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Frame of reference in a simple harmonic motion vertical spring

  1. Jan 2, 2014 #1
    I have doubts of how can I put my frame of reference in a simple harmonic motion vertical spring. Normally the books choose the origin in the equilibrium position and the positive distance (x>0) downward, and in this conditions Newton´s second law is: ma=-kx; but instead of putting the positive distance downward I want to put it upward, so the negative distance (x<0) is downward and in this conditions Hook´s law is going to be positive(because positive direction is upward) so Newton´s second law is: ma=kx
    I want you to tell me if the last expression is correct for the negative distance downward. I would appreciate your help
  2. jcsd
  3. Jan 2, 2014 #2

    Simon Bridge

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    Phew - for a moment you had me worried you wanted to work out the frame of reference with the coordinate system attached to the mass (i.e. non-inertial).

    Off what you actually want to know... I don't think you found the relation you want ... ma=kx says that the acceleration in the +x direction (upwards) is proportional to the displacement upwards ... so the higher the mass gets, the faster it goes. I think you need to to slow down as it goes higher?


    ##\vec{F}=-k\vec{x}## - because the force is always in the opposite direction to the displacement.
    Does not matter if you put +ve upwards or downwards.
    In this case, it's all 1D so ##\vec{x}=x\hat{\imath}## and we write:

    ##-kx\hat{\imath} = ma\hat{\imath}##

    ...and we can divide out the unit vectors and work in magnitudes.

    Except that there's still something wrong with this model: there's no gravity!
    The presence of gravity is what makes "up" and "down" special, otherwise it's the same as saying "forward" and "back" - gravity is what changes the equation.

    If +ve is up, then gravity is negative:
    ... you should be able to take it from there :)
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