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Show that the motion is simple harmonic

  1. Jan 6, 2014 #1
    1. The problem statement, all variables and given/known data

    A solid wooden cylinder of radius r and mass M. It's weighted at one end so that it floats upright in calm seawater, having density ρ the buoy is pulled down a distance x from it's equilibrium position and released.
    a- Show that the block will undergo s.h.m
    b- Determine the period of oscillations (the resistive effects of water are ignored)

    3. The attempt at a solution
    F=W-Fb
    g/ρ(ρ-ρw)=a

    I was trying to get a=-w^2 x
     
  2. jcsd
  3. Jan 6, 2014 #2

    rude man

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    If it's shm the wood block satisfies the equation mx'' + kx = 0. Find k (and of course assume a finite initial condition. I would pick x(0+). Your first equation F = W - Fb is germane. The second one I can't figure out.

    .
     
  4. Jan 6, 2014 #3
    I used Fnet=ma
    so W-Fb=ma
    ρVg-ρw*V*g=(M+m)a
    Vg(ρ-ρw)=(M+m)a
    ((M+m)/ρ)g(ρ-ρw)=(M+m)a
    Then I crossed out (M+m)
     
  5. Jan 6, 2014 #4

    rude man

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    What's V? You need to define all your parameters.
    Also pay attention to your parentheses. They don't make sense.

    Anyway, bottom line is:
    it's floating straight up in the water. If you push the cylinder down a distance x, what is the net force acting on it? That's your k.

    You don't need gravity in your equation. Gravity is a constant force. When x=0 it's already taken into account, and it doesn't vary with x. So the only force left is buoyancy. How does it vary with x?
     
  6. Jan 7, 2014 #5
    V is the total volume of the cylinder. ( I assumed that the cylinder is totally submerged in water; maybe that's not the case here
    I'm not sure I understand this why isn't the weight of the cylinder taken into account ??
    I think that as the block is pushed a distance x downwards, the bouyant force upwards increases (more water is being displaced now) however, I can't express this in a mathematical equation. Can I use Fb=ρblock *Areablock* x*g instead of ρ*Vsubmerged*g ??
     
  7. Jan 7, 2014 #6

    rude man

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    It is not totally submerged. If it was there'd be no up-and-down s.h.m.
    As I explained, it's a constant force canceled by the buoyancy of the floating cylinder. Since it's not a function of depth x it does not enter into the computation. However, if you had been asked to determine the position of the cylinder then you'd have had to use it.

    Why would ρblock have anything to do with the force as a function of depth x? What force pushes up on the block as you push it into the water? You yourself said 'buoyancy' which is correct. So express buoyancy force as a function of x, considering buoyancy = 0 at x = 0.

    PS another way of looking at it is to realize that as you push the block down into the water the pressure on the block's bottom increases while the pressure on top stays at atmospheric pressure.
     
  8. Jan 8, 2014 #7
    O.K I'll assume that the block is floating over water so that there's no submerged volume.
    I will use Fb=ma and prove that a="a constant"*x, if this is true the motion must be SHM right??
     
  9. Jan 8, 2014 #8

    rude man

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    No, a is not a constant. a is acceleration and it changes with the block's position.

    You're looking for an equation like F = ma = mx'' = -kx. Find k.

    There has to be some submerged volume, otherwise it would bob to the surface and then not face an opposing force at its bottom proportional to -x.

    Think of the block as sumberged about halfway & bobbing up and down with the top always above water and the bottom always below.
     
  10. Jun 20, 2017 #9
    This was my solution

    Let p represent density of the fluid
    Let V represent the volume of water being displaced by the cube
    Let s represent the length of one side of the cube
    let x represent represent how far the block is displaced into the fluid

    For this to be SHM it must satisfy the relation
    F = -kx (Restoring force is proportional to the displacement and always points towards the equilibrium position)


    The restoring force is the buoyant which is as follows
    F = Weight of Fluid Displaced
    = pVg
    = -p*(s^2)*x*g (since this is water p = 1.00 g/L)
    F = -([s^2]/g)x

    This satisfies the relation F = -kx where k = (s^2)/g
     
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