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Why change the coordinates?

  1. Feb 1, 2012 #1
    I have a question regarding a method I saw every now and then:
    Say I have a system containing of two masses, attached to one another by a spring. Each is attached to a wall by another spring.
    Now I wish to know the eigen vectors and eigen values( [itex]\omega[/itex]) of movement.

    I get:
    [itex]\ddot x_1= \frac {-2k} {m} x_1+\frac {k} {m} x_2[/itex]
    [itex]\ddot x_2= \frac {-2k} {m} x_2+\frac {k} {m} x_1[/itex]

    That is because the sysem is completely symmetric. Now, I get
    [itex]{\omega_1}^{2}= \frac {3k} {m}[/itex] and [itex] {\omega_2}^{2}= \frac {k} {m} [/itex]
    But if I try to find the eigen vectors using a matrix,

    |5 -1 |
    |-1 5 |
    As you can see, the only solution is the trivial one.

    So, what I saw was done in this case, was defining new coordinates:
    And now it works, and everything is fine.
    What I don't understand is why.

    How is it that when I move to y-s it is ok?
    Shouldn't I get the same result?
    Thank you!
  2. jcsd
  3. Feb 1, 2012 #2


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    Science Advisor
    Gold Member

    I'm not quite sure how you got your eigenvectors here. It's been a while but my recollection is that your matrix is something like [1 -1; 1 1]. Or maybe even that is wrong. With the original equation, you have two equations of second order so that's four unknowns for your coefficients that arise because each oscillator will be influenced by both eigenfrequencies. But the thing to note about the coordinate transform is that you have now gone to what are called normal coordinates. These normal coordinates are uncoupled and independent which allows you to write the equations of motion to be dependent upon a single eigenfrequency. So the uncoupled solutions makes this the more desirable method of analysis.
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