$A$ spring of force constant $k$ is cut into two pieces such that one piece is double the length of the other. Then the long piece will have a force constant of

$A$ spring has a certain mass suspended from it and its period for vertical oscillations is $T_1$. The spring is now cut into two equal halves and the same mass is suspended from one of the halves. The period of vertical oscillations is now $T_2$. The ratio $T_1 / T_2$ is

$A$ block with mass $M$ is connected by a massless spring with stiffness constant $k$ to a rigid wall and moves without friction on a horizontal surface. The block oscillates with small amplitude $A$ about an equilibrium position $x_0$. Consider two cases: $(i)$ when the block is at $x_0$; and $(ii)$ when the block is at $x = x_0 + A$. In both cases, a particle with mass $m$ is placed on the mass $M$. Which of the following statements are correct?

$A$ small mass $m$, attached to one end of a spring with a negligible mass and an unstretched length $L$, executes vertical oscillations with angular frequency $\omega_{0}$. When the mass is rotated with an angular speed $\omega$ by holding the other end of the spring at a fixed point, the mass moves uniformly in a circular path in a horizontal plane. Then the increase in length of the spring during the rotation is

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

$A$ block of mass $100 \,g$ is suspended vertically between two massless springs, each with a spring constant $k=1 \,N/m$. The block is hit from above to impart an impulse of $2 \,Ns$. Calculate the maximum displacement from the equilibrium position of the block. (Take $g=10 \,m/s^2$) (in $\,m$)