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How to find the equation for velocity of a mass on a spring?

  1. May 18, 2015 #1
    1. The problem statement, all variables and given/known data
    The problem is to use conservation of energy to determine the equation for the velocity at the equilibrium position of an oscillating mass on a vertical spring. The question says to use three types of energy. The question gives the equation, the problem is to solve for it. The equation is v=A(k/m)^.5

    2. Relevant equations
    Uelastic+Ugravitational+Ukinetic=Uelastic+Ugravitational+Ukinetic

    3. The attempt at a solution
    The initial position I chose was at the bottom of the mass's motion and the final at equilibrium, so inital kinetic and final elastic and gravitational are zero.

    mgh+.5kx^2=.5mv^2
    since h and x are the amplitude,
    mgA+.5kA^2=.5v^2
    simplifying gives,
    2gA+(k/m)A^2=v^2
    How do I get rid of the gravity term?
     
  2. jcsd
  3. May 18, 2015 #2

    TSny

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    If you are taking gravitational PE to be zero at the equilibrium position, then what should be the sign of the gravitational PE at the bottom of the motion?

    x represents the amount that the spring is stretched from the unstretched length. If the mass hangs at rest at the equilibrium position, then the spring is stretched. So, x is not equal to zero at the equilibrium position. Also, x is not equal to A at the bottom of the motion.
     
  4. May 18, 2015 #3
    Thanks for the help. But why wouldn't x be amplitude at the bottom?
     
  5. May 18, 2015 #4

    TSny

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    The spring already has some stretch at the equilibrium position. Let this stretch be xe.

    Then, at the bottom the spring will be stretched by x = xe + A.
     
  6. May 18, 2015 #5
    Would gravitational potential energy still be zero if I define xe to be 0?
     
  7. May 18, 2015 #6
    That depends. Are you redefining the axis so that Xe is the new zero?
     
  8. May 18, 2015 #7

    TSny

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    The formula PE = 1/2 k x2 for the spring assumes that x is the amount of stretch of the spring. Since the spring is stretched at the equilibrium position, you cannot let xe = 0. You should be able to determine xe in terms of m, g, and k. (Think about the mass hanging at rest at the equilibrium position.)

    For the gravitational PE you can use PE = mgy and you can choose y = 0 anywhere you want. So, you can choose y = 0 at the equilibrium position if you want. Then PE for gravity = 0 at the equilibrium position.
     
  9. May 18, 2015 #8
    I have it now. Thank you very much.
     
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