Under the influence of force $F_1$,a body oscillates with a period $T_1$,and due to another force $F_2$,the body oscillates with a period $T_2$. If both forces act simultaneously,what is the resultant period? (Consider the displacement is the same in all three cases.)

$A$ disc of radius $R$ and mass $M$ is pivoted at the rim and is set for small oscillations. If a simple pendulum has to have the same period as that of the disc,the length of the simple pendulum should be

$A$ 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

$A$ particle executes $S.H.M.$ along a straight line. The amplitude of oscillation is $2 \,cm$. When the displacement of the particle from the mean position is $1 \,cm$,the magnitude of its acceleration is equal to the magnitude of its velocity. The time period of oscillation is ........

The potential energy of a particle of mass $1\,kg$ in motion along the $x-$axis is given by $U = 4(1 - \cos 2x)\,J,$ where $x$ is in $meters$. The period of small oscillations (in $sec$) is

$A$ circular disc of mass $10 \; kg$ is suspended by a wire attached to its centre. The wire is twisted by rotating the disc and released. The period of torsional oscillations is found to be $1.5 \; s$. The radius of the disc is $15 \; cm$. Determine the torsional spring constant of the wire in $N \; m \; rad^{-1}$. (Torsional spring constant $\alpha$ is defined by the relation $J = -\alpha \theta$,where $J$ is the restoring couple and $\theta$ is the angle of twist).