Angular Frequency of a Piston with Ideal Gas

[phys3107_]

Homework Statement

A frictionless piston of mass m is a precise fit in the vertical cylindrical neck of a large container of volume V. The container is filled with an ideal gas and there is a vacuum above the piston. The cross-sectional area of the neck is A. Assuming that the pressure and volume of the gas change slowly and isothermally, determine the differential equation of motion for small displacements of the piston about its equilibrium position and hence calculate the angular frequency of oscillation.

$$m = 0.1 \rm{kg}$$
$$V = 0.1 \rm{m^{3}}$$
$$A = \pi*10^{-4} \rm{m^{2}}$$

Homework Equations

Ideal Gas: $$PV=nRT$$
Isothermal: $$P_{1}V_{1}=P_{2}V_{2}$$
Newton's 2nd: $$F=ma$$
Pressure: $$P=\frac{F}{A}$$

3. The Attempt at a Solution
As the process is isothermal:
$$P_{0}V_{0}=PV$$
If $$x$$ is the position of the piston and $$x_{0}$$ is the equilibrium position:
$$P_{0}Ax_{0}=PAx$$
So:
$$P=P_{0}\frac{x_{0}}{x}$$

At any point in time:
$$ma = PA$$

Substituting for the pressure into the above:
$$ma = \frac{P_{0}Ax_{0}}{x}$$
$$ma = \frac{P_{0}V_{0}}{x}$$

As a differential:
$$m\frac{\mathrm{d}^2 x}{\mathrm{d} t^2} - \frac{P_{0}V_{0}}{x} = 0$$

This is non-linear so I can't solve it for the angular frequency. Been looking at this for a while so have I made silly errors or am I just approaching this completely wrong?
Any help or hints appreciated :).

P.S. Sorry for the awkward formatting. How do I do inline equations?

 

[haruspex]

At any point in time:

ma=PA​

Gravity?

 

[Chestermiller]

As Haruspex was hinting at, the equation should be $$ma=A(P-P_0)$$Can you figure out why?

 

[phys3107_]

Sorry for the late reply, I decided to revisit it a day later after I realized the gravity blunder. I managed to get that equation, Chestermiller.
At any point in time:
$$ma = PA - mg$$
But we also know mg from equilibrium conditions:
$$mg = P_{0}A$$
So:
$$ma = A(P - P_{0})$$

So do I now just need to find P as a function of x?

 

[Chestermiller]

Sorry for the late reply, I decided to revisit it a day later after I realized the gravity blunder. I managed to get that equation, Chestermiller.
At any point in time:
$$ma = PA - mg$$
But we also know mg from equilibrium conditions:
$$mg = P_{0}A$$
So:
$$ma = A(P - P_{0})$$

So do I now just need to find P as a function of x?

You already had P as a function of x. You need to substitute $x=x_0+\delta$ in the equation $P=\frac{P_0x_0}{x}$, and then linearize with respect to $\delta$.

You should really write $$P=\frac{P_0V_0}{(V_0+A\delta)}$$and linearize with respect to $\delta$

 

[phys3107_]

Thank you although I'm not sure I understand what you mean by "linearize".

 

[haruspex]

Thank you although I'm not sure I understand what you mean by "linearize".

Chet means expand the expression into a linear form, c+dδ, by making an approximation that is valid for small δ.
The reference to small displacements in the question is a hint that you need to do this.

 

[Chestermiller]

You already had P as a function of x. You need to substitute $x=x_0+\delta$ in the equation $P=\frac{P_0x_0}{x}$, and then linearize with respect to $\delta$.

You should really write $$P=\frac{P_0V_0}{(V_0+A\delta)}$$and linearize with respect to $\delta$

$$\frac{1}{V_0+A\delta}=\frac{1}{V_0\left(1+\frac{A\delta}{V_0}\right)}\approx \left(\frac{1}{V_0}\right)\left(1-\frac{A\delta}{V_0}\right)$$

 

[phys3107_]

Ahh I see, I understand the approximation now. So to get to the differential equation I just need to substitute that into $ma = A(P - P_{0})$?

 

[Chestermiller]

Ahh I see, I understand the approximation now. So to get to the differential equation I just need to substitute that into $ma = A(P - P_{0})$?

Yes, but you also need to eliminate $P_0$ by using $P_0A=mg$, and you need to express the acceleration in terms of the second derivative of $\delta$

 

[phys3107_]

Yes, its clicked now, thank you for your help! The differential is give by:
$$\frac{\mathrm{d}^2 \delta}{\mathrm{d} t^2} + \frac{gA}{V_{0}}\delta = 0$$
So the solution for the angular frequency is:
$$\omega = \sqrt{\frac{gA}{V_{0}}}$$

Correct? :D

 

[Chestermiller]

Yes, its clicked now, thank you for your help! The differential is give by:
$$\frac{\mathrm{d}^2 \delta}{\mathrm{d} t^2} + \frac{gA}{V_{0}}\delta = 0$$
So the solution for the angular frequency is:
$$\omega = \sqrt{\frac{gA}{V_{0}}}$$

Correct? :D

Sure.

 

[Chestermiller]

Yes, its clicked now, thank you for your help! The differential is give by:
$$\frac{\mathrm{d}^2 \delta}{\mathrm{d} t^2} + \frac{gA}{V_{0}}\delta = 0$$
So the solution for the angular frequency is:
$$\omega = \sqrt{\frac{gA}{V_{0}}}$$

Correct? :D

Sure.