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Homework Help: Horizontal spring, momentum

  1. Nov 15, 2007 #1
    1. The problem statement, all variables and given/known data

    A block of mass m is attached to the end of a horizontally mounted spring as shown.
    The spring has a spring constant k and obeys Hooke’s law. The block is given an initial
    displacement xo, after which it oscillates back and forth without frictional effects.
    (a) Write an expression for the magnitude of the momentum p of the block as a function of its displacement x
    from its equilibrium position, given m, k, and xo.
    (b) What are the minimum and maximum values of the magnitude of the momentum, and where do they occur in the motion of the block?

    2. Relevant equations

    I know that the force of the spring on the block is F= -kx, and that the spring has stored the elastic PE= 1/2 kx^2, but i am not sure how to relate this to momentum. I do know that the rate of change (with respect to time) of the momentum is equal to the force, so is it equal to -kx?

    any help would be greatly appreciated.

    as for part b, im not sure how to attain this answer.
  2. jcsd
  3. Nov 15, 2007 #2
    Magnitude is directly related to velocity.
    If a mass is oscillating back and forth, where is velocity the greatest?

    Well, starting at the right hand +x side, [tex]V_{1} = 0[/tex]. When it reaches its maximum on the -x side, [tex]V_{2} = 0[/tex] as well. So, this means that velocity will be at a maximum when the mass passes through the center point. If you think about this, it makes sense because as the mass is accelerating towards the center, its velocity in acceleration are in the same direction. After it passes the midpoint, the spring is exerting a force OPPOSITE the motion and thus an OPPOSITE accleration so the mass slows down.

    Hope that helps. Ill check back later when Im not at work.
  4. Nov 15, 2007 #3
    Ah, and I forgot, when you displace the spring distance x, and you have PE of 1/2kx², this will be your KE @ 0 which will be 1/2mv² so you can relate x to v which you can relate to p
  5. Nov 15, 2007 #4
    ah i think i see what you are saying, but when you said "Magnitude is directly related to velocity." you meant momentum is directly related correct? Also, i do understand your second post, the relation between the PE and KE, but im still confused as to how i would express the momentum of the block as a function of its displacement from its equilibrium position. I know that the momentum is equal to velocity X mass. So do i just substitute p for mv in the KE equation?

    But i do see, that when the spring is stretched momentum is at its minimum, 0, because velocity is 0, as well as at the maximum compression of the spring, on the -x side, where velocity is again 0, making the momentum 0. So when the block is at its equilibrium velocity is greatest, and therefore momentum is greatest. But as i express this magnitude, is it just expressed as mv^2?
    Last edited: Nov 15, 2007
  6. Nov 15, 2007 #5
    When the mass is all the way right ( or left ) you have the relationship that PE+KE=0. So, 1/2kx²+1/2mv²=0. Now you can solve for v in terms of x and thusly write p in terms of x also.
  7. Nov 15, 2007 #6
    ohh ok i understand very well now. Thanks for all your help!!
  8. Nov 15, 2007 #7
    And dont worry about the negative -k that you come up with. Since values for k are always negative, a -k in an equation will actually be positive and youll get a positive energy value.
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