How to Derive the Oscillation Period of Water in a U-Tube?

[newtn46]

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?

 

[Beer-monster]

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?

 

[newtn46]

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.

 

[blacknovember]

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.

 

[newtn46]

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?

 

[ehild]

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

 

[shauric]

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

 

[pahr samsalah]

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