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Mechanics SHM question watch

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    I am having trouble with this SHM question. Here's my work so far:

    After releasing it, the object will move to the equilibrium position x = \alpha M . Therefore  k(\alpha M) = Mg , so k=\frac{g}{\alpha}.

    Now I consider a displacement y from equilibrium i.e. x = y, y<\alpha M.

    Using Newton's 2nd Law: Mg - ky = Ma. But after substituting for k, I am unable to obtain an equation in which acceleration is proportional to displacement...
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    (Original post by FXLander)
    I am having trouble with this SHM question. Here's my work so far:

    After releasing it, the object will move to the equilibrium position x = \alpha M . Therefore  k(\alpha M) = Mg , so k=\frac{g}{\alpha}.

    Now I consider a displacement y from equilibrium i.e. x = y, y<\alpha M.

    Using Newton's 2nd Law: Mg - ky = Ma. But after substituting for k, I am unable to obtain an equation in which acceleration is proportional to displacement...
    Couple of points:

    1) After release, the object will actually move past the equilibrium point before changing direction and commencing SHM about the equilibrium point. That said, I agree with you expression for k.

    2) When the object is displaced y below the equilibrium point the net force on the object is made up of the weight Mg acting downwards, and spring tension (Mg +ky) acting upwards. The Mg component of tension is the tension at the equilibrium point.
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    (Original post by old_engineer)
    Couple of points:

    1) After release, the object will actually move past the equilibrium point before changing direction and commencing SHM about the equilibrium point. That said, I agree with you expression for k.

    2) When the object is displaced y below the equilibrium point the net force on the object is made up of the weight Mg acting downwards, and spring tension (Mg +ky) acting upwards. The Mg component of tension is the tension at the equilibrium point.


    Ah I didn't realise that the 'new equilibrium' is when x = \alpha M. That's what they meant by 'equilibrium displacement'.
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    (Original post by FXLander)
    Ah I didn't realise that the 'new equilibrium' is when x = \alpha M. That's what they meant by 'equilibrium displacement'.
    Yes. If the mass was lowered gently to alphaM and released there, it would sit motionless in equilibrium (with no net force acting on it). At that position the spring tension will necessarily be Mg acting upwards.
 
 
 
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