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Oscillating mass on a spring - non-constant mass

  1. Jan 16, 2012 #1
    1. The problem statement, all variables and given/known data
    A 2.10 kg bucket containing 13.0 kg of water is hanging from a vertical ideal spring of force constant 130 N/m and oscillating up and down with an amplitude of 3.00 cm. Suddenly the bucket springs a leak in the bottom such that water drops out at a steady rate of 2.00 g/s.

    When the bucket is half full, find the rate at which the period is changing with respect to time.

    2. Relevant equations
    T=2[itex]\pi[/itex]sqrt(Ʃm/k)


    3. The attempt at a solution
    I know that I need to find T as a function of t, then take the derivative wrt t and evaluate it at the time when the bucket is half full.
    But I'm not sure how to set this up.. I tried:
    T=2[itex]\pi[/itex]sqrt[(m1+m2+Δmt)/k]
    where m1=2.1 kg, m2=13 kg, and Δm=0.02 kg/s
    but it seems to be the wrong set-up. Anyone have any ideas?

    Also, when the bucket is half full, is t=(half the mass of water)/(0.02 kg/s)=6.5/0.02=325 seconds? Is this the t that we should evaluate the derivative at? (assuming we figure it out first :tongue2:)

    Thanks for helping!
     
  2. jcsd
  3. Jan 16, 2012 #2

    Curious3141

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    Homework Helper

    You need to find T as a function of m, which you've already done. m consists of a fixed component (the bucket) and a decreasing component (the leaking water).

    Then, to get an expression for the rate of change of the period with respect to time, use the Chain Rule. You don't actually have to explicitly derive an expression for m in terms of t.

    Once you get the expression, it's as simple as figuring out the mass when the bucket is half full and putting this into the expression.

    BTW, if you're taking the tack of actually deriving an expression for m in terms of t, then remember that the rate of change of mass is NEGATIVE 0.002 kg/s (-0.002 kg/s) - note the sign and the number of zeros.

    Your method to work out the time when the bucket is half-full is correct in principle, but you have the rate of leakage too high by an order of magnitude, as mentioned above.
     
    Last edited: Jan 16, 2012
  4. Jan 16, 2012 #3
    Thanks for pointing out what I needed! I've figured it out, and I really appreciate the help :biggrin:
     
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