$A$ 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 executing a simple harmonic motion. Its maximum acceleration is $\alpha$ and maximum velocity is $\beta$. Then,its time period of vibration will be

$A$ metal ball released from height $h$ makes a perfectly elastic collision with the ground. The frequency of the periodic vibratory motion is $\quad(g = \text{acceleration due to gravity})$

$A$ circular disc of mass $10 \; kg$ is suspended by a wire attached to its centre. The wire is twisted by rotating the disc and released. The period of torsional oscillations is found to be $1.5 \; s$. The radius of the disc is $15 \; cm$. Determine the torsional spring constant of the wire in $N \; m \; rad^{-1}$. (Torsional spring constant $\alpha$ is defined by the relation $J = -\alpha \theta$,where $J$ is the restoring couple and $\theta$ is the angle of twist).

$9 \text{ kg}$ of solution is poured into a glass $U$-tube as shown in the figure. The tube's inner diameter is $2 \sqrt{\frac{\pi}{5}} \text{ m}$ and the solution oscillates freely up and down about its position of equilibrium $(x=0)$. The period of oscillation in seconds is (Given: $1 \text{ m}^3$ of solution has a mass $\rho=900 \text{ kg/m}^3$, $g=10 \text{ m/s}^2$, ignore frictional and surface tension effects).

$A$ 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?