$A$ ball is rolling without slipping in a spherical shallow bowl (radius $R$) as shown in the figure and is executing simple harmonic motion. If the radius of the ball is doubled,then the time period of oscillation

$A$ 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?

$A$ 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 $x = 2 \ m$).

$A$ cylindrical block of mass $M$ and area of cross-section $A$ is floating in a liquid of density $\rho$ with its axis vertical. When depressed a little and released, the block starts oscillating. The 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?

$A$ small spherical ball of radius '$r$' is rolling on a curved surface which is frictionless and has a radius of curvature '$R$'. Its motion is simple harmonic. Then its time period of oscillation is proportional to ($g=$ acceleration due to gravity).