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

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

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 the unstretched spring. Then $\omega$ 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

$A$ spring-block system in horizontal oscillation has a time period $T$. If the spring is cut into four equal parts and the block is re-connected with one of the parts,what will be the new time period of oscillation?

Five identical springs are used in the following three configurations. The time periods of vertical oscillations in configurations $(i)$,$(ii)$ and $(iii)$ are in the ratio