$A$ particle is executing a simple harmonic motion. Its maximum acceleration is $\alpha$ and maximum velocity is $\beta$. Then,its time period of vibration will 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 ............

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

In both figures shown below, there is a hole along the diameter of the Earth. In the first figure, a particle is released from point $A$ and it oscillates with a time period $T_1$. In the second figure, the same particle is released from point $B$ and it oscillates with a time period $T_2$. Then, [$O$ is the centre of the Earth]

$A$ particle of mass $m$ is executing oscillations about the origin on the $x-$axis. Its potential energy is $V(x) = k | x |^3$,where $k$ is a positive constant. If the amplitude of oscillation is $a$,then its time period $T$ is

$A$ particle executes linear $S.H.M.$ with amplitude $4 \ cm$. The magnitude of velocity and acceleration is equal when it is at $3 \ cm$ from the mean position. The time period of oscillation is: