$A$ body executes simple harmonic motion under the action of a force $F_1$ with a time period $(4/5) \ s$. If the force is changed to $F_2$,it executes $SHM$ with a time period $(3/5) \ s$. If both the forces $F_1$ and $F_2$ act simultaneously in the same direction on the body,its time period (in $s$) is:

$A$ rod of mass $M$ and length $2L$ is suspended at its middle by a wire. It exhibits torsional oscillations. If two masses each of $m$ are attached at a distance $L/2$ from its centre on both sides,the oscillation frequency reduces by $20\%$. The value of the ratio $m/M$ is close to:

$A$ uniform stick of mass $M$ and length $L$ is pivoted at its centre. Its ends are tied to two springs each of force constant $K$. In the position shown in the figure,the springs are in their natural length. When the stick is displaced through a small angle $\theta$ and released,the stick:

$A$ 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.

$A$ 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?

$A$ $U$-tube of a uniform bore has its arms vertical. The total length of the liquid in the two arms of the $U$-tube is $L$. The time period $T$ of the oscillation of the liquid column when it is displaced by $y$ is ($g$ = acceleration due to gravity).