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Vertical Mass-Spring Oscillator

  1. Oct 5, 2012 #1
    Hi,

    So, I have a doubt regarding the equations for vertical oscillations on a spring.

    My book says the net force on the block is: F = k(d+y) - mg.

    If we define d the distance at -kd = mg.

    I, don't understand, the reason being:

    When the block is moving downwards, if its performing simple harmonic motion, it is accelerating upwards. This means that the upward force, that is, that provided by the string, must exceed mg. In this case, the book's formula holds.

    However, as soon as the block goes up through its equilibrium position (the one after the mass was hung) the acceleration should be downwards meaning mg is greater than the force provided by the string. Shouldn't the equation, thus read:

    F = mg - k(d+y)

    Thanks in advance,

    Peter G.
     
  2. jcsd
  3. Oct 5, 2012 #2

    BruceW

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    Homework Helper

    I agree with you, and the book is wrong, As long as k is positive, and:
    [tex]F=m \frac{d^2 y}{dt^2} [/tex]
    Then your equation F = mg - k(d+y) is the correct one, because the spring force must always be opposite to the displacement. Also F= -mg -k(d+y) would also be correct (where the y axis has simply been inverted, so that gravity is now pointing in negative y direction).

    Edit: also, F= -mg -k(d+y) is a 'nicer' equation than your equation F = mg - k(d+y) since for your equation, y=0 is not the point of equilibrium, but both equations are correct really, its just a different choice of coordinate system.
     
    Last edited: Oct 5, 2012
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