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Find period of vibration of liquid

  1. May 8, 2014 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    A liquid (density ρ) is poured into a bent tube such that the two halves of the bent form angles α and β with the horizontal. The length of liquid column is l. If one of the liquid levels is depressed and released, the levels begin to vibrate. Find the period of vibration. Neglect capillary forces and the viscosity of the liquid.


    3. The attempt at a solution

    Let the cross-sectional area of bottom-most part of the tube that separates the two sides be S, left part be S1 and right part be S2.
    S1 = S sec α
    S2= S sec β
    Let the equilibrium height of liquid surface be h. when the liquid is depressed, let the left surface of liquid descend a distance Δx1 and the right part ascend a distance Δx2.
    Since volume is constant,
    S1Δx1=S2Δx2
    sec α Δx1 = sec β Δx2
    Let the force on the right part of the liquid by the left part be F1 and that on left part by the right part be F2 at the bottommost point

    [itex]F_1=(P_0 + \rho g (h-Δx_1) ) S \\
    F_2 = (P_0 + \rho g (h+Δx_2) ) S \\
    F_1 - F_2 = - \rho g S (Δx_1+Δx_2)\\
    \ddot{x} =\dfrac{-\rho gS}{m} \left(\dfrac{\sec \beta}{\sec \alpha}+1 \right) x[/itex]

    But this equation gives me the incorrect angular frequency.
     

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  3. May 8, 2014 #2

    TSny

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    From Newton's 3rd law, the magnitude of F1 must equal the magnitude of F2.

    You can't use the standard formulas for pressure in a static fluid. (You can't even use Bernoulli's equation in its standard form because the fluid is not in steady motion.)

    You might try using Newton's second law applied to the fluid as a whole. Think about the net gravitational force acting along the direction of motion of the fluid.

    Or you can try using energy concepts.
     
  4. May 8, 2014 #3

    utkarshakash

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    potential energy of the left part = m1 g(h-Δx1)
    potential energy of the right part= m2 g(h+Δx2)
    K.E. of left part = 1/2 m1 v1^2
    K.E. of right part = 1/2 m2 v2^2

    sec α v1 = sec β v2

    I'm not sure if what I've done above is correct or not.
     
  5. May 8, 2014 #4

    TSny

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    I'm not sure of your notation here.

    Since only differences in PE matter, you can take the PE to be zero in the equilibrium position (as shown in the left figure). Then you need to find an expression for the PE when the fluid is displaced down a distance S along the tube on the left side, as shown in the right-hand figure.

    Some things to think about:

    If the fluid on the left moves a distance S along the tube from equilibrium as shown, how far dos the fluid move along the tube from equilibrium on the right?

    How do the masses mA and mB shown in the figure compare?

    If the fluid on the left is moving with speed V, what is the speed of the fluid on the right?
     

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  6. May 9, 2014 #5

    utkarshakash

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    I'm finding it difficult to relate the cross-sectional areas of two parts.
     
  7. May 9, 2014 #6

    TSny

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    You can assume the tube has the same cross section everywhere. Note that in going from the left figure to the right figure it's as though there's been a decrease in mass, mA, of fluid in the left section of the tube and an increase in mass, mB, in the right section.
     
  8. May 9, 2014 #7

    utkarshakash

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    If they have same cross-sectional area, shouldn't then they travel the same distance?
     
  9. May 9, 2014 #8

    TSny

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    Yes, they travel the same distance S along the tube. Try to find an expression for the increase in PE of the system in terms of S. Hint: The displacement of the fluid is equivalent to moving the mass mA in the figure to mB as far as PE is concerned.
     

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  10. May 10, 2014 #9

    utkarshakash

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    ΔPE = (mb S(sin β) - ma S(sin α))g

    I have one more doubt. If cross-sectional area is same and they travel the same distance S, then shouldn't ma=mb ?
     
  11. May 10, 2014 #10

    haruspex

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    Yes, the masses are the same.
     
  12. May 10, 2014 #11

    utkarshakash

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    The kinetic energy will be mv2.

