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Velocity of a Spring- SHM

  1. Dec 9, 2007 #1
    1. The problem statement, all variables and given/known data
    At which point does the mass on a vibrating spring have its largest velocity? Smallest velocity?


    2. Relevant equations
    Hooke's Law? See below.


    3. The attempt at a solution

    I know the mass on a vibrating spring has the greatest acceleration when the greatest amount of force is acting on the mass. So at the two endpoints of the motion, where the spring is either stretched or compressed the most. We know this because of Hooke’s Law. Hooke’s Law= the further the spring is displaced from its equilibrium position (x) the greater the force the spring will exert in the direction of its equilibrium position (F - restoring force). And the acceleration is smallest at the midpoint of its motion, so equilibrium. But what about velocity? Smallest velocity would be at the endpoints of motion too, because for that one instant it's not moving?

    Edit: Afterthought- would it possibly have something to do with potential or kinetic energy at a certain point in its motion?
     
    Last edited: Dec 9, 2007
  2. jcsd
  3. Dec 10, 2007 #2
    There are no specific values, but it's more of an in general type of question. :)
     
  4. Dec 10, 2007 #3
    What do you mean by vibrating spring?

    EDIT: Yeah that's why I deleted my last comment XD
     
  5. Dec 10, 2007 #4
    Think of the position versus time curve for a mass on a spring (cosine function). The derivative of this function, the change in position over time will give you the velocity versus time curve (an inverted sine curve). We know that a sine function will have maximum values when the phase is some multiple of [tex]\pi[/tex] /2. We also know that the corresponding position of the mass undergoing SHM is at equilibrium when the phase is some multiple of [tex]\pi[/tex]/2. Therefore the velocity is greatest when the mass is at the equilibrium position.
    You can also look at the energy conservation expressions for SHM. Since the total energy is given by 1/2kdx^2 + 1/2mv^2 = 1/2kA^2 = 1/2m(vmax)^2, we can deduce that when the velocity is greatest, the potential energy of the spring is zero, which occurs only at the equilibrium position.
    It's probably most useful to look at position and velocity versus time graphs for SHM, but i hope this helps.
     
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