A piston of mass 20kg cylindrical in shape has a base area of 10 cm². It slides  smoothly inside a long cylinder with gas of certain mass on the other side. The  cylinder is kept with its axis horizontal. The piston is at equilibrium for length of  20cm of the gas column. If the piston is disturbed from its equilibrium, it executes  simple harmonic motion. Then the time period of oscillation of the cylinder is  (Atmospheric pressure =  ${10}^{5}Pa$ )

The restoring force responsible for simple harmonic motion is the excess pressure.  Considering the process to be isothermal,
$PV=\left(P+\Delta P\right)\left(V-\Delta V\right)$ . Simplifying this $0=\left(\Delta P.V\right)-\left(P.\Delta V\right)$  (Neglecting $\Delta P.\Delta V)$.
Here, $V=Ah$ and $\Delta V=Ax$  
Substituting these, we get
$\Delta P=\frac{\left(P.\Delta V\right)}{V}$  i.e.  $\Delta P=\frac{\left(P.Ax\right)}{Ah}$
By using the definition of force, $F=\Delta P.A$ ,
$\Delta P.A=F=\frac{P.Ax}{h}$ .
This can be now equated to restoring force $F=kx$  
Therefore, $kx=\frac{P.Ax}{h}$ ,
Or, $k=\frac{P.A}{h}$ . 
For a simple harmonic motion, we know that  $k=m{\omega }^2$ 
Therefore, $m{\omega }^2=\frac{P.A}{h}$ 
Or, ${\omega }^2=\frac{P.A}{mh}$ , i.e. $\omega =\sqrt{\frac{P.A}{mh}}$ ,
Using the relation  $\omega =\frac{2\pi }{T}$ , we get
$\frac{2\pi }{T}=\sqrt{\frac{P.A}{mh}}$ 
Or  $T=2\pi \sqrt{\frac{mh}{P.A}}$. Here $\mathrm{m}=20\mathrm{kg},\mathrm{h}=20\mathrm{cm}=0.2\mathrm{m},10{\mathrm{cm}}^2={10}^{-3}{\mathrm{m}}^2,P={10}^5\mathrm{Pa}$
Substituting these values we get,
$T=2\pi \sqrt{\frac{20\times 0.2}{{10}^5\times {10}^{-3}}}$

$T=2\pi \sqrt{\frac{4}{{10}^2}}=2\times \pi \times \frac{2}{10}=\frac{2\pi }{5}s$