Angular Frequency of a Piston with Ideal Gas

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1. Feb 1, 2017

phys3107_

1. The problem statement, all variables and given/known data
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.

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

2. Relevant 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?

Last edited: Feb 1, 2017
2. Feb 1, 2017

Gravity?

3. Feb 2, 2017

Staff: Mentor

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

4. Feb 2, 2017

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?

5. Feb 2, 2017

Staff: Mentor

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 $\delta$

Last edited: Feb 2, 2017
6. Feb 2, 2017

phys3107_

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

7. Feb 2, 2017

haruspex

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.

8. Feb 2, 2017

Staff: Mentor

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

9. Feb 2, 2017

phys3107_

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

10. Feb 2, 2017

Staff: Mentor

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 $\delta$

11. Feb 2, 2017

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

12. Feb 2, 2017

Staff: Mentor

Sure.

13. Feb 2, 2017

Sure.