Exploring Soap Bubble Oscillations: A Comparison of Force and Energy Approaches

First law? :sleeping:Yeah, I realized a little bit later that the last part was a mistake. I think I better go to sleep and deal with this in the morning 😅Some final ramblings before I doze off... in the isothermal case$$E = 8 \pi r^2 \gamma + \frac{1}{2}m_b \dot{r}^2$$and ##\dot{E} = \dot{Q} = -\dot{W}=P \frac{dV}{dt} = P\frac{dV}{dr} \frac{dr}{dt}##, whilst in the adiabatic case,$$E =
  • #1
etotheipi
Homework Statement
Determine the period of oscillation of a soap bubble with surface tension ##\gamma##, as well as an equilibrium radius of ##r_0##, equilibrium thickness of ##h## and density of ##\rho##. The soap bubble is surrounded by vacuum, and contains a gas that undergoes isothermal changes.
Relevant Equations
##P(r_0) = \frac{4\gamma}{r_0}##
I have solved it with a force approach, but would like to know how to do it via an energy approach. For starters, here is the force approach. Consider a small, approximately circular, surface element of mass ##m## such that the angle from the centre to the edge of this element is ##\alpha##, which we keep constant as the bubble expands and compresses. In the radial direction it is acted upon by a force due to the internal pressure, of magnitude ##f_p(r)##, and a force due to the surface tension, of magnitude ##f_{st}(r)##. We have$$f_p(r) - f_{st}(r) = m\ddot{r}$$Now let ##r = r_0 + \varepsilon##, such that a first order Taylor expansion gives $$\left[f'_p(r_0)- f'_{st}(r_0) \right] \varepsilon = m\ddot{\varepsilon}$$Since ##\frac{dV}{dr} = 4\pi r^2 = \frac{3V}{r}##, and because the changes are isothermal, we have that ##\frac{dP}{dr} = -\frac{P}{V} \frac{dV}{dr} = -\frac{3P}{r}##. We also have that the area of the element, ##A = \pi (r\alpha)^2 \implies \frac{dA}{dr} = 2\pi \alpha^2 r##. So$$f'_p(r_0) = P(r_0)\frac{dA}{dr}(r_0) + A(r_0)\frac{dP}{dr}(r_0) = - 4\pi \gamma \alpha^2$$We also know that ##f_{st}(r) = 2 \times 2\pi r \alpha \gamma \times \sin{\alpha} \approx 4\pi \alpha^2 \gamma r##, which means that $$f'_{st}(r_0) = 4\pi \alpha^2 \gamma$$Finally we can insert this into the equation of motion, to obtain $$\ddot{\varepsilon} = -\frac{8\gamma}{h \rho r_0^2} \varepsilon$$which agrees with the answer in the solution manual. I wanted to now solve it with an energy approach, but am struggling. So far I have written the total energy of the configuration (with ##m_b## the total mass of the bubble) as the sum of the free energy of the two surfaces and the kinetic energy of the bubble:$$E = 2 \times \gamma A + \frac{1}{2}m_b \dot{r}^2 = 8 \pi r^2 \gamma + \frac{1}{2}m_b \dot{r}^2$$Differentiating w.r.t. time to express conservation of energy would seem to imply$$16 \pi \gamma r \dot{r} + m_b \dot{r} \ddot{r} = 0 \implies \ddot{r} = -\frac{16 \pi \gamma}{m_b} r$$which is evidently incorrect, since the bubble would just collapse in on itself. So I hoped someone could help out with the energy approach. Thanks ☺
 
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  • #2
What about the energy changes in the contained gas?
 
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  • #3
haruspex said:
What about the energy changes in the contained gas?

I had assumed that since the process is isothermal, and that the energy of a fixed mass of an ideal gas depends only on temperature, that there would be no such energy changes. That said, there must be a term of some sort pertaining to the gas, it's just I can't think of it at the moment :wink:
 
  • #4
etotheipi said:
I had assumed that since the process is isothermal, and that the energy of a fixed mass of an ideal gas depends only on temperature, that there would be no such energy changes. That said, there must be a term of some sort pertaining to the gas, it's just I can't think of it at the moment :wink:
Your force method includes a force from the pressure of the gas. As the bubble expands, this is doing work.

