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Transformations of energy in oscillatory motion

  1. Feb 29, 2016 #1
    1. The problem statement, all variables and given/known data
    I thought that in the situation where mass is attached to the string and put into motion (vertically), energy goes between 1/2 kA^2 and 1/2 mV^2 like here - http://imgur.com/RU4P6UW

    However in student's book I saw the table, which said that potential energy of gravity is also changing, which is logical, but I didn't see it included in equations anywhere for above situation.

    For example here it isn't included: http://www.acs.psu.edu/drussell/Demos/SHO/mass.html

    2. Relevant equations
    Does it mean that 1/2 kA^2 already covers potential energy of gravity? I can't see any other explanation that would work, but I don't understand why would that be, because 1/2 kA^2 is supposed to be potential energy of oscillatory motion, and in horizontal situation equations are the same without any change in potential energy of gravity.

    Maybe the equations are incomplete? If that's the case, then how should they look like?

    3. The attempt at a solution
    (1st link)
     
  2. jcsd
  3. Feb 29, 2016 #2

    Simon Bridge

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    The standard derivation for mass-on-a-spring SHM assumes no gravity, or friction, or any other forces besides the spring.
    The spring is also modelled as an ideal massless spring that always obeys Hook's law.
    IRL you need to account for these things.

    You can use a free-body diagram to help you see what the equation's look like if you include gravity...
    You may want to make the near-Earth approximation for gravitational potential energy and force and take care to pick a suitable position for zero potential energy.
    http://www.ux1.eiu.edu/~cfadd/1150/15Period/Vert.html
     
  4. Mar 1, 2016 #3
    Achh!!! I think I got it! I had calculus physics quite a while ago, and your clue about free-body diagram reminded me that work done is just integral of force over distance. In this situation force of gravity F = mg just changes equilibrium position, and the net force is still F = kx, where x is the distance from new equilibrium position. Given that, integral of force over distance results in the same gains of potential energy 1/2 kx^2, so it is all ok :)

    Did I got it correctly?
     
  5. Mar 3, 2016 #4

    Simon Bridge

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    Pretty much - you can also do it from Newton's laws:

    If you measure position down from the attachment point ...
    - when the spring is not stretched the mass is at position ##y=u##
    - when at equilibrium, the mass is lower down at ##y=e##
    - stretch the spring to ##y=y_0> e## and let it go ... at ##t>0##, ##y(t)## satisfies:
    ##mg-k(y-u) = m\ddot y## (1)

    But the equilibrium position is determined by ##k(e-u)=mg \implies ku = ke-mg## ... substitute into (1) above to give:
    ##-k(y-e)=m\ddot y## (2)

    which is the same equation as you get for the horizontal mass and spring, with e replacing u.
    The solution looks like: ##y(t) = e + (y_0-e)\cos\omega t##
     
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