$A$ particle is placed at the lowest point of a smooth wire frame in the shape of a parabola,lying in the vertical $xy-$ plane having equation $x^2 = 5y$ ($x, y$ are in meters). After a slight displacement,the particle is set free. Find the angular frequency of oscillation in $rad/s$ (take $g = 10 \ m/s^2$).

$A$ 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

$A$ solid cube of side $l$ is made to oscillate about a horizontal axis passing through one of its edges. Its time period will be

$A$ body of mass $0.01 \ kg$ executes simple harmonic motion $(S.H.M.)$ about $x = 0$ under the influence of a force $F$ as shown in the graph (where $F$ is in $N$ and $x$ is in $m$). The period of the $S.H.M.$ is ... $s$. (Given: The graph is a straight line passing through the origin with slope $k = 4 \ N/m$ calculated from $F = 8 \ N$ at $x = 2 \ m$).

$A$ 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?

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]