(Hard to believe it would not be adiabatic, though.)
 
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  • #5
haruspex said:
Your force method includes a force from the pressure of the gas. As the bubble expands, this is doing work.

Right, so in that case there is also heat flowing into the bubble during the expansion and out of the bubble during the compression, in order for ##\frac{dU}{dt} = 0##. That would be the missing piece... however it doesn't seem easy at all to calculate (maybe we need to look at things like the heat equation...). Would you say that this means an energy approach is basically impractical here?
 
  • #6
etotheipi said:
Right, so in that case there is also heat flowing into the bubble during the expansion and out of the bubble during the compression, in order for ##\frac{dU}{dt} = 0##. That would be the missing piece... however it doesn't seem easy at all to calculate (maybe we need to look at things like the heat equation...). Would you say that this means an energy approach is basically impractical here?
You have PV constant, a radius increase dr does work ##4\pi r^2P.dr##, and you know how V depends on r.
It's all the same as your force calculation, just integrated up to get energy.
 
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  • #7
haruspex said:
You have PV constant, a radius increase dr does work ##4\pi r^2P.dr##, and you know how V depends on r.
It's all the same as your force calculation, just integrated up to get energy.

Ah, right yes, I'm being dumb. I'll scribble some stuff out but I think I see what you are getting at. :doh:

haruspex said:
(Hard to believe it would not be adiabatic, though.)

Yes, quite. In fact the problem comes with the statement "somewhat artificial and non-realistic setup, but still a valid exercise".

I think the energy equation I wrote in its current form would apply to the adiabatic case.
 
  • #8
etotheipi said:
I think the energy equation I wrote in its current form would apply to the adiabatic case
No, you had no work done by the gas. Whether isothermal or adiabatic there will be work done by the gas; the difference is that in the isothermal case something is balancing that with a heat flow (from a vacuum?!).
 
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  • #9
haruspex said:
No, you had no work done by the gas. Whether isothermal or adiabatic there will be work done by the gas; the difference is that in the isothermal case something is balancing that with a heat flow (from a vacuum?!).

Yeah, I realized a little bit later that the last part was a mistake. I think I better go to sleep and deal with this in the morning 😅
 
  • #10
Some final ramblings before I doze off... in the isothermal case$$E = 8 \pi r^2 \gamma + \frac{1}{2}m_b \dot{r}^2$$and ##\dot{E} = \dot{Q} = -\dot{W}=P \frac{dV}{dt} = P\frac{dV}{dr} \frac{dr}{dt}##, whilst in the adiabatic case,$$E = 8 \pi r^2 \gamma + \frac{1}{2}m_b \dot{r}^2 + U_{gas}$$ $$\dot{E} = 0 = 16 \pi \gamma r \dot{r} + m_b \dot{r} \ddot{r} + \dot{U}_{gas}$$with ##\dot{U}_{gas} = \dot{W}## by the bubble
 
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  • #11
I am from India , and the notations written here are confusing for me, i also had a doubt in a similar question but i am getting the wrong answer by force method, the wordings of the question to be precise were: A soap bubble of radius ‘r’ and wall thickness ‘h’ is made in vacuum by blowing an ideal diatomic gas in it. The surface tension of the soap solution is ‘T’ and density of the soap solution is ‘Rho’. Assume that the heat capacity of the soap film is much greater than that of the gas in the bubble...…

I started this way
pressure inside soap bubble: 4T/r, as outside pressure is zero, and excess pressure is same as the pressure inside soap bubble
so P.r= constant, or P.V^(1/3)= constant, not a isothermal process in this case
then if due to oscillations the radius of the whole sphere increases by a small amount "dr" then
change in pressure inside soap bubble, delta P=4T/r+dr -4T/r = approximately -4T.dr/(r^2) [i approximated r+dr=r, at last part] and hence
restoring force =delta P*Area(4pie*r^2)
restoring force =M*a
(here M= mass of air bubble, a=acceleration of oscillations), M= Rho*Area*h
equating both a= -(4T/Rho*h*r^2)*dr
from here angular frequency of oscillations=square root of (4T/Rho*h*r^2)
but the answer given is square root of (8T/Rho*h*r^2)
 
