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Masses on a spring

  1. Nov 6, 2016 #1
    1. The problem statement, all variables and given/known data

    When a mass m sits at rest on a spring, the spring is compressed by a distance d from its undeformed length. Supposed instead that the mass is released from rest when it barely touches the undeformed spring. Find the distance D that the spring is compressed before it is able to stop the mass. Does D=d. If not, why not.

    2. Relevant equations


    3. The attempt at a solution

    This problem really confused me. I got the right answer, but I don't know how to explain the difference. Basically, I started by doing:
    k*d=m*g, thus k=mg/d

    Ug1=Ug2+Uspring
    Ug1-Ug2=Uspring
    Mg*D-mg*0=(1/2)(mg/d)D^2
    1=D/2d
    D=2d

    This is what I calculated should happen for the dropping, but I don't understand why it goes further conceptually and hence can't answer why D does not equal d. Any tips/explanations? I couldn't find anything in the chapter of the book that dealt with anything like this.
     
  2. jcsd
  3. Nov 6, 2016 #2

    Charles Link

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    When you let go of the mass, you will get simple harmonic motion where the velocity is zero at the top as well as at the bottom. There is a place right in the middle where the acceleration is zero but the speed is maximum=going upward or downward. When the acceleration is zero the force from the spring balances the force of gravity. If you stop the mass at this middle point, it would remain stationary. If you then decide to pull up on the stationary system (balanced at the middle point), until the spring has no tension, you have added energy to the system. It is this energy that shows up as the kinetic energy of the system (so that the mass is moving) as it passes the equilibrium point. Hopefully this explanation is somewhat helpful. ## \\ ## Additional item: With a mass and a spring, you get simple harmonic motion at the same frequency with a gravitational field present that you would without. The gravitational field causes a displacement of the equilibrium position. In this case, the initial conditions were such that it required a raising from the euilibrium position to reach that starting point, so the result would be simple harmonic motion that is symmetric about the equilibrium position. ## \\ ## Without the gravitational field, the first position would be the equilibrium position. With the gravitational field, the system simply doesn't go to the new equilibrium position and remain at rest. If you were to introduce damping, it would eventually come to a stop at the new equilibrium position, but without any damping it simply will oscillate around it, essentially continuously in the case of zero damping.
     
    Last edited: Nov 7, 2016
  4. Nov 6, 2016 #3

    Simon Bridge

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    What is the difference between a mass being placed at a position and being dropped from above that position?
    What are the energy and momentum considerations?
     
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