$A$ particle of mass $m$ is located in a one-dimensional potential field where the potential energy is given by: $V(x) = A(1 - \cos px)$, where $A$ and $p$ are constants. The period of small oscillations of the particle is

The potential energy of a particle of mass $10 \ g$ as a function of displacement $x$ is $(50 x^2 + 100) \ J$. The frequency of oscillation is

$A$ 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?

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