Oscillation of a liquid in U -tube

[Vibhor]

I am facing difficulty in understanding a key point . Hence I am putting it here instead of Introductory Physics section .

Here is a part of the solution provided by the teacher ,but it doesn't look convincing to me.

Similar solution is provided here http://www.phy.duke.edu/~dtl/136126/36h2_sho.html

Solution - The liquid of length l is in the U-tube .If a compressive force is applied in one arm and it displace water by depth y in the same arm.

In the result difference in the liquid level in the tube = 2y.

The weight of the liquid present in this level 2y provides the necessary restoring force. So when the compressing force is removed the liquid tries to come back to its equilibrium position but due to inertia overshoots this position and thus starts oscillating.

Restoring force = - (2yAρ)g

It tries to bring the liquid back to its equilibrium position and thus acts on the whole of liquid present in the U-tube. Thus restoring acceleration is given by

If a compressive force is applied in one arm and it displace water by depth y in the same arm.

In the result difference in the liquid level in the tube = 2y.

The weight of the liquid present in this level 2y provides the necessary restoring force. So when the compressing force is removed the liquid tries to come back to its equilibrium position but due to inertia overshoots this position and thus starts oscillating.

Restoring force = - (2yAρ)g

It tries to bring the liquid back to its equilibrium position and thus acts on the whole of liquid present in the U-tube.

Line marked in red is what I don't understand . I feel that the restoring force acts only on the fluid of length l-2y ,whereas the solution says that the restoring force acts on the entire liquid .

Could somebody please explain this to me ?

Many thanks

 

[olivermsun]

The restoring force has to accelerate the entire length of fluid as a unit, so it's causing the fluid of length l - y (it's low by y in one arm and high in one arm by y, right?) to move back up and it's pushing the fluid of l + y down.

 

[Vibhor]

Right .

But isn't the restoring force like a pushing force on the column of length l-2y . In that case how can this force due to column length 2y exert force on itself ?

 

[Delta2]

Its the Earth that exerts force on the liquid as a whole. The net force from the Earth to the whole liquid is 2yAρg for any displacement y from the equilibrium position. Thats because that's the "extra weight" of the liquid in one side of the U tube relative to the other, the other two weights of magnitute (l-y)Aρg are canceled out.

Btw its a bit unclear the liquid has total length l or 2*l? It doesn't change anything in the analysis anyway just If its l then the height is l/2+y in one column and l/2-y in the other and the two weights that cancell out are (l/2-y)Aρg in this case.

 

[ehild]

Delta is right. It is gravity that exerts force on the whole liquid.
Let be the length of the whole liquid L.
See picture. The column of liquid in the right tube is y1 high and its mass is m₁.That in the left tube is y₂ and its mass is m₂. The mass in the horizontal part is m₃. The forces on m₁ are m₁g downward and the push N₁ from m₃ upward. The force on m₂ is m₂g downward and N₂ upward. (The normal forces are equal to the pressure at the interfaces multiplied by the cross section A. ) The horizontal part is pushed by N1 to the left and by N2 to the right. It can move only horizontally, its weight is balanced by force of the wall. The liquid is incompressible, it moves as a whole, with acceleration a. So $N_1-m_1 g=m_1\ddot y_1 = -m_1 a$ and $N_2-m_2 g=m_2 \ddot y_2=m_2 a$, For the horizontal part, $N_1-N_2= m_3 a$. Combine the equations : $(m_1+m_2+m_3) a=(m_1-m2)g$, or $La=(y_1-y_2) g$. If Δy is the displacement from the original level, $y_1-y_2=2Δy$. $2a=-(\ddot y_1-\ddot y_2)=-d^2(Δy)/dt^2$. The final equation is $Ld^2(Δy)/dt^2+2gΔy=0$.


ehild

 

[Delta2]

Well ehild gives a very good analysis, but i think a simple intuitive explanation maybe better: that the weights of the two equally lengthed (L/2-y) parts are not exactly canceled out as i say in my first post, however they cannot change the kinetic state of the fluid because when we have two equally lengthed parts on the left and right side of the tube, we don't observe fluid motion.

 

[Orodruin]

Delta is right. It is gravity that exerts force on the whole liquid.

This is not strictly true. There are internal forces due to pressure within the liquid. The entire external force that does work is gravity. The basic point is that the net force has to accelerate all of the liquid.

If familiar with the Lagrange or Hamilton formalisms, they are conceptually much simpler to use in this scenario.

 

[Vibhor]

The horizontal part is pushed by N1 to the left and by N2 to the right. It can move only horizontally, its weight is balanced by force of the wall.

ehild

How is normal force acting in horizontal direction on the horizontal part of the fluid ? Doesn't it act vertically upwards on the liquid in the vertical section of the tube ?

 

[ehild]

The name "normal force" was for analogy. It is pressure multiplied by the area of a surface/interface. The pressure does not have direction. The pressure times area is a force, perpendicular to the surface. At the same place, it is the same in any directions.

ehild

 

[Vibhor]

The name "normal force" was for analogy. It is pressure multiplied by the area of a surface/interface. The pressure does not have direction. The pressure times area is a force, perpendicular to the surface. At the same place, it is the same in any directions.

ehild

Right .

In that case I think the solution given in the OP is correct .

What do you and Delta² think about the solution provided by my teacher and the one given in the link in the OP ?

 

[ehild]

Are not all solutions the same?

ehild

 

[Delta2]

Well i ve to say that ehild's solution gives better understanding of the inner workings of this situation. Because just stating that the net restoring force due to gravity is - (2yAρ)g and that the weights from the two L/2-y parts "cancell out" is very intuitive and kind of vague. While for example if you follow ehilds' post from equation
$(m_1+m_2+m_3) a=(m_1-m2)g$,
one can understand clearly and formally why the two weights of m1 and m2 cancell out.