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

$A$ rectangular block of mass '$m$' and cross-sectional area '$A$' floats on a liquid of density '$\rho$'. It is given a small vertical displacement from equilibrium, it starts oscillating with frequency ($g$ = acceleration due to gravity).

$A$ 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 $4 \,kg$ moves simple harmonically such that its potential energy $U$ varies with position $x$ as shown in the figure. The period of oscillations 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?

The potential energy of a particle of mass $10 \ g$ as a function of displacement $x$ is $(50 x^2 + 100) \ J$. The frequency of oscillation is