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    I have done part a already. Here's the MS



    All I want to know, for part B, is why have they done T2 - T1. Surely it would be T1-T2 for the resultant force? Since the particle will move in the direction of T1 when let go. Thanks.
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    I think it doesn't make a difference which they use, it just changes the sign on the fraction in the first half of the second line.
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    It does change the sign. I get a positive whereas they get a -ve. If you dont get a negative sign in this, then you havent proved SHM. So, I need to know why they used T2-T1 instead of vice versa.
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    T2-T1 = -(T1-T2). It shouldn't make a difference to the presence of SHM.
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    (Original post by marcusmerehay)
    T2-T1 = -(T1-T2). It shouldn't make a difference to the presence of SHM.
    But that -ve sign is vital. The formula for SHM is:

     a = -w^2 x

    If I don't have that -ve sign, how am I supposed to show SHM?
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    (Original post by 2710)
    But that -ve sign is vital. The formula for SHM is:



    If I don't have that -ve sign, how am I supposed to show SHM?
    Initially, you're measuring the displacement from equilibrium, x, from left to right (because it starts to the right of the equilibrium point). Thus, as the net restoring force acting upon the mass is directed toward the left (toward the equilibrium point), you state that the force is the negative of the mass times the acceleration. This means you have -(T1 - T2) = T2 - T1.

    It's not 'proof' in the nice sense of the word, but looking at the solutions you've provided, I can't think of a better answer.
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    (Original post by 2710)
    But that -ve sign is vital. The formula for SHM is:

     a = -w^2 x

    If I don't have that -ve sign, how am I supposed to show SHM?
    it will be
    (T1-T2)=-ma
    the -ma specify the particle is accelerating in the opposite direction towrds the centre

    multiply through by -1 you get
    -T1+T2=ma
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    (Original post by james.h)
    Initially, you're measuring the displacement from equilibrium, x, from left to right (because it starts to the right of the equilibrium point). Thus, as the net restoring force acting upon the mass is directed toward the left (toward the equilibrium point), you state that the force is the negative of the mass times the acceleration. This means you have -(T1 - T2) = T2 - T1.

    It's not 'proof' in the nice sense of the word, but looking at the solutions you've provided, I can't think of a better answer.
    Ah I get it. SO either the force of the displacement has to be -ve/+ve, because if they were the same it would imply that they are in the same direction, when the force is actually moving in the opposite direction. Amirite?

    Thanks
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    (Original post by 2710)
    All I want to know, for part B, is why have they done T2 - T1. Surely it would be T1-T2 for the resultant force? Since the particle will move in the direction of T1 when let go. Thanks.
    Look at the direction in which you are measuring 'x'.

    T2 acts in the same direction, so its contribution to \ddot{x} is positive.
    T1 acts in the opposite direction, so its contribution to \ddot{x} is negative.
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    No problem.
 
 
 
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