$A$ spring is stretched by $0.2 \ m$ when a mass of $0.5 \ kg$ is suspended to it. The time period of the spring when the $0.5 \ kg$ mass is replaced with a mass of $0.25 \ kg$ is (Acceleration due to gravity $= 10 \ m \ s^{-2}$) (in $s$)

In the arrangement given in the figure,if the block of mass $m$ is displaced,the frequency is given by:

Two bodies of masses $1\, kg$ and $4\, kg$ are connected to a vertical spring,as shown in the figure. The smaller mass executes simple harmonic motion of angular frequency $25\, rad/s$,and amplitude $1.6\, cm$ while the bigger mass remains stationary on the ground. The maximum force exerted by the system on the floor is ..... $N$ (take $g = 10\, m/s^2$).

$A$ spring with a force constant $180 \text{ N/m}$ is loaded with a mass $0.2 \text{ kg}$. The amplitude of oscillations is $4 \text{ cm}$. When the mass comes to the equilibrium position, its velocity is: (in $\text{ m/s}$)

$A$ mass $m$ is suspended from a spring of force constant $k$ and just touches another identical spring fixed to the floor as shown in the figure. The time period of small oscillations is

Two particles $A$ and $B$ of masses $m$ and $2m$ are suspended from massless springs of force constants $K_1$ and $K_2$. During their oscillation, if their maximum velocities are equal, then the ratio of amplitudes of $A$ and $B$ is