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Tray oscillation problem

  1. Nov 4, 2008 #1
    I think I solved the problem but tell me if I made any errors

    1. The problem statement, all variables and given/known data

    A tray of mass m, when suspended from a spring attached to the ceiling, stretches the spring by a distance Δx. A glob of peanut butter with mass M is dropped from a height h onto the tray, which is at rest, and sticks there.

    a) What is the spring constant k of the spring in terms of the variables given above?
    b) Find the maximum extension of the spring below its initial equilibrium position before the tray had been attached.

    2. Relevant equations

    x = Acos(wt + @)
    v = -Awsin(wt + @)
    w = sqrt (k/mass) etc.

    3. The attempt at a solution

    (a) Since the spring stretches Δx when tray is attached,
    kΔx = mg, k = Δx/g

    (b) The butter of mass M drops height h, so the velocity just before it sticks to the tray is
    sqrt(2gh). And using conservation of momentum during the very short time period it sticks to the tray, M * sqrt(2gh) = (M + m)Vi, Vi = (M*sqrt(2gh))/(M + m)
    Vi can be found from this and it is the v at time 0.
    v(0) = -Awsin(@)
    0 = x(0) = Acos(@)

    A = sqrt((v(0)/-w)^2 + 0^2) = v(0)/w = Vi/w

    So the maximum extension is (M*sqrt(2gh))/(M + m) / (sqrt(k/(M+m))) + Δx

    Is this correct?
    Is there a faster way to do this?
     
  2. jcsd
  3. Nov 5, 2008 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi kky1638! :smile:

    Your vi is right, but I'm not following what you've done after that. :confused:

    (why are you using angles?)

    Hint: use conservation of energy …

    that is, KE + gravitational PE + spring energy. :smile:
     
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