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    If there is a spring attached to a ceiling hanging downwards with a particle attached to the bottom performing shm, what direction is the acceleration of the particle at maximum extension assuming that downwards is positive?
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    (Original post by economist29)
    If there is a spring attached to a ceiling hanging downwards with a particle attached to the bottom performing shm, what direction is the acceleration of the particle at maximum extension assuming that downwards is positive?
    Acceleration in SHM is always towards the equilibrium position.

    So, in this case it will be upwards, and negative if downwards is positive.
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    thanks! so am i right in thinking that if upwards was now positive, -(T-MG)=MA following the rule that F=-KX ?
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    (Original post by economist29)
    thanks! so am i right in thinking that if upwards was now positive, -(T-MG)=MA following the rule that F=-KX ?
    I can't follow the jump from your last question to this - confusing.

    -(T-MG)=MA seems to be coming from F=ma, in which case it would be T-MG = MA, taking up as positive. There's no mention of displacement in there, so I can't see the connection to F= -KX.
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    Name:  Screen Shot 2018-02-21 at 15.02.20.png
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Size:  21.9 KB I'm still not sure why acceleration is downwards even though the equilibrium is above T. Surely the equation of motion is T-MG=MA because acceleration should be upwards not down?

    Is it because if downwards is now positive, then acceleration upwards is -A so the equation of motion upwards is T-MG=-MA ?
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    (Original post by economist29)
    Name:  Screen Shot 2018-02-21 at 15.02.20.png
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Size:  21.9 KB I'm still not sure why acceleration is downwards even though the equilibrium is above T. Surely the equation of motion is T-MG=MA because acceleration should be upwards not down?
    I thougth T was a force, the tension. So, what does equilibrium is above T mean? Don't confuse the equilibrium position with the position where the spring is at rest with no weight attached.

    As long as the spring is stretched, T will act upwards, regardless of whether the mass is above or below the equilibrium position.


    Taking downwards as x positive.

    \ddot{x} is the acceleration in the direction of x positive, i.e. downwards.

    Using F=ma

    mg acts in the direction of x positive.
    T acts in the opposite direction. So,

    mg-T = m\ddot{x}

    Note that \ddot{x} will itself be negative or positive depending on the position of the mass. If the mass is above the equilibrium position then it is positive. If it is below the equilibrium position, T will be much greater (greater than mg actually), and \ddot{x} will be negative.
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    Thank you so much!
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