A uniform solid cylinder of length 0.50 m and density 0.8 g cm^-3 floats upright in a liquid of density 1.0 g cm^-3 in a lift. The lift accelerates…

In the lift's frame the effective gravity is $g_{\text{eff}}=g+a=12.5\ \mathrm{m\,s^{-2}}$. Equilibrium submersion depth $h$ follows from $\rho_c L=\rho_l h$ (the effective gravity cancels), so $h=0.8\times0.50=0.40\ \mathrm{m}$. If the cylinder is displaced by $x$, the net restoring force is $\rho_l A g_{\text{eff}}x$ while its mass is $\rho_cAL$, so $\omega^2=\dfrac{\rho_l g_{\text{eff}}}{\rho_c L}=\dfrac{g_{\text{eff}}}{h}$. Thus $T=2\pi\sqrt{h/g_{\text{eff}}}=2\pi\sqrt{0.40/12.5}=1.12\ \mathrm{s}$. Using $g$ alone gives $2\pi\sqrt{0.40/10}=1.26\ \mathrm{s}$, using $L$ in place of $h$ together with $g$ gives $2\pi\sqrt{0.50/10}=1.41\ \mathrm{s}$, and using $g-a$ gives $1.45\ \mathrm{s}$. The result does not depend on the cross-sectional area, and doubling the length of the cylinder would scale the period by $\sqrt2$ because $h$ would double. A free-falling lift ($a=-g$) would give no restoring force at all, which is a useful limiting check on the factor $g+a$.