Oscillation of free surface of water in parallelepiped container

[charie]

Homework Statement

it was assigned as by the lecturer.

Relevant Equations

conservation of energy

[Mentor Note: Two duplicate threads merged...]

in container with dimensions L×D, rests water of height H and density ρ. we disturb the water along L dimension, and accept an oscillation is caused on the free surface of the water, which maintains its flatness, so that the central of mass of the water moves up-down and left-right, and the water does not move in the extreme positions but only has change in potential energy due to gravity, while in the middle positions where the surface is horizontal, all the energy gets converted to kinetic energy. say the central of mass is in position $(x_o,y_o)$ in equilibrium, and $(x,y)$ during the oscillation. how do we show that $mg(y-y_o)$ is proportional to $(x-x_o)^2$, and find cyclic frequency of the oscillation of the water?

i have tried to prove said proportionality using conservation of energy as follows: for every column of height $(H+a)$, it is $dm=σ*a*dx$, and $m=σ*L*H$, where $σ=ρ*D$. also, let K be the kinetic energy and U dynamic potential energy. then, $K={1 \over 2 }*m*u^2$, $U=mg(y-y_o)$. because ${dE \over dt}=0$ (we assumed we have simple harmonic oscillations) and $E=K+U$, we find that $(y-y_o)~u^2$. the main problem here is what assumptions would i be correct to make about the relationship of $(x-x_o)$ and velocity u? how would i compute frequency?

thank you for reading.

 

[berkeman]

Welcome to PF.

Can you upload a diagram of this scenario? Use the "Attach files" link below the Edit window to upload a PDF or JPEG diagram.

Also, is this for schoolwork? What is the context of the question?

 

[charie]

Welcome to PF.

Can you upload a diagram of this scenario? Use the "Attach files" link below the Edit window to upload a PDF or JPEG diagram.

Also, is this for schoolwork? What is the context of the question?

hello, i am new here and I don't know much about the site yet, thank you for information. it is a question my professor gave in the last lecture. i will upload the diagram shortly.

 

[berkeman]

Okay, great. I'll move this to the schoolwork forums now.

Also, for schoolwork, we require you to post the Relevant Equations that you think are applicable, and to show us your best efforts to start working the problem.

We prefer that math is posted here using LaTeX. You can learn more about that in the "LaTeX Guide" link below the Edit window. I'll also send you a Private Message (PM) with more hints about how to use LaTeX.

 

[berkeman]

what exactly do you mean by my best efforts? I've made various efforts with different thought processes but they are not very easy to read, would that beok to post?

Yeah, it means to post the math that you've been working on (preferably using LaTeX, but clear images can work at first as you learn LaTeX). Thanks.

 

[erobz]

latex: for inline equations use

## [latex code]##

, not $

For centered equations use:

 $$ [latex code] $$

see LaTeX Guide

 

[charie]

latex: for inline equations use

## [latex code]##

, not $

For centered equations use:

 $$ [latex code] $$

see LaTeX Guide

thank you for the help!

 

[haruspex]

I have doubts about the validity of the model. Seems to make the assumption that at any instant all the water at a given horizontal displacement from the centre is moving with the same horizontal velocity. But in reality, as H increases, there is less movement at depth.

Glossing over that, consider a vertical slice through the tank, width dx, displacement x from the centre line. Let the angle of the surface be $\theta(t)$.
In terms of $\dot\theta$ etc., can you get an equation for the velocity of the water in the slice?

Disclaimer: I am heavily jetlagged, so I have not checked my approach works. There may be simpler ways.

 

[kuruman]

We know (or can easily show) that if the tank accelerates with constant acceleration $a$ along a straight line, the fluid level will be inclined at angle $\varphi$ relative to the horizontal given by $$\tan\varphi=\frac{a}{g}.$$ We are told that the water surface "maintains its flatness" throughout the motion. Therefore it is safe to assume that each intermediate inclination corresponds to an intermediate acceleration. So we can model this as having the tank oscillate horizontally with frequency $\omega$ and amplitude $A$.

Then the tangent of the maximum angle of inclination is $$\frac{y_{\text{max}}}{L}=\tan\varphi_{\text{max}}=\frac{a_{\text{max}}}{g}=\frac{\omega^2A}{g}\implies \omega^2=\frac{g\,y_{\text{max}}}{AL}.$$ Now to find $A$, I think it is safe to say that at $x=A$ the gravitational potential energy change of the fluid of mass $m$ from the equilibrium rectangular configuration to the non-equilibrium trapezoidal configuration is equal to $\frac{1}{2}m\omega^2A^2.$

Disclaimer: I have not attempted this and I hope it does not lead to something like $1=1$ but it would be my first approach.

 

[haruspex]

I have doubts about the validity of the model. Seems to make the assumption that at any instant all the water at a given horizontal displacement from the centre is moving with the same horizontal velocity. But in reality, as H increases, there is less movement at depth.

Glossing over that, consider a vertical slice through the tank, width dx, displacement x from the centre line. Let the angle of the surface be $\theta(t)$.
In terms of $\dot\theta$ etc., can you get an equation for the velocity of the water in the slice?

Disclaimer: I am heavily jetlagged, so I have not checked my approach works. There may be simpler ways.

Update:

My idea was to find the KE of each vertical element as a function of $x, \theta, \dot\theta$ and integrate to find the whole KE. The GPE is a simple function of $\theta$, so energy conservation gives an ODE in $\theta, t$.
However, for the whole KE I am getting an expression involving $\ln(H+L\tan(\theta)/2)$.
We could make a small angle approximation, but it still doesn’t look promising.

 

[haruspex]

it is safe to assume that each intermediate inclination corresponds to an intermediate acceleration.

It corresponds to an intermediate constant acceleration, yes, but does it work for a varying one?

 

[kuruman]

It corresponds to an intermediate constant acceleration, yes, but does it work for a varying one?

We are told that the surface maintains its flatness. I think that this is a hint to assume that the surface inclination adjusts instantly to the acceleration as it varies from zero to a maximum value. If the tank executes simple harmonic motion, we have the acceleration and hence the angle as a linear function of position.

 

[haruspex]

We are told that the surface maintains its flatness. I think that this is a hint to assume that the surface inclination adjusts instantly to the acceleration as it varies from zero to a maximum value. If the tank executes simple harmonic motion, we have the acceleration and hence the angle as a linear function of position.

With a surface angle perfectly adjusted to one horizontal acceleration, the water is at its lowest potential. If that acceleration changes, it is no longer at lowest potential. The readjustment releases energy which will persist as slop.
I don't read it as the hint you suggest.

It seems clear that the answer will have the form $\omega=f(H,L)\sqrt g$. The density will be irrelevant.

Provided $L<2H$, I suspect the answer would not differ greatly from that of a solid semicylindrical body rocking back and forth in a smooth cylindrical channel of diameter L.

One handy feature of the problem is that it lends itself to kitchen verification - if the flatness can be achieved.

 

[haruspex]

Just thought of another line…
With the surface at angle $\theta$, what is the net force exerted by the end walls? What relationship does that give between the horizontal displacement of the mass centre and its horizontal acceleration?

Edit: @charie, were you able to get an answer using the above?