Frequency of Oscillations of a Bead Resting on a Gas

[penguin46]

Homework Statement

A glass flask with a long narrow neck, of small cross-section area 𝐴, is
filled with 1 mol of a dilute gas with Cp/Cv = gamma, at temperature 𝑇. A glass
bead of mass 𝑚 fits snugly into the neck and can slide along the neck
without friction. Find the frequency of small oscillations of the beads about
its equilibrium position.

Relevant Equations

P=F*A, Cp = dU/dT Cv = dH/dT

Hi, this is a question from my textbook that I can't quite make sense of. I don't really know where even to begin, to be honest. The only thing I can think of is that the bead is at equilibrium, thus the force of gravity is equal to the pressure exerted by the gas. I don't know where the heat capacity would come in. Possibly something to do with How much the gas gets compressed during a small perturbation? Any help would be appreciated.

 

[TSny]

The only thing I can think of is that the bead is at equilibrium, thus the force of gravity is equal to the pressure exerted by the gas.

If the bead is not oscillating, then it will be in equilibrium. In that case, the net force on the bead will be zero. But, pressure is not a force, so it is not quite right to say that the force of gravity is equal to the pressure of the gas when the bead is in equilibrium.

Of course, you are interested in the case where the bead is oscillating. So, it is not in equilibrium. When the bead is displaced from the equilibrium position, think about why the net force on the bead will not be zero.

I don't know where the heat capacity would come in. Possibly something to do with How much the gas gets compressed during a small perturbation?

The problem statement suggests that it's the ratio $C_p/C_v$ that's important. This is a hint as to the type of process that the gas undergoes when the bead is oscillating. Is the process isothermal? isobaric? adiabatic? something else?

 

[penguin46]

It should be non-zero force because the pressure depends on volume, right?

I would say it should be adiabatic, because it is probably moving too fast to be isothermal.

 

[TSny]

It should be non-zero force because the pressure depends on volume, right?

I would say it should be adiabatic, because it is probably moving too fast to be isothermal.

Yes. I think you are probably expected to assume adiabatic.

Can you think of a way to obtain an expression for the net force on the bead if the bead is displaced a small distance from the equilibrium position?

It's not clear from the wording of the problem whether the neck of the flask is vertical or horizontal or in some other orientation.

 

[penguin46]

well the net force would be equal to pressure x area, and so we would calculate how much the pressure changes based on a small displacement?

it is vertical

 

[TSny]

well the net force would be equal to pressure x area, and so we would calculate how much the pressure changes based on a small displacement?

Yes. Should you worry about the force on the bead created by the external atmospheric pressure?

Edit: Does the net force on the ball include the force of gravity?

it is vertical

Ok

 

[penguin46]

Yes. Should you worry about the force on the bead created by the external atmospheric pressure?

i have no idea, none of this makes any sense. How am i supposed to know whether the question writer wants me to take that into account? i have thought about this question for like 3 hrs now and still don't even know where to start

 

[TSny]

How am i supposed to know whether the question writer wants me to take that into account?

I just wanted you to realize that the force on the bead that is due to the external atmospheric pressure is a significant force.

So, if the bead is sitting at rest in the neck in equilibrium, there are three forces acting on the bead that add up to a net force of zero. Can you list these three forces? If the bead displaces a bit from equilibrium, which of these three forces remain unchanged and which force changes?

 

[penguin46]

force of gravity, outside pressure, inside pressure. only inside pressure changes

 

[TSny]

force of gravity, outside pressure, inside pressure. only inside pressure changes

OK. Again, pressure is not a force. But you have the right idea.

Suppose the bead makes a small displacement from the equilibrium position. If $dP$ is the corresponding change in pressure of the gas inside, how would you express the $net$ force on the bead when it is at this displaced position?

 

[penguin46]

A*dP

 

[TSny]

A*dP

Yes. (I hope you are clear on why this simple expression for the net force on the bead does not include the force from the external atmospheric pressure or the weight of the bead.)

Let the small displacement of the bead from equilibrium be denoted $dy$. (We're representing small quantities as differentials.)

For simple harmonic motion, the net force and the displacement obey Hooke's law, $F_{net} = -k x$ for some "force constant" $k$. See if you can show that for this problem the net force $AdP$ is related to $dy$ by Hooke's law and discover the force constant for this problem. This is where the adiabatic nature of the process is important.

 

[penguin46]

yeah sorry dude i give up

 

[TSny]

yeah sorry dude i give up

Maybe it will help to take a break from it.

