$A$ body of mass $0.04 \text{ kg}$ executes simple harmonic motion $(SHM)$ about $x=0$ under the influence of force $F$ as shown in the graph. The time period of the motion is:

$A$ particle executes linear simple harmonic motion with an amplitude of $2 \ cm$. When the particle is at $1 \ cm$ from the mean position,the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is

$A$ pendulum is formed by pivoting a disc. What distance from the center of mass should it be pivoted for the minimum time period while performing $SHM$?

$A$ particle of mass $5 \times 10^{-5} \ kg$ is placed at the lowest point of a smooth parabola $x^2 = 40y$ ($x$ and $y$ in $m$). If it is displaced slightly such that it is constrained to move along the parabola,the angular frequency of oscillation (in $rad/s$) will be approximately:

$A$ light hollow cube of side length $10 \ cm$ and mass $10 \ g$ is floating in water. It is pushed down and released to execute simple harmonic oscillations. The time period of oscillations is $y \pi \times 10^{-2} \ s$,where the value of $y$ is (Acceleration due to gravity,$g = 10 \ m/s^2$,density of water $\rho = 10^3 \ kg/m^3$)

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]