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Energy in a spring system

  1. Jan 17, 2009 #1
    The spring constant of a helical spring is 28Nm^-1. A 0.40 kg mass is suspended from the
    spring and set into simple harmonic motion of amplitude 60mm.

    (i) the static extension produced by the 0.40 kg mass,

    (ii) the maximum potential energy stored in the spring during the first oscillation.

    The markscheme is shown below:


    Id the question, and get the correct answer to both parts. However, I do not understand the second method of working out the second part of the question - why does one have to add "mgA" in the final part? I presume it is due to E = mgh, however do not understand the reasining behind the fact that it must be added - when the mass is at the equilibrium position, i thought all of its energy is a) kinetic energy, and b) eleastic potential energy.

  2. jcsd
  3. Jan 18, 2009 #2
    The object has a potential energy difference between equilibrium and the lowest point. If the reference point is in fact the lowest point, at the lowest point the potential energy is W = 0 but at the equilibrium it is W = mgA, where A is amplitude or distance between equilibrium and the lowest point. If, while at equilibrium, we cut the spring, the object will start a free-fall and, if we don't take into account friction between the object and air (or the medium in which the oscillations occured), the potential energy, which it had at equilibrium, will have turned to kinetic energy at the point, where once was the lowest point of these oscillations.

    I hope that helps.
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