$A$ ring is suspended from a point $S$ on its rim as shown in the figure. When displaced from equilibrium,it oscillates with a time period of $1 \, s$. The radius of the ring is ..... $m$ (take $g = \pi^2$).

$A$ body of mass $0.01\, kg$ executes simple harmonic motion $(S.H.M.)$ about $x = 0$ under the influence of a force shown in the graph. The period of the $S.H.M.$ is .... $s$

$A$ uniform thin ring of radius $R$ and mass $m$ is suspended in a vertical plane from a point on its circumference. Its time period of oscillation is ........

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 $\text{ s}$)

$A$ 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,and it starts oscillating with frequency $n$. The frequency $n$ is equal to (where $g$ is the acceleration due to gravity):

$A$ particle of mass $m$ moves in a one-dimensional potential energy $U(x) = -ax^2 + bx^4$,where $a$ and $b$ are positive constants. The angular frequency of small oscillations about the minima of the potential energy is equal to