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Frequency of girl bobbing in swimming pool

  1. Apr 3, 2012 #1
    1. The problem statement, all variables and given/known data
    A girl with mass m kg steps into her inflatable ring with horizontal cross sectional area Am^2 and jumps into the pool. After the first splash, what is the frequency of the girl bobbing up and down?


    2. Relevant equations
    I assume that we need the extra force,F_e, after the buoyant force and the weight cancel. Archimedes F_b = mg

    We can then use Newton 2, F=ma, where a=x".
    x" = sqrt(F_e/m)

    f=1/[2π√(F_e/m)]

    3. The attempt at a solution
    The above ω will be for the frequency.

    I am not sure if the above is right and I do not know how to solve for the F_e

    Thanks!
     
  2. jcsd
  3. Apr 3, 2012 #2

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    Hello kevlar94,

    Welcome to Physics Forums!

    Just so we are both on the same page, do you really mean that the cross sectional area is a function of the mass squared? Maybe the problem statement is written that way, but I just want to be sure.

    In other words, do you really mean that

    [tex] \mathrm{Area} = Am^2 [/tex]

    where [itex] m [/itex] is the mass and [itex] A [/itex] is an some constant of proportionality?
    Well, you can say Fb = mg if nothing is accelerating, and is in static equilibrium. But that's not the case when the girl is bobbing up and down.

    Perhaps you mean to say that Fe = Fb - mg?
    Where did the square root come from? Newton's second law doesn't contain a square root.

    [tex] \sum_i \vec F_i = m \ddot{z} [/tex]

    (since the motion in this problem is always in the up/down direction, I chose to use the variable [itex] z [/itex] to represent the position. You could just as easily use the variable [itex] x [/itex] to represent position if you want to though.)
    You'll have to show me where that came from.
    There are two forces involved. The weight of the girl and the buoyant force. You've already figured out the weight is mg.

    The buoyant force is equal to the weight of the water that is displaced. The weight of the water is proportional to g and the density of water ρ. It's also proportional to the volume of the water that is displaced.

    The cross sectional area is already given in the problem statement. You just need to throw in the vertical displacement to find the volume of displaced water.

    Plug those back into Newton's second law and you'll get an equation containing both [itex] z [/itex] and [itex] \ddot{z} [/itex]: an ordinary, second order differential equation that you can solve.
     
    Last edited: Apr 3, 2012
  4. Apr 3, 2012 #3
    Thanks for the help.

    Sorry, the prompt says the girl is 25kg and the inflatable ring has a horizontal cross-sectional area of 0.7m^2.

    Yes, that is what I meant. I skipped a step I should have mentioned.

    The square root is from the formula for frequency using k(from hookes law) from my book.

    So F_r = F_b - mg = ρ(displaced water)(V)g

    I start with the assumption that the ring goes dx distance into the water. Which results in dV=Adx

    dF_r=ρ*g*A*dx

    Since I am using hookes law for a linear oscillator, F=kx or dF=kdx

    so dF_r = kdx= ρ*g*A*dx which gives a k value of ρ*g*A -- when the dx cancels

    Using k I can solve for ω using √(k/m) and then solve for frequency using ω=2π*f

    Using the above values gives ω= √(9800*.7)/25 = 16.56

    f=16.56/(2π) = 2.637Hz

    Does that look correct?
     
  5. Apr 4, 2012 #4

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    Oh, 'm' is meters (not mass). Okay, I understand now.
    Ah, Hooke's law. I never thought of that. The buoyant force is proportional to the displacement. So yes, Hooke's law and associated equations will work just fine. That should save you from having to solve the differential equation -- essentially from re-deriving the ω = √(k/m) formula.
    That's what I got (out to the first three significant figures anyway). Good job. :approve:
     
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