$A$ spring has a certain mass suspended from it and its period of 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 of $T_2 / T_1$ is

$A$ block is placed on a frictionless horizontal table. The mass of the block is $m$ and springs are attached on either side with force constants $K_1$ and $K_2$. If the block is displaced a little and left to oscillate,then the angular frequency of oscillation will be

Consider two identical springs each of spring constant $k$ and negligible mass compared to the mass $M$ as shown. Fig. $1$ shows one of them and Fig. $2$ shows their series combination. The ratio of the time period of oscillation of the two $SHM$ is $\frac{T_b}{T_a} = \sqrt{x}$,where the value of $x$ is (Round off to the Nearest Integer).

The variation of potential energy of a harmonic oscillator is as shown in the figure. The spring constant is

$A$ bar of mass $m$ is suspended horizontally on two vertical springs of spring constant $k$ and $3k$. The bar bounces up and down while remaining horizontal. Find the time period of oscillation of the bar (Neglect mass of springs and friction everywhere).

Two particles $A$ and $B$ of equal masses are suspended from two massless springs of spring constants $K_{1}$ and $K_{2}$ respectively. If the maximum velocities during oscillations are equal,the ratio of the amplitude of $A$ and $B$ is