Time period of piston performing SHM Suppose I have a closed piston-cylinder system which contains an ideal gas,and it is being compressed adiabatically by the piston which has a vertical orientation,and in doing so,the piston is actually undergoing SHM. Now if I use a similar piston in a horizontal orientation (keeping in mind the dimensions of the cylinder), would the time period of the SHM be affected? Would anything change in this orientation? The equation $pV^\gamma=\text{constant}$ would still be valid, but would anything change at all?
 A: It is straightforward to show that if $$pV^\gamma =\text{constant}$$
then
$$\frac{dp}{dV}=-\gamma \frac pV$$
Therefore for small changes, $\Delta V$, in volume, giving rise to small changes, $\Delta p$, in pressure, we have to first order,
$$\frac{\Delta p}{\Delta V}=-\gamma \frac pV$$
$p$ and $V$ are the equilibrium pressure and volume. If the gas cylinder is vertical with the (heavy) piston at the top, $p$ will be greater and $V$ will be smaller than if the cylinder were horizontal, so $p/V$ will be greater and $|\Delta p|$ will be greater for a given $\Delta V$. This implies that the restoring force per unit displacement of the piston will be greater, so the time period will be smaller. The piston's inertia is, of course, the same in both cases. 
Edit
The above treatment leads leads to a periodic time given by
$$\tau=2 \pi \sqrt{\frac{mV}{\gamma p A^2}}$$
$m$ is the mass of the piston and $A$ is its cross-sectional area.
And this very equation appears in my trusty copy of Zemansky, in his treatment of Rüchhardt's determination of $\gamma$.
Note that a smaller $V$ and larger $p$, as when the system is vertical with the piston on top, will indeed, according to this equation, give a smaller $\tau$.
We can easily work out how much smaller. For the horizontal and vertical cases respectively we have
$$\tau_0=2 \pi \sqrt{\frac{mV_0}{\gamma p_0 A^2}}\ \ \text{and}\ \ \tau_1=2 \pi \sqrt{\frac{mV_1}{\gamma p_1 A^2}}$$
Dividing the second by the first
$$\frac{\tau_1}{\tau_0}=\sqrt{\frac{V_1 p_0}{V_0 p_1}}$$
But if the temperature at equilibrium is the same in both cases
$$p_1 V_1=p_0 V_0$$
So eliminating $V_{1}/V_{0}$ and tidying,
$$\frac{\tau_1}{\tau_0}=\frac{p_0}{p_1}=\frac{p_0}{p_0+(mg/A)}$$
Here, $p_0$ is atmospheric pressure, $m$ and $A$ are the piston's mass and  cross-sectional area.
A: The equilibrium position of the piston would change due to the weight of piston. However it would not interfere with time period or frequency as it is acting throughout the process. In the second position equilibrium position would be different. For more accuracy you could also take into account the change in external pressure in the two cases . Yet it will also have no effect on time period but only on equilibrium position.
