Maximum amplitude(in $cm$) of $SHM$ so block A will not slip on block $B , K =100 N / m$

An assembly of identical spring-mass systems is placed on a smooth horizontal surface as shown. At this instant, the springs are relaxed. The left mass is displaced to the/left and theiright mass is displaced to the right by same distance and released. The resulting collision is elastic. The time period of the oscillations of system is

A body of mass $0.01 kg$ executes simple harmonic motion $(S.H.M.)$ about $x = 0$ under the influence of a force shown below : The period of the $S.H.M.$ is .... $s$

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)

A $100 \,g$ mass stretches a particular spring by $9.8 \,cm$, when suspended vertically from it. ....... $g$ large a mass must be attached to the spring if the period of vibration is to be $6.28 \,s$.

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