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Frequency of a mass bobbing up and down in water

  1. Nov 16, 2016 #1
    1. The problem statement, all variables and given/known data
    This was a test question I had today but basically, initially the mass is at rest as the buoyant force opposes the force of gravity. Then we push it down X meters and let it go. This can be described by SHM. We are also given the density of water, a cross sectional area of the mass, and the actual mass, M.

    2. Relevant equations
    A=area (not amplitude)
    X_0= equilibrium position
    X_new= furthest value it can be bobbed up or down
    Archimede's Principle
    3. The attempt at a solution
    Net force=mg-A*(X_0+X_new)*rho_w*g note that this is zero at the equilibrium point where X_new equals 0.
    ma=mg-A*(X_0+X_new)*rho_w*g
    a=g-(A*(X_0+X_new)*rho_w*g)/m

    The general form of a SHM equation is x(t)=Acos(ωt) which can be solve to x''(t)=-ω^2Acos(ωt)

    I assumed ωt to be 0 so as we are using the amplitude. We know the amplitude to be X_0+X_new however I chose X_0 to be zero for simplicity (I know this isn't correct as the volume would be zero but this was just the position). Thus, the amplitude is now X_new. I then set the acceleration equal to x''(t) to solve for ω:

    g-(A*X_new*rho_w*g)/m=-ω^2X_new

    ω=sqrt(-g/X_new+(A*rho_w*g)/m)

    If anyone can confirm this, that'd be awesome!
     
    Last edited: Nov 16, 2016
  2. jcsd
  3. Nov 16, 2016 #2

    ehild

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    Homework Helper
    Gold Member

    The differential equation contains the constant term mg. The solution for X_new is SHM, but X_0 is not zero. It is the depth in equilibrium. You have to find it with the condition a=0, X_new=0.
     
  4. Nov 17, 2016 #3

    gneill

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    Staff: Mentor

    SHM occurs when the restoring force obeys Hookes law, f = -kx. If you can show that the restoring force for the bob has this form and find k, then you can invoke the standard expression for the frequency of a mass-spring system.
     
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