  • #12
gourvi said:
I am from India , and the notations written here are confusing for me, i also had a doubt in a similar question but i am getting the wrong answer by force method, the wordings of the question to be precise were: A soap bubble of radius ‘r’ and wall thickness ‘h’ is made in vacuum by blowing an ideal diatomic gas in it. The surface tension of the soap solution is ‘T’ and density of the soap solution is ‘Rho’. Assume that the heat capacity of the soap film is much greater than that of the gas in the bubble...…

I started this way
pressure inside soap bubble: 4T/r, as outside pressure is zero, and excess pressure is same as the pressure inside soap bubble
so P.r= constant, or P.V^(1/3)= constant, not a isothermal process in this case
then if due to oscillations the radius of the whole sphere increases by a small amount "dr" then
change in pressure inside soap bubble, delta P=4T/r+dr -4T/r = approximately -4T.dr/(r^2) [i approximated r+dr=r, at last part] and hence
restoring force =delta P*Area(4pie*r^2)
restoring force =M*a
(here M= mass of air bubble, a=acceleration of oscillations), M= Rho*Area*h
equating both a= -(4T/Rho*h*r^2)*dr
from here angular frequency of oscillations=square root of (4T/Rho*h*r^2)
but the answer given is square root of (8T/Rho*h*r^2)
You have considered the change in pressure for a small change in radius, but the force from the surface tension will change too.
 
  • #13
haruspex said:
You have considered the change in pressure for a small change in radius, but the force from the surface tension will change too.
thanks for the quick reply, but i didn't consider the force due to surface tension, i only multiplied the pressure difference by Area of action of the radial restoring force, i may not be getting your point, so could you elaborate further
 
  • #14
gourvi said:
thanks for the quick reply, but i didn't consider the force due to surface tension, i only multiplied the pressure difference by Area of action of the radial restoring force, i may not be getting your point, so could you elaborate further
You need to consider what the net restoring force is when the radius increases by a small amount. The pressure will drop because of the increased volume, but the force from surface tension will also increase because of the increased area.

I have another concern but it is late here. Will take a look tomorrow.
 
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  • #15
ok for more information, we were also asked the the molar heat capacity of the gas as the first part, i am posting the official question as well as the provided solution , why did you comment that i treated the system as isothermal, i am genuinely intrigued as solutions posted by the institute are incorrect sometimes
 

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  • #16
haruspex said:
You need to consider what the net restoring force is when the radius increases by a small amount. The pressure will drop because of the increased volume, but the force from surface tension will also increase because of the increased area.

I have another concern but it is late here. Will take a look tomorrow.
ok thanks i am struggling at this for days and just want to understand it, doesn't matter how long it takes
 
  • #17
gourvi said:
ok for more information, we were also asked the the molar heat capacity of the gas as the first part, i am posting the official question as well as the provided solution , why did you comment that i treated the system as isothermal, i am genuinely intrigued as solutions posted by the institute are incorrect sometimes
In your original solution you wrote the equilibrium equation ##P=4\frac Tr=4(\frac{4\pi}{3V})^\frac 13## and deduced that as it oscillates the pressure varies as V-1/3. That does not follow. Oscillation means it moves away from equilibrium, so ##P=4\frac Tr## does not apply.

You need to consider some small change in radius, x, how that affects the two pressures, and the consequent rate of change in x.

Let P=P(r). It helps to think of the surface tension as exerting a pressure Q(r)=4T/r. At equilibrium this balances the gas pressure, hence P(r0)=Q(r0)=4T/r0.
But when r increases to r0+x this pressure changes by Q'x, and the gas pressure by P'x. This leads to a net restoring pressure -(P'-Q')x.
The mass per unit area of the film is ##\rho h##, so ##\rho h\ddot x=(P'-Q')x##.

I would say a realistic model is adiabatic, so ##P∝V^{-\gamma}##, so ##P'=-3\gamma P/r##. Meanwhile, Q'=-Q/r.
##\rho h \ddot x=-(3\gamma-1)\frac {4T}{r_0^2}x##.
If you want to take it as isothermal then set ##\gamma=1##.

