How Does a Liquid Oscillate in a U-Shaped Tube?

[qazxsw11111]

Homework Statement

A liquid is placed in a U-Shaped tube. Initially both are at the same level. Total length of liquid is L. But later one side is displaced from equilibrium by X.(Difference of both sides is 2x now). Show that the water will oscillate harmonically in the form of a=-w squared x. Hence derive an expression for period of oscillation in terms of L.

Homework Equations

Pressure= hpg
F=ma

The Attempt at a Solution

Not really sure about this question...

 

[Raze2dust]

You should show at least some steps..or at least tell us any idea you might have about how to approach the problem, so we know you have spent some time on it. :)

 

[Moetasim]

I would first clarify the parameters and assumptions of the situation. From the given information, it can be assumed that the liquid in question is incompressible and that the U-shaped tube is being held in a vertical position. Additionally, the displacement of one side by X can be assumed to be a small displacement, as larger displacements may result in non-harmonic oscillations.

To begin, we can use the equation F=ma to analyze the forces acting on the liquid in the tube. Since the liquid is in equilibrium initially, the forces must also be in equilibrium. The forces acting on the liquid are the weight of the liquid (mg) and the pressure exerted by the liquid (hpg), where h is the height of the liquid and g is the acceleration due to gravity. Since the two sides of the tube are at the same level, the pressure exerted by the liquid on both sides is equal. Therefore, we can write the equation mg=hpg, or h=m/g.

When one side of the tube is displaced by X, the forces are no longer in equilibrium and the liquid will begin to oscillate. The displacement of one side by X results in a difference in height of 2X between the two sides of the tube. This difference in height, or h, can now be expressed as h=(m/g)+X. This difference in height will change as the liquid oscillates, resulting in a changing pressure and therefore a changing force acting on the liquid.

To derive the equation for the period of oscillation, we can use the equation F=ma again. The force acting on the liquid is now the weight of the liquid (mg) minus the changing pressure due to the changing height (hpg). We can express this as F=mg-hpg. Since the mass of the liquid remains constant, we can rewrite this as F=m(g-hp). We can also substitute h=(m/g)+X, giving us F=m(g-(m/g+X)pg). Simplifying this, we get F=m(g-(mg/g^2+Xg)). Since g^2 is much larger than Xg, we can ignore Xg and rewrite the equation as F=m(g-mg/g^2). Simplifying further, we get F=m(g-1/g).

Now, we can use the equation F=ma to find the acceleration of the liquid. Since we know the force acting on the liquid and