$A$ uniform circular disc of mass $12 \ kg$ is held by two identical springs. When the disc is slightly pressed down and released,it executes $S.H.M.$ of period $2 \ s$. The force constant of each spring is (nearly) (Take $\pi^2=10$) (in $Nm^{-1}$)

$A$ force of $6.4 \, N$ stretches a vertical spring by $0.1 \, m$. The mass that must be suspended from the spring so that it oscillates with a period of $\left( \frac{\pi}{4} \right) \, s$ is ... $kg$.

The time period of a mass suspended from a spring is $T$. If the spring is cut into four equal parts and the same mass is suspended from one of the parts,then the new time period will be

For a simple harmonic motion in a mass-spring system shown,the surface is frictionless. When the mass of the block is $1\,kg$,the angular frequency is $\omega_1$. When the mass of the block is $2\,kg$,the angular frequency is $\omega_2$. The ratio $\omega_2 / \omega_1$ is

$A$ mass '$m_1$' is suspended from a spring of negligible mass. The spring is pulled slightly in the downward direction and released; the mass performs $S.H.M.$ with a period '$T_1$'. If the mass is increased by '$m_2$',the time period becomes '$T_2$'. The ratio $\frac{m_2}{m_1}$ is

$A$ spring stretches by $2 \text{ mm}$ when it is loaded with a mass of $200 \text{ g}$. From the equilibrium position,the mass is further pulled down by $2 \text{ mm}$ and released. The frequency associated with the system and the maximum energy in the spring are . . . . . . $\text{Hz}$ and . . . . . . $\text{J}$,respectively. (Take $g = 10 \text{ m/s}^2$)