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 $\text{ s}$)

Two light identical springs of spring constant $k$ are attached horizontally at the two ends of a uniform horizontal rod $AB$ of length $l$ and mass $m$. The rod is pivoted at its centre $O$ and can rotate freely in a horizontal plane. The other ends of the two springs are fixed to rigid supports as shown in the figure. The rod is gently pushed through a small angle $\theta$ and released. The frequency of the resulting oscillation is

$A$ 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

$A$ body executes $SHM$ under the influence of one force and has a period $T_1 \, s$. The same body executes $SHM$ with period $T_2 \, s$ when under the influence of another force. When both forces act simultaneously and in the same direction,the time period of the same body is:

$A$ number of holes are drilled along a diameter of a disc of radius $R$. To get the minimum time period of oscillations,the disc should be suspended from a horizontal axis passing through a hole whose distance from the centre should be:

$A$ uniform rod of mass $m$ and length $l$ is suspended about its end. The time period of small angular oscillations is ..........