( ) corresponding to 1 2 and 2 = ( 2 ) corresponding to 6 1 Thus we obtain four complex solutions 1 = 1( 賊 ) 26 = 2(6 賊 6) Thus the real solutions are 1, 1, 26, 26 A general solution is obtained by taking a linear combination of all the above solutions. = 1(1 + 1) + (226 + 26) with arbitrary constants 1, 2, 1, 2 which can be defined from the initial conditions. From the structure of the vectors 1 and 2, one can write the final solutions as- Joint initiative of IITs and IISc Funded by MHRD Page 10 of 17
- 3. NPTEL Physics Mathematical Physics - 1 1 = 1 + 1 + 226 + 226 2 = 21 + 21 26 26 These solutions represent the displacement of the coupled masses. Joint initiative of IITs and IISc Funded by MHRD Page 11 of 17