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N-coupled oscillator- both series, and parallel.

by SteveDB
Tags: ncoupled, oscillator, parallel, series
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SteveDB
#1
Nov3-05, 09:37 PM
P: 17
Hi all.
Ok, the title is a bit deceptive, as that is typically used to denote a string type oscillator system. In this case, I'm referring to multiple masses, and multiple springs.
To be specific, we're doing a system with 5 masses, and 13 springs. This means that there are 2 to 4 springs connected either in parallel, or in series to each of the masses.
From what I remember-- parallel springs are treated as:
1/k_eff = 1/k1 + 1/k2+ ... + 1/kn. Where n is the nth spring, and k is the spring constant.
And series springs are just k_eff = k1 + k2 + ... + kn. (the sum of all spring constants.)
Did I get this backwards, or is it correct?
Next, because we're doing a 5 mass system, we'll have a 5 x 5 matrix for the M matrix. Where the masses will be the diagonal elements of the M matrix. The K matrix however will be more complicated. Which is why I'm here.
It seems to me-- from previous lectures, and homework problems-- that the only elements of the K matrix that will have any non-zero values will be the diagonal, and the elements one space to the left, and one space to the right of the diagonal elements. Is this correct?
E.g. for a 2 x 2 matrix, the elements would be
k_11 -k_12
-k_21 k_22
In this case, the k_12 element = k_21. The reason would be due to a non-restoring force acting ON x_2, etc...
If we then expand this matrix out to n x n-- anything larger than a 2 x 2, all other elements--- e.g. sample 3 x 3 mat.
k_11 -k_12 0
-k_21 k_22 -k_23
0 -k_32 k_33
As you can see, only the diagonal, and "immediate" off-diagonal elements are non-zero.
My question-- does this play out in ALL oscillator matrices? I have asked my prof., but he has yet to answer.
In my specific 5 x 5 K matrix, this would in fact give me 13 elements, accounting for the 13 springs of the system.
The general equation to solve this system would be:
M^-1 K A = omega^2 A
where M, and K are n x n matrices, A is a column vector (and can be used to obtain our eigenvectors/normal modes), and omega^2 is the eigenvalue to the system-- can be multiple eignevalues.
A clarification of this aspect of a multiple oscillator system would be deeply appreciated.
Thanks.
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elhinnaw
#2
Nov4-05, 05:33 AM
P: 13
Well the k relations can be diagnosed with a free body diagram. If you have two springs in parallel, each spring only has to support a fraction of the overall weight, so each spring feels less of an effective force, displacing the springs a smaller amount. A smaller overall displacement means a larger overall spring constant meaning

kparallel = k1 +k2

When the springs are in series, each spring has to displace the weight of the object (as well as the springs), so each spring will extend the amount it would in just a simple spring problem, adding up to an overall longer length, meaning the overall k is smaller, hence

1/kseries = 1/k1 + 1/k2

By no means is that a mathematical proof but it will remind you of the relation.

As far as your K matrix question goes, the reason your 2x2 matrix looks the way it does is because if you multiplied the matricies out you would get the correct springs matched up with the correct masses ie you would not have a mass multiplied by a spring constant it wasnt connected to. If you have masses on a string the matricies will be similar as the coupled harmonic oscillator but if you have 5 masses and 13 springs I dont think it would be so similar.


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