I'll take a look at the official solution later.
 
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  • #18
Ok, I have looked at the official answer and straightaway it is wrong. As I wrote, ##P∝V^{-1/3}## is a statement about equilibrium values, i.e. if you had a number of bubbles of the same material and thickness, all at equilibrium, that would be the relationship between their various pressures and volumes. It is not a statement about how pressure and volume vary as a given bubble is disturbed from equilibrium.

It does appear you are expected to take it as adiabatic, with ##\gamma=7/5##. So my answer becomes ##\rho h \ddot x=-\frac{64\sigma}{5r^2}x##.
 
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  • #19
This is all very good, one thing I would like to add is to numerically check the relevant quantities after one obtains the symbolic solution. If the period ##t_{osc}## is very small in the sense that heat conducted ##Q = k A \frac{\Delta T}{thickness} t_{osc}## through the soap surface can be negligible compared with the work done in the process, then adiabatic approximation is much more appropriate.

haruspex said:
In your original solution you wrote the equilibrium equation ##P=4\frac Tr=4(\frac{4\pi}{3V})^\frac 13## and deduced that as it oscillates the pressure varies as V-1/3. That does not follow. Oscillation means it moves away from equilibrium, so ##P=4\frac Tr## does not apply.

You need to consider some small change in radius, x, how that affects the two pressures, and the consequent rate of change in x.

Let P=P(r). It helps to think of the surface tension as exerting a pressure Q(r)=4T/r. At equilibrium this balances the gas pressure, hence P(r0)=Q(r0)=4T/r0.
But when r increases to r0+x this pressure changes by Q'x, and the gas pressure by P'x. This leads to a net restoring pressure -(P'-Q')x.
The mass per unit area of the film is ##\rho h##, so ##\rho h\ddot x=(P'-Q')x##.

I would say a realistic model is adiabatic, so ##P∝V^{-\gamma}##, so ##P'=-3\gamma P/r##. Meanwhile, Q'=-Q/r.
##\rho h \ddot x=-(3\gamma-1)\frac {4T}{r_0^2}x##.
If you want to take it as isothermal then set ##\gamma=1##.

I'll take a look at the official solution later.
 

Related to Exploring Soap Bubble Oscillations: A Comparison of Force and Energy Approaches

1. What causes a soap bubble to oscillate?

A soap bubble oscillates due to the surface tension of the soap film. The surface tension creates a force that pulls the bubble inward, while the air pressure inside the bubble pushes outward, causing the bubble to expand. As the bubble expands, the surface tension decreases, allowing the bubble to contract. This cycle of expansion and contraction creates the oscillations.

2. How do the properties of the soap solution affect the oscillations of a soap bubble?

The properties of the soap solution, such as the concentration of soap and the type of soap used, can affect the surface tension of the soap film and therefore impact the oscillations of a soap bubble. A higher concentration of soap can decrease the surface tension, leading to larger and faster oscillations. Different types of soap can also have varying effects on the surface tension, resulting in different oscillation patterns.

3. What factors can influence the frequency of oscillations in a soap bubble?

The frequency of oscillations in a soap bubble can be influenced by several factors, including the size of the bubble, the thickness of the soap film, and the air pressure inside the bubble. A larger bubble will have a lower frequency of oscillations, while a thinner soap film will result in a higher frequency. Additionally, a higher air pressure inside the bubble will also lead to a higher frequency of oscillations.

4. How do external forces affect the oscillations of a soap bubble?

External forces, such as wind or vibrations, can disrupt the oscillations of a soap bubble. These forces can cause the bubble to move or burst, altering the oscillation pattern. However, if the external force is applied consistently, it can also create a new frequency of oscillation in the bubble.

5. Can the oscillations of a soap bubble be used for scientific purposes?

Yes, the oscillations of a soap bubble can be used for various scientific purposes, such as studying the properties of soap films and surface tension. The frequency and pattern of the oscillations can also be used to measure the viscosity of a liquid or the elasticity of a material. Soap bubbles are also commonly used in demonstrations and experiments to explain concepts such as resonance and harmonics.

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