Thermodynamics and simple harmonic motion

[vladimir69]

Homework Statement

A horizontal piston-cylinder system containing n mole of ideal gas is surrounded by air at temperature $T_{0}$ and pressure $P_{0}$. If the piston is displaced slightly from equilibrium, show that it executes simple harmonic motion with angular frequency $\omega=\frac{AP_{0}}{\sqrt{MnRT_{0}}}$, where A and M are the piston area and mass, respectively. Assume the gas temperature remains constant.

Homework Equations

PV=nRT
F=Ma
P=F/A
x=amount by which piston moves
L=length of cylinder
F = force on piston

The Attempt at a Solution

$$V_{0}=AL$$
$$V=A(L-x)$$
$$F_{net}=A(P_{0}-P)$$
$$L=\frac{nRT_{0}}{P_{0}A}$$
$$P=\frac{ALP_{0}}{V}$$
Popping all this into the mix gives
$$F_{net}=\frac{-P_{0}^2A^2x}{nRT_{0}-P_{0}Ax}$$
$$M\frac{d^2x}{dt^2}=\frac{-P_{0}^2A^2x}{nRT_{0}-P_{0}Ax}$$
If I get rid of the $P_{0}Ax}$ term then we arrive at the correct answer but not sure why one should throw away this term

Thanks

 

[ehild]

The denominator is equal to $$nRT_0 (1-x/L)$$

It is assumed that the piston is only slightly displaced from equilibrium, so that x/L<<1. That means that its second or higher power can be neglected with respect to it.
We often apply this method in Physics when working with small quantities: use Taylor expansion and keep linear terms.
Your expression of force can be written in terms of (x/L),

$$F=AP_0(1-\frac{1}{1-x/L})$$

Expanding the the fraction with respect to x/L :

$$F=AP_0(1-(1+x/L+(x/L)^2+(x/L)^3+...))<br />$$

Omitting all terms but linear, we get:

$$F=-AP_0(x/L)<br />$$

ehild

 

[vladimir69]

ok
thanks for that