(N/A) Volume of the air chamber $= V$
Area of cross-section of the neck $= a$
Mass of the ball $= m$
The pressure inside the chamber is equal to the atmospheric pressure.
Let the ball be depressed by $x$ units. As a result of this depression,there is a decrease in the volume and an increase in the pressure inside the chamber.
Decrease in the volume of the air chamber,$\Delta V = a x$
Volumetric strain $= \frac{\text{Change in volume}}{\text{Original volume}} = \frac{\Delta V}{V} = \frac{a x}{V}$
Bulk Modulus of air,$B = \frac{\text{Stress}}{\text{Strain}} = \frac{-p}{\frac{a x}{V}}$
In this case,stress is the increase in pressure. The negative sign indicates that pressure increases with a decrease in volume. Thus,$p = \frac{-B a x}{V}$.
The restoring force acting on the ball is $F = p \times a = \frac{-B a x}{V} \cdot a = \frac{-B a^2 x}{V} \dots (i)$
In simple harmonic motion,the equation for restoring force is $F = -k x \dots (ii)$,where $k$ is the force constant.
Comparing equations $(i)$ and $(ii)$,we get $k = \frac{B a^2}{V}$.
The time period of oscillations is $T = 2 \pi \sqrt{\frac{m}{k}} = 2 \pi \sqrt{\frac{V m}{B a^2}}$.