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Math Help - Classical Mechanics

  1. #1
    Senior Member Educated's Avatar
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    Classical Mechanics

    I'm starting on classical mechanics in university (second year) and I skipped first year physics. I do have quite a good mathematical knowledge but I'm not getting this question:


    Verify by direct substitution that the function \phi(t) = A \sin (\omega t) + B \sin(\omega t) is a solution to the second order differential equation \ddot{\phi} = -\omega ^2 \phi
    (Since this solution involves two arbitary constants - the coefficients of the sine and cosine functions - it is in fact the general solution)



    So I was just thinking of substituting the first equation into the second to get:

    \ddot{\phi} = -\omega ^2 (A \sin (\omega t) + B \sin(\omega t))

    And then distribute and integrate twice to get the differential?
    I'm really confused about the question.
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  2. #2
    MHF Contributor

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    Re: Classical Mechanics

    No, substitute on the left side also. In other words differentiate twice and see if the two sides of the equation are really equal.
    Thanks from Educated and topsquark
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  3. #3
    Senior Member Educated's Avatar
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    Re: Classical Mechanics

    Actually I just realised typed in the equation wrong. The first one is supposed to be: \phi(t) = A \sin (\omega t) + B \cos(\omega t) but I don't think it really matters.


    So something like this??

    \phi(t) = A \sin (\omega t) + B \cos(\omega t) \\EQUATION 1

    \dot{\phi}(t) = A \omega \cos (\omega t) - B \omega \sin(\omega t)

    \ddot{\phi}(t) = - A \omega ^2 \sin (\omega t) - B \omega^2 \cos(\omega t) \\EQUATION 2



    So I substitute equation 1 onto the right hand side and equation 2 onto the left hand side of \ddot{\phi} = -\omega ^2 \phi?

    That'll give me:


    - A \omega ^2 \sin (\omega t) - B \omega^2 \cos(\omega t) = -\omega ^2 (A \sin (\omega t) + B \cos(\omega t))

    And yes if I distribute the righthand side, it is clear that they are equal.
    So that's the answer?
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