    Total energy = [itex]msg(\sin \beta - \sin \alpha) + m v^2 \\
    \rho A S^2 g(\sin \beta - \sin \alpha) + \rho ASv^2 [/itex]

    If I differentiate this wrt t and set it to zero I get


    [itex]2g(\sin \beta - \sin \alpha)S+ 2S \ddot{S} + v^2 = 0 [/itex]

    This doesn't look right to me. The equation still contains v^2. Ugh! :uhh:
     
  13. May 10, 2014 #12

    utkarshakash

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    Thanks for the reply.
     
  14. May 10, 2014 #13

    TSny

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    A couple of things to consider here.

    1. Getting all the signs correct is a little tricky. Note that your expression gives no change in PE if α = β. But that can't be right.

    2. The change in PE is due to the loss of mA on the left and the gain in mB on the right. It's as though you remove mA and lift it up and place it at mB. When considering the change in PE of the chunk of mass, you need to think about the change in vertical height of the center of mass of the chunk.
     
  15. May 10, 2014 #14

    TSny

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    Besides correcting the PE term (see post #13), you also need to fix the KE term. All of the fluid in the tube moves.
     
    Last edited: May 10, 2014
  16. Jul 4, 2014 #15
    I would like to solve this problem using force concepts .Please guide me .

    Suppose the length of the left column is l1 and that of the right column is l2 .When left column goes down along the length by x ,the right column moves up along the length by x.

    Component of gravitational force on left tube ( downwards along the length of left tube ) = F1 = ρA(l1-x)gsinα

    Component of gravitational force on right tube ( downwards along the length of right tube ) = F2 = ρA(l2+x)gsinβ

    Net gravitational force acting along the direction of motion of the fluid = F2-F1

    F2-F1 = ρA(l2+x)gsinβ - ρA(l1-x)gsinα = ρAl2gsinβ+ρAxgsinβ - ρAl1gsinα+ρAxgsinα

    From equilibrium situation , ρAl2gsinβ = ρAl1gsinα

    Thus , F2-F1 = ρAxgsinβ+ρAxgsinα = ρAxg(sinβ+sinα)

    I don't know if this is what you were suggesting . What should I do next ?

    Many Thanks
     
  17. Jul 4, 2014 #16

    TSny

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    Good work on getting the net force. Newton's second law might be useful.
     
  18. Jul 4, 2014 #17
    That does give correct answer , but I haven't understood few things .

    1) But Newton's second law is in vector form .Here component of gravitational force along the length on both the left and right parts act downwards at angle 180 - (α+β ) . We didn't consider this fact .Rather added them algebraically . Why ?

    2) What about the component of force perpendicular to the motion of fluid ? Is it cancelled by the force exerted by tube ? Does the tube exert only normal force ?
     
  19. Jul 4, 2014 #18

    TSny

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    1) The net force that you calculated was the tangential component of the net force, which is the force that causes the fluid to accelerate along the tube. We can assume negligible friction from the tube.

    2) Yes, the component of gravity perpendicular to the tube is balanced out by the normal force from the tube. The exception is at the bottom of the tube where the fluid must go around the bend. The normal force and gravitational force provide the centripetal force here.
     
  20. Jul 4, 2014 #19
    Thanks .

    But if we calculate the pressure at the bottom due to the left tube and that due to right tube just as in the case of static fluid and consider the pressure difference, we arrive at the same correct result.

    Suppose P1 be the pressure at the bottom due to left tube , P1=ρg(l1-x)sinα
    P2 be the pressure at the bottom due to right tube , P2=ρg(l2+x)sinβ

    Then F2-F1 = (P2-P1)A = ρAxgsinβ+ρAxgsinα ,same result as in post#15 .

    Why application of wrong concept is giving us right result :confused:?
     
  21. Jul 4, 2014 #20

    TSny

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    How do you plan to use this expression for F2-F1 to get the period of oscillation? What does this "net force" F2-F1 act on?
     
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