The springs in the figure are identical,but the length of the spring in $A$ is three times that in $B$. The ratio of the time periods $T_A/T_B$ is:

In figure $(A)$,mass '$2m$' is fixed on mass '$m$' which is attached to two springs of spring constant $k$. In figure $(B)$,mass '$m$' is attached to two springs of spring constant '$k$' and '$2k$'. If mass '$m$' in $(A)$ and $(B)$ are displaced by distance '$x$' horizontally and then released,then the time periods $T_{1}$ and $T_{2}$ corresponding to $(A)$ and $(B)$ respectively follow the relation.

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

$A$ body of mass $m$ is suspended to an ideal spring of force constant $k$. The expected change in the position of the body due to an additional force $F$ acting vertically downwards is

$A$ particle is suspended from a vertical spring which is executing $S.H.M.$ of frequency $5 \ Hz$. The spring is unstretched at the highest point of oscillation. What is the maximum speed of the particle? (Take $g = 10 \ m/s^2$)

$A$ mass $m$ is suspended from a spring of negligible mass. The spring is pulled a little and then released,so that the mass executes $S.H.M.$ with a time period $T$. If the mass is increased by $m_0$,the time period becomes $\frac{5T}{4}$. The ratio $\frac{m_0}{m}$ is