Oscillation of fluids in a U tube

[Rob2024]

Homework Statement

A U-tube manometer consists of a uniform diameter cylindrical tube that is bent into a U shape. It is originally filled with water that has a density ρw. The total length of the column of water is L. Ignore surface tension and viscosity.

The water is displaced slightly so that one side moves up a distance x and the other side lowers a distance x. Find the frequency of oscillation.

Relevant Equations

$U = \frac{1}{2} k x^2 = \frac{1}{2} m v^2$

We can relate the maximum speed of the fluid with the displaced energy of the fluid. Imagine a small block of fluid of height h is moved from left to the right. Then $$m g h = \frac{1}{2} M v^2$$ where $m = \rho A h$, $M = \rho A L$. Therefore $v = \sqrt{2g/L} h$, from here we know the angular frequency is $\sqrt{2g/L}$, therefore the frequency is $f = \frac{1}{2\pi} \sqrt{2g/L}$.

What troubles me is that what if we moved 2h amount of fluid from left to the right, then the energy equation becomes $\rho A 2h g 2h = \frac{1}{2} M v^2$, then the frequency becomes $f = \frac{1}{2\pi} \sqrt{8g/L}$. Normally we can avoid this when velocity is related to amplitude of a given point. However in fluid, all the points are moving at the same velocity because of continuity. What is the resolution for the 2nd approach that does not give correct answer.

 

[ergospherical]

The strategy is to write down an equation of energy conservation, and then use that to (hopefully) derive something resembling simple harmonic motion. We can denote the displacement of the water from its equilibrium position along the tube by $x$.

Let's take the potential energy at equilibrium ($x=0$) to be zero, which you can always do. If the water is displaced by $x$ from equilibrium, then you've effectively moved a small block of fluid of mass $dm = A \rho_w x$ from the left to the right, raising its height by $x$ in the process. So the potential energy at displacement $x$ is $U(x) = A \rho_w g x^2$. The total energy:$$E = \frac{1}{2} M \dot{x}^2 + A \rho_w g x^2$$$E$ is constant, so you can take the derivative with respect to time $t$ and derive an equation for SHM...

 

[Rob2024]

Yes, that's exactly what I am thinking. The problem I am having is that if I lifted $2h$ amount of fluid, then the equation seems to give me incorrect answer. Doesn't every point in the fluid move at the same velocity? Therefore I could use either $\dot{x}$ or $\dot{0.5 x}$ to find the KE of the fluid?

 

[TSny]

What troubles me is that what if we moved 2h amount of fluid from left to the right, then the energy equation becomes $$\rho A 2h g 2h = \frac{1}{2} M v^2$$,

I think this is right.

then the frequency becomes $$f = \frac{1}{2\pi} \sqrt{8g/L}$$.

How are you getting $f$ from $v$? I think this is where you are making a mistake.

 

[Rob2024]

Because every point in the fluid moves at the same speed, $\dot{h} = v$.

 

[TSny]

Let's go back to where you had a small block of height h.

Then $m g h = \frac{1}{2} M v^2$ where $m = \rho A h$, $M = \rho A L$. Therefore $v = \sqrt{2g/L} h$

Ok.

from here we know the angular frequency is $\sqrt{2g/L}$

This is correct, but how did you arrive at this expression for the angular frequency?

 

[Rob2024]

Using the spring-obj analogy $U = \frac{1}{2} k h^2 = \frac{1}{2} m v^2$, $f = \frac{1}{2 \pi} \sqrt {\frac{{k}{m}} $

 

[TSny]

Using the spring-obj analogy $U = \frac{1}{2} k h^2 = \frac{1}{2} m v^2$, $f = \frac{1}{2 \pi} \sqrt {\dfrac{k}{m}}$

Note that $h$ on the left side of the first equation is the amplitude, $A$, of the simple harmonic motion, and $v$ on the right side is the maximum speed of the motion. So, the energy equation may be written $$\frac{1}{2} k A^2 = \frac{1}{2} m v_{max}^2.$$ When you consider the case of moving a block of water of height $2h$ from one side of the tube to the other, what is the amplitude, $A$, of the motion?

 

[Rob2024]

That may be the problem, how do we define the amplitude in this case? Why do we have to use the end point of the fluid instead of the center of mass of the part of the fluid deviating from the equilibrium position? Every point in the fluid moves at the same speed.

 

[TSny]

Yes, every point moves at the same speed. Every point moves in simple harmonic motion with the same amplitude. Thus, the amplitude can be obtained from the motion of any point you choose. So, you can use a point at the surface of the fluid on one side of the tube.

