$A$ 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 $SHM$?

$A$ body executes $SHM$ under the influence of one force and has a period $T_1 \, s$. The same body executes $SHM$ with period $T_2 \, s$ when under the influence of another force. When both forces act simultaneously and in the same direction,the time period of the same body is:

$A$ uniform thin ring of radius $R$ and mass $m$ is suspended in a vertical plane from a point on its circumference. Its time period 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?

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 $m$ 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