$A$ $100 \,g$ 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 $6.28 \,s$?

Two bodies $M$ and $N$ of equal masses are suspended from two separate massless springs of force constants $k_1$ and $k_2$ respectively. If the two bodies oscillate vertically such that their maximum velocities are equal,the ratio of the amplitude of $M$ to that of $N$ is

$A$ body executes $SHM$ under the action of force $F_1$ with time period $T_1$. If the force is changed to $F_2$,it executes $SHM$ with time period $T_2$. If both the forces $F_1$ and $F_2$ act simultaneously in the same direction on the body,its time period is:

Two massless springs of force constants $K_1$ and $K_2$ are joined end to end. The resultant force constant $K$ of the system is

$A$ spring-mass system is oscillating horizontally. What will be the effect on the time period if the spring is made to oscillate vertically?

As shown in the figure,two blocks of masses $m_1$ and $m_2$ are connected to a spring of force constant $k$. The blocks are slightly displaced in opposite directions to $x_1$ and $x_2$ distances and released. If the system executes simple harmonic motion,then the angular frequency of oscillation of the system $(\omega)$ is: