Oscillations of a Piston in a cylinder containing a trapped gas

[Varak]

A frictionless cylinder of cross-sectional area A contains a gas that is trapped by a piston of mass m that fits the cylinder tightly but is free to move up and down.

It is open to atmospheric pressure (PA) on one end.

The piston is slightly displaced and when released oscillates about its equilibrium position.

Find the frequency of small oscillations and show that the oscillations are approximately harmonic with frequency w^2=gamma(g/H+P0A/mH), where H is the height of the trapped gas.

The hint is that it is an adiabatic compression where PV^gamma = constant and gamma=5/3. We're supposed to write P(y) and expand in a power series.
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Here is what I have so far.

The three forces on the piston are -mg - PA + P(y)A

P0 = Initial Pressure
V0 = Initial Volume = HA

P0*V0^gamma = P(y)*V(y)^gamma

P0*HA^gamma = P(y)*((H+y)A)^gamma

P(y)=(P0*H^gamma)/(H+y)^gamma

I'm guessing this is where I need to expand into a power series, but I'm not sure how to do so. Any help would be appreciated.

Also, if someone could tell me how to input equations in a better format, I'd appreciate it. I've seen many posts with Mathcad-like text which is much easier to read, but, since I'm a new user, I don't know how to input all the special characters.

 

[sic]

yes, you take the first order approximation:
dP(y)/dy=Whatever
hence
dP(y)=Whatever*dy
F=dP(y)*A=Whatever*A*dy (harmonic oscillator equation)
and k/m=omega^2=Whatever*A/m
now pay attention that
PO=PA+mg/A

 

[quicknote]

Thanks for your help!

First, let's rewrite the equation PV^gamma = constant as P = k/V^gamma, where k is a constant. This will make it easier to work with in terms of the variables given in the problem.

Next, we can use the fact that the gas is trapped by the piston to relate the volume and displacement of the piston. Since we know that the piston is free to move up and down, we can assume that the volume of the gas is equal to the area of the cylinder times the displacement of the piston, or V(y) = A(H+y).

Substituting this into our equation, we get P(y) = k/(A(H+y))^gamma.

Now, we can expand this equation using the binomial theorem, which states that (1+x)^n = 1 + nx + (n(n-1)x^2)/2! + (n(n-1)(n-2)x^3)/3! + ... for any real number x. In our case, our variable is y, so we can rewrite our equation as P(y) = k/(A(H(1+y/H)))^gamma. Using the binomial theorem, we get:

P(y) = k/(A(H(1+y/H)))^gamma = k/(A(H(1+gy/H)))^gamma = k/(A(H+gy))^gamma = k/[(AH)^gamma(1+gy/H)^gamma] = k/[(AH)^gamma(1+gamma(gy/H))] = k/[(AH)^gamma(1+gamma(gy/H)+ (gamma(gy/H))^2/2! + ...]

We can simplify this further by noting that gamma = 5/3, so our equation becomes:

P(y) = k/[(AH)^gamma(1+5gy/3H+ (25g^2y^2)/(9H^2) + ...)]

Since we are only interested in small oscillations, we can neglect the higher order terms in the expansion and only consider the first two terms. Thus, our equation becomes:

P(y) = k/[(AH)^gamma(1+5gy/3H)]

Now, we can use this equation to determine the frequency of small oscillations. The restoring force on the piston is given by F = P(y)A - PA = P(y)A - P0A = A(P(y) -