The ends of a rod of length $l$ and mass $m$ are attached to two identical springs. The rod is free to rotate about its centre $O$. The rod is depressed slightly at end $A$ and released. The time period of the resulting oscillation is

$A$ body of mass $m$ is situated in a potential field $U(x) = U_0 (1 - \cos \alpha x)$,where $U_0$ and $\alpha$ are constants. Find the time period of small oscillations.

$9 \text{ kg}$ of solution is poured into a glass $U$-tube as shown in the figure. The tube's inner diameter is $2 \sqrt{\frac{\pi}{5}} \text{ m}$ and the solution oscillates freely up and down about its position of equilibrium $(x=0)$. The period of oscillation in seconds is (Given: $1 \text{ m}^3$ of solution has a mass $\rho=900 \text{ kg/m}^3$, $g=10 \text{ m/s}^2$, ignore frictional and surface tension effects).

$A$ solid cube of side $l$ is made to oscillate about a horizontal axis passing through one of its edges. Its time period will be

$A$ circular arc of mass $m$ is connected with the help of two massless strings as shown in the figure in a vertical plane. About point $P$,small oscillations are given in the plane of the arc. The time period of the oscillations of $SHM$ will be

The relation between the force ($F$ in newton) acting on a particle executing simple harmonic motion and the displacement of the particle ($y$ in metre) is $500 F + \pi^2 y = 0$. If the mass of the particle is $2 \text{ g}$, the time period of oscillation of the particle is (in $\text{ s}$)