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.