How Does Acceleration Affect Spring Stretch in an Adiabatic Piston System?

[AdityaDev]

Homework Statement

An adiabatic piston of mass m equally divides an insulated container of volume V₀ and length l filled with Helium.The initial pressures on both sides of the piston is P₀ and the piston is connected to a spring of constant k. The container starts moving with acceleration a. Find the stretch in spring when acceleration of piston becomes a. Assume displacement of piston << l.

Homework Equations

For adiabatic process, ##PV^{\gamma}##=constant

The Attempt at a Solution

using above equation, if piston is displaced by x towards left, ##P_1=P\frac{V^{\gamma}}{V^{\gamma}-2Ax}##
similarly, for right portion, ##P_2=P\frac{V^{\gamma}}{V^{\gamma}+2Ax}##
Now A=V/l
substituting, ##P_1=P(1-2x/l)^{\gamma}=P(1-\frac{2x\gamma}{l})##
similarly, ##P_2=P(1+2x/l)^{\gamma}=P(1+\frac{2x\gamma}{l})##
Now ##\Delta P=4\gamma P/l##
and from Newton's law, ##\Delta PA+kx=ma##
so ##4x\gamma \frac{P}{l}\frac{V}{l}+kx=ma##
so $$x=\frac{ma}{K+\frac{4P_0V_0\gamma}{l^2}}$$
But answer given is: $$x=\frac{ma}{K+\frac{8P_0V_0\gamma}{l^2}}$$

 

[Delta2]

There seems to be some mistakes or typos in the algebra involved. The denominator in the first 2 equations should be ##({V+-2Ax})^\gamma## . I seem to get ##P_1=P(1-2x/l)^{-\gamma}## and similar for P2.

 

[haruspex]

substituting ##P_1=P(1-2x/l)^{\gamma}##

Do you mean ##P(1-2x/l)^{-\gamma}##?

 

[AdityaDev]

sorry... ##P_1=(1-2x/l)^{-\gamma}=(1+2x\gamma/l)##
Answer is still same.

 

[Delta2]

You use the approximation ##(1-2x/l)^{-\gamma}\approx(1+2x\gamma/l)## which i think its valid only if the exponent is positive and 2x<<l.

 

[AdityaDev]

yes 2x<<l.
also, ##(1-x)^{n}=1-\frac{n}{1!}x+\frac{n(n+1)}{2!}x^2-##.This expansion is true for both positive and negative rational numbers and for negetive integers.
so, if x is small, then you can neglect all terms starting from 3rd term which gives you ##1-nx##

 

[haruspex]

I'm inclined to agree with your answer, it's 4, not 8.

 

[Delta2]

Well the only "easy" thing i see and could fix the result is that the V0 might refer to the volume of the half container, so that it is ##A=2V_0/l##. Maybe check the excersice description again?

 

[AdityaDev]

Well the only "easy" thing i see and could fix the result is that the V0 might refer to the volume of the half container, so that it is ##A=2V_0/l##. Maybe check the excersice description again?

The question is correct and i checked it again. The answer given is wrong.