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 <br /> nRT_0 (1-x/L)<br />

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})<br />

Expanding the the fraction with respect to x/L :

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

Omitting all terms but linear, we get:

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

ehild

 

[vladimir69]

ok
thanks for that