$A$ 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:

$A$ particle of mass $m$ is executing oscillations about the origin on the $X-$axis. Its potential energy is $U(x) = k|x|^3$,where $k$ is a positive constant. If the amplitude of oscillation is $a$,then its time period $T$ is:

$A$ body of mass $0.01 \ kg$ executes simple harmonic motion $(S.H.M.)$ about $x = 0$ under the influence of a force $F$ as shown in the graph (where $F$ is in $N$ and $x$ is in $m$). The period of the $S.H.M.$ is ... $s$. (Given: The graph is a straight line passing through the origin with slope $k = 4 \ N/m$ calculated from $F = 8 \ N$ at $x = 2 \ m$).

$A$ particle of mass $m$ is executing oscillations about the origin on the $x-$axis. Its potential energy is $V(x) = k | x |^3$,where $k$ is a positive constant. If the amplitude of oscillation is $a$,then its time period $T$ is

$A$ 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?

$A$ 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: