Weight of the cork = Weight of the liquid displaced by the floating cork

Let the cork be depressed slightly by x. As a result, some extra water of a certain volume is displaced. Hence, an extra up-thrust acts upward and provides the restoring force to the cork.

Up-thrust = Restoring force F = Weight of the extra water displaced

F = –(Volume depressed × Density × g)

Volume depressed= Area × Distance through which the cork is depressed

Volume = Ax

$\therefore \mathrm{F}=-{\mathrm{Ax\rho }}_{\mathrm{l}}\mathrm{g} ... \left(\mathrm{i}\right)$

$\mathrm{In} \mathrm{equilibrium}:$
$\mathrm{F}=\mathrm{kx}$
$\mathrm{k}=\frac{\mathrm{F}}{\mathrm{x}}$
$\mathrm{where} \mathrm{k} \mathrm{is} \mathrm{a} \mathrm{constant}.$
$\mathrm{k}=\frac{\mathrm{F}}{\mathrm{x}}=-{\mathrm{A\rho }}_{\mathrm{l}}\mathrm{g} ...\left(\mathrm{ii}\right)$
$\mathrm{The} \mathrm{time} \mathrm{period} \mathrm{of} \mathrm{the} \mathrm{oscillations} \mathrm{of} \mathrm{the} \mathrm{cork}:$
$\mathrm{T}=2\mathrm{\pi }\sqrt{\frac{\mathrm{m}}{\mathrm{k}}} ...\left(\mathrm{iii}\right)$
$\mathrm{where},$
$\mathrm{m}=\mathrm{Mass} \mathrm{of} \mathrm{the} \mathrm{cork}$
$\mathrm{m}=\mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{cork} \times \mathrm{Density}$
$\mathrm{m}=\mathrm{Base} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{cork} \times \mathrm{Height} \mathrm{of} \mathrm{the} \mathrm{cork} \times \mathrm{Density} \mathrm{of} \mathrm{the} \mathrm{cork}$
$\mathrm{m}= \mathrm{Ah\rho }$
$\mathrm{Hence}, \mathrm{the} \mathrm{expression} \mathrm{for} \mathrm{the} \mathrm{time} \mathrm{period} \mathrm{becomes}:$
$\mathrm{T}=2\mathrm{\pi }\sqrt{\frac{\mathrm{Ah\rho }}{{\mathrm{A\rho }}_{\mathrm{l}}\mathrm{g}}}=2\mathrm{\pi }\sqrt{\frac{\mathrm{h\rho }}{{\mathrm{\rho }}_{\mathrm{l}}\mathrm{g}}}$