Consider an air chamber of volume V with a long neck of uniform area of cross section A, and a frictionless ball of mass m fitted smoothly in the neck at position C, the pressure of air below the ball inside the chamber is equal to the atmospheric pressure. Increase the pressure on the ball by a little amount P, so that the ball is depressed to position D, where CD = y.There will be decrease in volume and hence increase in pressure of air inside the chamber. The decrease in volume of the air inside the chamber, ΔV A= yVolumetric strain $= \frac{\text{change in volume}}{\text{original volume}}$ $= \frac{\Delta V}{V}= \frac{-A y}{V}$∴ Bulk modulus of elasticity B, will be$B= \frac{\text{stress (or increase in pressure)}}{\text{volumetric strain}}$ $= \frac{-P}{A y V}= \frac{-P V}{A y}$Here, negative sign shows that the increase in pressure will decrease the volume of air in the chamber.$\therefore P= \frac{-B A y}{V}$Due to this excess pressure, the restoring force acting on the ball is$F=P \times A= \frac{-B A y}{V} \cdot A= \frac{-B A^{2}}{V} y$Clearly, $F \propto y$, Negative sign shows that the force is directed towards equilibrium position. If the applied increased pressure is removed from the ball, it will start executing linear SHM in the neck of chamber with C as mean position. In SHM, the restoring force, $F = -k y$Comparing (i) and (ii), we have$k = \frac{B A^2}{V}$, which is the spring factor.Now, inertia factor = mass of ball $= m$.As, $T=2\pi \sqrt{ \frac{\text{inertia factor }}{\text{spring factor}} }$ $=2\pi \sqrt{ \frac{m}{\frac{B A^2}{V}} }$ $= \frac{2\pi}{A} \sqrt{ \frac{m V}{B} }$$\therefore$ Frequency, $\upsilon = \frac{1}{T}= \frac{A}{2\pi} \sqrt{ \frac{B}{m V} }$