The volume of the air chamber = V
Area of the 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 would be a decrease in the volume and an increase in the pressure inside the chamber.
The decrease in the volume of the air chamber, ΔV = ax
Volumetric strain = $\frac{\mathrm{Change} \mathrm{in} \mathrm{volume}}{\mathrm{Original} \mathrm{volume}}$

$\Rightarrow \frac{∆\mathrm{V}}{\mathrm{V}}=\frac{\mathrm{ax}}{\mathrm{V}}$
$\mathrm{Bulk} \mathrm{Modulus} \mathrm{of} \mathrm{the} \mathrm{air}, \mathrm{B}=\frac{\mathrm{Stress}}{\mathrm{Strain}}=\frac{-∆\mathrm{P}}{{\displaystyle \frac{\mathrm{ax}}{\mathrm{V}}}}$
$\mathrm{In} \mathrm{this} \mathrm{case}, \mathrm{stress} \mathrm{is} \mathrm{equal} \mathrm{to} \mathrm{the} \mathrm{increase} \mathrm{in} \mathrm{the} \mathrm{pressure}.$
$\mathrm{The} \mathrm{negative} \mathrm{sign} \mathrm{indicates} \mathrm{that} \mathrm{pressure} \mathrm{increases} \mathrm{with} \mathrm{a} \mathrm{decrease} \mathrm{in} \mathrm{volume}.$
$∆\mathrm{P}=\frac{-\mathrm{Bax}}{\mathrm{V}}$
$\mathrm{The} \mathrm{restoring} \mathrm{force} \mathrm{acting} \mathrm{on} \mathrm{the} \mathrm{ball},$
$\mathrm{F}=∆\mathrm{P}\times \mathrm{a}$
$=\frac{-\mathrm{Bax}}{\mathrm{V}}.\mathrm{a}$
$=\frac{-{\mathrm{Ba}}^2\mathrm{x}}{\mathrm{V}}$
$\mathrm{Comparing} \mathrm{it} \mathrm{with} \mathrm{the} \mathrm{standard} \mathrm{equation} \mathrm{of} \mathrm{SHM}:$
$\mathrm{F} = –\mathrm{kx}, \mathrm{where} \mathrm{k} \mathrm{is} \mathrm{the} \mathrm{spring} \mathrm{constant}.$
$\mathrm{k}=\frac{{\mathrm{Ba}}^2}{\mathrm{V}}$
$\mathrm{Time} \mathrm{period},$
$\mathrm{T}=2\mathrm{\pi }\sqrt{\frac{\mathrm{m}}{\mathrm{k}}}$
$=2\mathrm{\pi }\sqrt{\frac{\mathrm{Vm}}{{\mathrm{Ba}}^2}}$