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Spring-Mass system with friction

  1. Aug 3, 2011 #1
    1. The problem statement, all variables and given/known data
    A block shown in the drawing is acted on by a spring with spring constant k, and a weak friction force of constant magnitude f . The block is pulled distance x0 from equilibrium and released. It oscillates many times before coming to a halt.

    (a) Show that the amplitude decreases by same amount in each cycle of oscillation.

    (b) Find the number of cycles n, the mass oscillates before coming to rest.

    2. Relevant equations
    Conservation of energy

    3. The attempt at a solution
    For (a) I used conservation of energy. Let xi be the amplitude of a cycle (with energy Ei) and xf be the amplitude (with energy Ef)

    [itex]E_{i}=\frac{1}{2}kx_{i}^{2}[/itex]
    [itex]E_{f}=\frac{1}{2}kx_{f}^{2}+f(x_{o}+x_{f})[/itex]

    Yielding

    [itex]x_{f}=x_{i}-\frac{2f}{k}[/itex]

    Showing that the amplitude decreases by 2f/k between subsequent cycles.

    For (b) I calculated the distance x for the mass to come to rest at:

    [itex]\frac{1}{2}kx_{o}^{2}=fx \implies x=\frac{kx_{o}^{2}}{2f}[/itex]

    Meanwhile, the total distance covered by the block using the result from (a),

    [itex]x=x_{0}+x_{0}-\frac{2f}{k}+x_{0}-\frac{4f}{k}+(...)+x_{0}-\frac{2(n-1)f}{k}[/itex]

    for n cycles. Simplifying,

    [itex]x=nx_{0}-\frac{2f}{k}(1+2+(...)+n-1)[/itex]

    If I equate those expressions, I don't get the right answer. Any hints?
     
  2. jcsd
  3. Aug 3, 2011 #2
    It looks like your "initial" and "final" energies are not a whole cycle but only half a cycle apart. Can you see why?
     
  4. Aug 3, 2011 #3
    So a cycle is when the mass returns to the same side? If that is so, I can't see how talking about amplitude makes any sense, as the mass does not even go x0 on the other side of the equilibrium point, but rather goes [itex]x_{0}-\frac{2f}{k}[/itex] as I showed.
     
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