$A$ 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).

$A$ drinking straw is dipped in a pan of water to a depth $d$ from the surface (see figure). Water is sucked into it up to an initial height $h_0$ and then left to oscillate. As a result,its height $y$ from the surface of the water varies periodically. Ignoring damping,the equation for $y$ is ($g$ is the acceleration due to gravity):

$A$ rectangular block of mass $m$ and area of cross-section $A$ floats in a liquid of density $\rho$. If it is given a small vertical displacement from equilibrium,it undergoes simple harmonic motion with a time period $T$. Then:

$A$ 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:

$A$ uniform rod of length $L$ and mass $M$ is pivoted at the centre. Its two ends are attached to two springs of equal spring constants $k$. The springs are fixed to rigid supports as shown in the figure,and the rod is free to oscillate in the horizontal plane. The rod is gently pushed through a small angle $\theta$ in one direction and released. The frequency of oscillation is

$A$ 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: