$A$ particle of mass $m$ undergoes oscillations about $x=0$ in a potential given by $V(x) = \frac{1}{2} k x^2 - V_0 \cos \left(\frac{x}{a}\right)$,where $V_0, k, a$ are constants. If the amplitude of oscillation is much smaller than $a$,the time period is given by

$A$ particle is performing a linear simple harmonic motion of amplitude $A$. When it is midway between its mean and extreme position,the magnitudes of its velocity and acceleration are equal. What is the periodic time of the motion?

$A$ uniform rod of length $L$ and mass $M$ is pivoted at the centre. Its two ends are attached to two springs of equal spring constants $k$. The springs are fixed to rigid supports as shown in the figure,and the rod is free to oscillate in the horizontal plane. The rod is gently pushed through a small angle $\theta$ in one direction and released. The frequency of oscillation is

$A$ disc of radius $R$ and mass $M$ is pivoted at the rim and is set for small oscillations. If a simple pendulum has to have the same period as that of the disc,the length of the simple pendulum should be

$9 \text{ kg}$ 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).

$A$ uniform rod of length $2.0 \, m$ is suspended through an end and is set into oscillation with small amplitude under gravity. The time period of oscillation is approximately .... $\sec$.