$A$ mass at the end of a spring executes harmonic motion about an equilibrium position with an amplitude $A$. Its speed as it passes through the equilibrium position is $V$. If extended $2A$ and released,the speed of the mass passing through the equilibrium position will be

An assembly of identical spring-mass systems is placed on a smooth horizontal surface as shown. Initially,the springs are relaxed. The left mass is displaced to the left while the right mass is displaced to the right and released. The resulting collision is elastic. The time period of the oscillations of the system is:

When a block of mass $m$ is suspended separately by two different springs having time periods $t_1$ and $t_2$,respectively. If the same mass is connected to the series combination of both springs,then its time period is given by:

$A$ block of mass $M$ hangs from a spring and oscillates vertically with an angular frequency $\omega$. If the block is removed from the spring,when it is in equilibrium position,the spring shortens by

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

$A$ block of mass $M$ rests on a piston executing $S.H.M.$ with a period of $1 \,s$. The amplitude of oscillations, such that the mass is separated from the piston, is (acceleration due to gravity, $g=10 \,m/s^2$, $\pi^2=10$)