When a particle of mass $m$ is attached to a vertical spring of spring constant $k$ and released, its motion is described by $y ( t )= y _{0} \sin ^{2} \omega t ,$ where $'y'$ is measured from the lower end of unstretched spring. Then $\omega$ is

A body of mass $m$ is attached to one end of a massless spring which is suspended vertically from a fixed point. The mass is held in hand, so that the spring is neither stretched nor compressed. Suddenly the support of the hand is removed. The lowest position attained by the mass during oscillation is $4\,cm$ below the point, where it was held in hand.

$(a)$ What is the amplitude of oscillation ?

$(b)$ Find the frequency of oscillation.

A uniform cylinder of length $L$ and mass $M$ having cross-sectional area $A$ is suspended, with its length vertical, from a fixed point by a massless spring, such that it is half submerged in a liquid of density $\sigma $ at equilibrium position. When the cylinder is given a downward push and released, it starts oscillating vertically with a small amplitude. The time period $T$ of the oscillations of the cylinder will be

A mass $m$ is suspended from the two coupled springs connected in series. The force constant for springs are ${K_1}$ and ${K_2}$. The time period of the suspended mass will be

The frequency of oscillation of the springs shown in the figure will be

Consider two identical springs each of spring constant $k$ and negligible mass compared to the mass $M$ as shown. Fig. $1$ shows one of them and Fig. $2$ shows their series combination. The ratios of time period of oscillation of the two $SHM$ is $\frac{ T _{ b }}{ T _{ a }}=\sqrt{ x },$ where value of $x$ is

(Round off to the Nearest Integer)