Sorry, but we are not allowed to just give away the solution. The general idea is that if you know the basic equation that relates $P$ and $V$ for an adiabatic process of an ideal gas, then you can use it to relate $dP$ to $dV$. And $dV$ is related to $dy$ and $A$. So, you can discover how $dP$ is related to $dy$, etc.

 

[penguin46]

i don't know that equation

 

[TSny]

See anything familiar here?

 

[Fred Wright]

Spoiler

I have the following suggestions:
First, I think the reference to the adiabatic constant implies that T is constant throughout the experiment. Now assume we have an ideal gas in a tube of length $L$ with cross-sectional area $A$. From the ideal gas law,
$$
P=\frac{F_{gas}}{A}=\frac{nRT}{V}
$$
$$
F_{gas}=A\frac{nRT}{V}
$$
Now take the variable $z$ to be positively increasing from the top of the tube towards its bottom. The volume of the tube is,
$$
V=A(L-z)
$$
We have for the force exerted by the gas in the tube,
$$
F_{gas}(z)=\frac{nRT}{(L-z)}=\frac{nRT}{L(1-\frac{z}{L})}
$$
We assume the equilibrium position of the bead $z_0$ is not very far from the top of the long tube such that $z_0 << L$ and we take the first order approximation of $F_{gas}(z)$,
$$
F_{gas}(z)\approx \frac{nRT((1+ \frac{z}{L})}{L}
$$
There are three forces acting on the bead $F_{gas}$, $mg$, and the force from atmospheric pressure $\frac{P_{atmosph}}{A}$. We can now write a force equation for the system,
$$
m\ddot{z} =mg + \frac{P_{atmosph}}{A} - F_{gas}(z)
$$
At the equilibrium position,
$$
\frac{nRT((1+ \frac{z_0}{L})}{L}=mg + \frac{P_{atmosph}}{A}
$$
and the differential equation becomes,
$$
m\ddot{z}=\frac{nRT((1+ \frac{z_0}{L})}{L}-\frac{nRT((1+ \frac{z}{L})}{L}=-\frac{nRT(z-z_0)}{L^2}
$$
Make the substitution $z'=z-z_0$ and the differential equation becomes,
$$
\ddot{z'} + \frac{nRT}{mL^2}z'=0
$$
[/SPOILER

 

[TSny]

Spoiler

First, I think the reference to the adiabatic constant implies that T is constant throughout the experiment.

For an ideal gas undergoing a reversible, adiabatic process we have $TV^{\gamma-1} = \rm const$. Thus, any change in $V$ must be accompanied by a change in $T$.

Now, whether or not the adiabatic assumption for this problem is a good assumption, I don't know. But I'm guessing that as an academic exercise, they want the student to make this assumption.

 

[Fred Wright]

For an ideal gas undergoing a reversible, adiabatic process we have $TV^{\gamma-1} = \rm const$. Thus, any change in $V$ must be accompanied by a change in $T$.

Now, whether or not the adiabatic assumption for this problem is a good assumption, I don't know. But I'm guessing that as an academic exercise, they want the student to make this assumption.

Your right of course (my bad),
$$
T=\frac{C}{V^{\gamma -1}}
$$
The force exerted on the bead by the gas in the tube is,
$$
F_{gas}(z)=\frac{CAnR}{V^{\gamma}}=\frac{CA^{1-\gamma}nR}{L^{\gamma}(1-\frac{z}{L})^{\gamma}}=\frac{CA^{1-\gamma}nR}{L^{\gamma}}\sum_{k=0}^{\infty}
\begin{pmatrix}
\gamma \\
k
\end{pmatrix}\left( \frac{z}{L} \right)^k
$$
$\gamma$ is not an integer and taking the first order approximation the force becomes,
$$
F_{gas}(z)\approx \frac{CA^{1-\gamma}nR}{L^{\gamma}}
\left [1+\frac{\Gamma (1+ \gamma)}{\Gamma (\gamma)}\left( \frac{z}{L} \right)\right ]
$$
where the factorial is replaced by the Gamma function. I felt compelled to post this, out of embarrassment, to amend my silly misinterpretation of the problem. Evidently the problem requires a degree of mathematical sophistication beyond what is expected in an introductory course in thermodynamics.

 

[TSny]

The adiabatic law for an ideal gas can be expressed as $PV^{\gamma} = \rm const$. To work this problem, you only need to relate $dP$ to $dV$ since we assume "small oscillations". So, the math is not bad.