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A Simple Harmonic Motion Question

  1. Jun 13, 2016 #1
    1. The problem statement, all variables and given/known data
    Capture.JPG
    As in the given picture, the cylinder is drowned (not completely drowned as in partially drowned) in water. The cylinder is attached with a spring which has the spring constant of 200 N/m. The spring has attached to a unmovable point in the ceiling. The weight of the cylinder is 5 kg and the weight of the spring is inconsiderate. If the cylinder pulled slightly down then will there be a simple harmonic motion and if so what is the Period?

    Important note:
    The height of the cylinder is not given.
    The density of the cylinder is also not given.
    The density of the water is 1000 kg/m^3.

    2. Relevant equations

    The hook's law -> F= -kx where F=force
    k= the spring constant
    x= is the displacement


    3. The attempt at a solution
    Capture.JPG Capture.JPG

    The cylinder is moving in accelerated motion. Let's take the acceleration as a.
    Applying the Newton's Law:
    F=ma regarding upward motion
    Where,

    F=T-50+U

    ma= 5a

    Where T= Tension
    U= Up thrust

    Hence:

    T-50+U=5a

    -200x-50+U=5a

    U=Vdg

    Where V= Volume of the cylinder under water
    d= Density of the cylinder
    g= gravitation acceleration

    This is where I am stuck. I cannot identify the volume of the cylinder under water since it is changing during the motion. What should I do next...?
     
  2. jcsd
  3. Jun 13, 2016 #2

    haruspex

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    Are you not told the radius of the cylinder?
     
  4. Jun 14, 2016 #3
    Didn't I mention it?
    The radius is 6 cm
     
  5. Jun 14, 2016 #4

    haruspex

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    Ok. When the cylinder is displaced vertically by x (upward positive?) what is the change in the buoyancy force?
     
  6. Jun 14, 2016 #5
    That is the point where I cant understand any further. Please explain...
     
  7. Jun 14, 2016 #6

    haruspex

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    What is the change in the volume of water displaced? (If the radius of the beaker is not given then you will have to assume it is much wider than the cylinder.)
     
  8. Jun 14, 2016 #7
    How can I calculate it? when there is no height of the cylinder is given...?
     
  9. Jun 14, 2016 #8

    haruspex

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    We only want the change in the submerged volume for a small change in x. The height of the cylinder will not matter.
     
  10. Jun 14, 2016 #9
    So that explains it...
    But how can I identify the change in submerged volume?
    Please elaborate...
     
  11. Jun 14, 2016 #10

    haruspex

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    If you push a vertical cylinder radius r into water to a depth of x, what volume is submerged?
     
  12. Jun 14, 2016 #11
    That would be
    Volume = π*r2*x
    Then what?
     
  13. Jun 14, 2016 #12

    haruspex

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    Right. If the submerged volume increases by that much, how much does the buoyancy increase by?
     
  14. Jun 14, 2016 #13
    U = Vdg
    U = π*r2*x*1000*10
    U = 10000πr2x

    Then what?
     
  15. Jun 14, 2016 #14

    haruspex

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    It's better not to plug in numbers until the end. For now, keep it as ##A\rho_wgx##, where A is the cross-sectional area of the cylinder, i.e. ##\pi r^2##.
    Next is to write out the force and acceleration equation for the vertical movement of the cylinder. If the cylinder is height x above its equilibrium position, what is the net force acting on it?
     
  16. Jun 15, 2016 #15
    What's the radius/diameter of the beaker?
     
  17. Jun 15, 2016 #16

    haruspex

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    See post #3
     
  18. Jun 15, 2016 #17
    The radius of the beaker affects how much the water level moves up and down when the cylinder moves.

    Edit: I guess it doesn't matter in the end.
     
    Last edited: Jun 15, 2016
  19. Jun 15, 2016 #18

    haruspex

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    Sorry, I should have said "see post #6". Yes, I agree that the radius of the beaker matters too, but if it's not given we will have to take it as being effectively infinite.
     
  20. Jun 16, 2016 #19
    The problem was solved... The radius of the beaker does not matter nor the height of the cylinder and the density of it.
    The method to find the answer is first we have to draw a step of equilibrium of the cylinder.
    Then we have to note that the submerged volume of cylinder is h at the equilibrium stage. Whereas the displacement of the spring is noted as c.
    Then we can see
    F = -kx
    F = T
    x = c
    T = -kc
    U = Ahdg
    Considering equilibrium of the cylinder:
    F= ma
    T + U -mg = 0
    -kc + Ahdg -mg =0

    Then we go to the step where we pull the cylinder for x distance
    Then,
    Tx = -k (c+x)
    F= ma for up
    Tx + Ux - mg = ma
    Ux = A(h+x)dg
    -k (c+x) + A(h+x)dg - mg = ma

    -kc -kx + Ahdg +Axdg -mg =ma

    -kx + Axdg = ma

    The final equation is

    a = -(Adg - k)/m * x

    Hence the ω2 = (Adg - k)/m
    Then ω2 = 2π/t
    A = π*0.062 m2 = 0.0036π m2
    d = 1000 kgm-3
    g = 10 ms-2
    k = 200 Nm-1
    m = 5 kg
    Then t (period of the SHM) can be found...

    Thanks...
     
    Last edited: Jun 16, 2016
  21. Jun 16, 2016 #20

    haruspex

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    It does. Suppose the cylinder has radius r and the beaker has radius R. If the cylinder is displaced upward by x then the spring tension reduces by kx, but the volume of cylinder submerged reduces by ##\pi x\frac{R^2r^2}{R^2-r^2}##.
     
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