$A$ $2\, kg$ block moving with $10\, m/s$ strikes a spring of constant $\pi^2\, N/m$ attached to a $2\, kg$ block at rest kept on a smooth floor. The time for which the moving block remains in contact with the spring will be ... $\sec$.

The total spring constant of the system as shown in the figure will be

Show that the oscillations due to a spring are simple harmonic oscillations and obtain the expression for the period of oscillation.

$A$ mass '$m$' attached to a spring oscillates with a period of $3 \ s$. If the mass is increased by $0.6 \ kg$,the period increases by $3 \ s$. The initial mass '$m$' is equal to (in $kg$)

$A$ body of mass $4.9 \, kg$ hangs from a spring and oscillates with a period $0.5 \, s$. On the removal of the body, the spring is shortened by (take $g=10 \, m/s^2, \pi^2=10$)

$A$ 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: