$A$ rectangular block of mass $m$ and cross-sectional area $A$ floats in a liquid of density $\rho$. If we give it a small vertical displacement from equilibrium,it undergoes $SHM$ with a time period $T$. Then:

$A$ cylindrical log of wood of height $h$ and area of cross-section $A$ floats in water. It is pressed and then released. Show that the log would execute $SHM$ with a time period $T = 2\pi \sqrt{\frac{m}{A\rho g}}$,where $m$ is the mass of the body and $\rho$ is the density of the liquid.

$A$ $U$-tube of uniform bore of cross-sectional area $A$ has been set up vertically with open ends facing up. Now $M$ grams of a liquid of density $d$ is poured into it. The column of liquid in this tube will oscillate with a period $T$ such that

$A$ particle of mass $m$ undergoes oscillations about $x=0$ in a potential given by $V(x) = \frac{1}{2} k x^2 - V_0 \cos \left(\frac{x}{a}\right)$,where $V_0, k, a$ are constants. If the amplitude of oscillation is much smaller than $a$,the time period is given by

In both figures shown below, there is a hole along the diameter of the Earth. In the first figure, a particle is released from point $A$ and it oscillates with a time period $T_1$. In the second figure, the same particle is released from point $B$ and it oscillates with a time period $T_2$. Then, [$O$ is the centre of the Earth]

$A$ body executes simple harmonic motion under the action of a force $F_1$ with a time period $T_1 = \frac{4}{5} \ s$. If the force is changed to $F_2$,it executes $SHM$ with a time period $T_2 = \frac{3}{5} \ s$. If both the forces $F_1$ and $F_2$ act simultaneously in the same direction on the body,what will be its new time period in seconds?