$A$ 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

Two springs with force constants $k_1$ and $k_2$ have the same mass attached to their ends. Both objects have the same maximum velocity. What is the ratio of their amplitudes?

$A$ body of mass $64 \ g$ is made to oscillate turn by turn on two different springs $A$ and $B$. Spring $A$ and $B$ have force constants $4 \ N/m$ and $16 \ N/m$ respectively. If $T_{1}$ and $T_{2}$ are the periods of oscillation of springs $A$ and $B$ respectively,then $\frac{T_{1}+T_{2}}{T_{1}-T_{2}}$ will be:

$A$ 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 $\,m$)

$A$ light spring is suspended with mass $m_1$ at its lower end and its upper end is fixed to a rigid support. The mass is pulled down a short distance and then released. The period of oscillation is $T$ seconds. When a mass $m_2$ is added to $m_1$ and the system is made to oscillate, the period is found to be $\frac{3}{2} T$. The ratio $\frac{m_1}{m_2}$ 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?