Water oscillations in a U-tube

In summary: A)(w^2)(h^2-x^2))-(.5(lpA)(w^2)(x^2))In summary, the problem involves a confined column of liquid in a U-tube and the task is to construct expressions for potential and kinetic energies and show that the column will oscillate with a period of pi*root(2l/g). To solve this, one can use a Lagrangian approach and consider the force acting on the liquid in opposite direction to its position. This results in a differential equation for simple harmonic motion, and by solving for the frequency and time period, one can obtain the expressions for potential and kinetic energies. By showing that the total
  • #1
newtn46
7
0

Homework Statement


Consider a column of liquid (density ρ) of length ℓ confined in a U-tube of uniform cross-sectional area A. Suppose that the water level on one side is pushed down a small amount and then released.

1) Construct expressions for the potential and kinetic energies of the liquid.
2) Hence show that the column will oscillate with a period of pi*root(2l/g)


Homework Equations


PE=m*g*h
KE=m*V^2/2


The Attempt at a Solution


I am able to do part 2) using forces, but I should use energy. Therefore I should show that the total energy is constant and can be written as:
E=0.5*S*x^2+0.5*I*x^2
I've found an expression for the potential energy, it depends on the displacement, which is good. However I don't know how to find a relationship between kinetic energy and displacement. I know it should depend on the square of the displacement, but I am not able to find the equation. Any ideas?
 
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  • #2
Since (assuming no drag) the liquid is displaced by a distance h every half an oscillation couldn't you work out the velocity as distance/time?
 
  • #3
No, actually the liquid also has some acceleration, so I can't do that. I have to show that kinetic energy is a function of x^2.
 
  • #4
I believe you should use a energy argument.

1. Construct a differential eq. from cons. of energy expression. (Drawing a diagram helps)
2. Solve.(Should be SH considering small displacements)
3. ect.

The most difficult part is getting the PE and KE right...this is what the diagram is useful for.

By no means a solution, this is just my thoughts.
 
Last edited:
  • #5
Thanks for the answer. I know what I have to do, but I don't know how to construct the expression for the kinetic energy. Any hints?
 
  • #6
You know the expression for the kinetic energy of an object with mass m and speed v, don't you? The object is the mercury. The velocity is the time derivative of the displacement.

ehild
 
  • #7
newtn46 said:

Homework Statement


Consider a column of liquid (density ρ) of length ℓ confined in a U-tube of uniform cross-sectional area A. Suppose that the water level on one side is pushed down a small amount and then released.

1) Construct expressions for the potential and kinetic energies of the liquid.
2) Hence show that the column will oscillate with a period of pi*root(2l/g)


Homework Equations


PE=m*g*h
KE=m*V^2/2


The Attempt at a Solution


I am able to do part 2) using forces, but I should use energy. Therefore I should show that the total energy is constant and can be written as:
E=0.5*S*x^2+0.5*I*x^2
I've found an expression for the potential energy, it depends on the displacement, which is good. However I don't know how to find a relationship between kinetic energy and displacement. I know it should depend on the square of the displacement, but I am not able to find the equation. Any ideas?

you may use lagrangian
 
  • #8
the above question is simply based on simple harmonic motion as the force acting on the liquid is acting in opposite direction to position of the liquid level and the force is proportional to the displacement of the fluid.
so,
f=2pAxg x is instaneous distance of the liquid level from mean or rest position
now,
mass of liquid column is lpA
therefore,
a=f/m= 2pAgx/lpA
now in simple harmonics
a=w^2x w=frequency of vibration
therefore,
w^2x=2pgAx/lpA
therefore,
w=(2g/l)^.5
therefore time period T is
T= pi*root(2l/g)
kinetic energy=.5(lpA)(w^2)(h^2-x^2) h is the small displacement by which water is
pushed down
potential energy=.5(lpA)(w^2)(x^2)
SINCE YOU KNOW TOTAL ENERGY AND TOTAL ENERGY IS CONSERVED THEREFORE
kinetic energy =TOTAL ENERGY-potential energy
 

1. What causes water oscillations in a U-tube?

The water oscillations in a U-tube are caused by the phenomenon of capillary action, which is the ability of water to "climb" up narrow tubes due to the cohesive forces between water molecules.

2. Why does the water level in one side of the U-tube rise while the other side falls?

This is due to the difference in surface tension and adhesion forces between the water and the walls of the U-tube. The side with higher surface tension will have a higher water level, while the side with higher adhesion will have a lower water level.

3. How does the diameter of the U-tube affect the water oscillations?

The diameter of the U-tube affects the rate of oscillations. A narrower diameter will result in faster oscillations, while a wider diameter will result in slower oscillations.

4. Can the length of the U-tube impact the water oscillations?

Yes, the length of the U-tube can impact the water oscillations. A longer U-tube will have a slower rate of oscillations compared to a shorter U-tube.

5. What factors can cause the water oscillations to stop or become unstable?

The water oscillations can stop or become unstable due to external factors such as air currents, temperature changes, and vibrations. Changes in the diameter or length of the U-tube can also affect the stability of the oscillations.

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