$A$ uniform disc of radius $R$ and mass $M$ is free to oscillate about the axis $A$ as shown in the figure. For small oscillations, the time period is . . . . . . . ($g$ is acceleration due to gravity)

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 $\text{ s}$)

$A$ ring is hung on a nail. It can oscillate,without slipping or sliding $(i)$ in its plane with a time period $T_{1}$ and,$(ii)$ back and forth in a direction perpendicular to its plane,with a period $T_{2}$. The ratio $\frac{T_{1}}{T_{2}}$ will be

$A$ pendulum is formed by pivoting a disc. What distance from the center of mass should it be pivoted for the minimum time period while performing $SHM$?

$A$ 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 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 .... $s$