From $\frac 1 2 k A^2 = \frac 1 2 m v_{max}^2$, you can see that $$v_{max} = \sqrt{\frac k m} A = \omega A$$ You can use this to find $\omega$ from $v_{max}$ and $A$. I believe you have already found the correct expressions for $v_{max}$ for the two cases you considered in your first post. To get $\omega$ for these two cases, you need to use the appropriate value for $A$ for each case. [Hopefully using $A$ for the amplitude will not cause confusion with using $A$ for the area of the tube.]

 

[Rob2024]

That's the problem. It seems obvious when you choose the end point of the fluid to measure amplitude. As you say, the amplitude can be obtained from the motion of any point, if I use CM of the part of the fluid away from equilibrium as amplitude, then I am halving the amplitude but the maximum speed remains the same leading to a different answer.

 

[ergospherical]

But if you were to "start" the oscillations by lifting a portion of height $2h$ from the left to the right side, then the amplitude of the resulting oscillations will be double the amplitude had you only lifted a portion of height $h$ to start off with. The former system also has $2^2$ times the energy, so double the maximum speed.

 

[Rob2024]

I I am starting to see what's wrong with using CM of the deviated fluid to define GPE. The problem is that that part of the fluid is shrinking and expanding as the fluid oscillates. Therefore such an object is not well defined and that's why there seems to be a contradiction of velocity. To do this properly, we should use the CM of the entire fluid which is well defined. If the initial CM height (base is at 0) is $y_0$. It can be shown from CM definition if we lift one side by $h$, the CM of the entire fluid rises by $h^2/L$. Then $Mg h^2/L = \frac{1}{2} Mv^2$ leads to the correct answer and there is no ambiguity. Can we say it's a coincidence that using end of the fluid as amplitude (a popular method) yields correct answer? I haven't thought through that part clearly yet.

 

[ergospherical]

In my previous notation the energy of the whole system is $E = \frac{1}{2} M \dot{x}^2 + (Mg/L) x^2$, with $M$ the total mass of the water and $\dot{x}$ the common speed. The potential energy of the system relative to the equilibrium position ($x=0$) is $U(x) = (Mg/L) x^2$. You could derive that by considering how much the gravitational energy would increase by if you lifted a parcel of water of height $x$ from one side to the other. If you differentiate the energy equation with respect to $t$, you will get $\ddot{x} = -(2g/L)x$ and then you are done.

 

[erobz]

If we give each column a fluid height $x,y$ we should be able to say via energy conservation:

$$ \frac{1}{2} \rho A g \left[ ( y + dy )^2 - y^2 + (x+dx)^2-x^2 \right] + \frac{1}{2} \rho A \left[ y \left( ( \dot y + d \dot y )^2 - \dot y^2 \right) + x \left( ( \dot x + d \dot x )^2 - \dot x^2 \right) \right] = 0 $$

Then use the constraint that the total mass of the system is constant $M = \rho A ( y + x )$ and its derivatives to eliminate one of the variables $x,y$ in terms of the other to get the ODE.

 

[Rob2024]

I realized the real problem troubling me was the definition of amplitude, not necessarily how much fluid we lift or conservation of energy. That's resolved now. Thanks for your help.

 

[Chestermiller]

Another way to do this problem, under the assumption of an inviscid fluid, is to recognize that the sum of the potential energy and the kinetic energy of the fluid does not change with time. So the time derivative of this total energy is zero.

 

[Chestermiller]

Let the U-tube consist of a horizontal section below and two vertical sections in which the liquid equilibrium height in each vertical section is h. Then at time t, the height of liquid in the left hand vertical tube is $h-\delta$ and the height of liquid in the right hand section is $h+\delta$. Then taking the datum of gravitation potential energy as the elevation of the horizontal section joining the two vertical tubes, we have that the total potential energy of the fluid is $$PE=\rho gA\frac{(h-\delta)^2}{2}+\rho gA\frac{(h+\delta)^2}{2}=\rho gA(h^2+\delta^2)$$and the kinetic energy of the fluid is $$KE=\rho AL\left(\frac{d\delta}{dt}\right)^2/2$$So the total energy is $$E=\rho gA(h^2+\delta^2)+\rho AL\left(\frac{d\delta}{dt}\right)^2/2$$Differentiating this with respect to time and setting the result equal to zero gives $$2g\delta \frac{d \delta}{dt}+L\frac{d\delta}{dt}\frac{d^2\delta}{dt^2}=0$$ or $$2g\delta +L\frac{d^2\delta}{dt^2}=0$$