A force of $6.4\, N$ stretches a vertical spring by $0.1 \,m$. The mass that must be suspended from the spring so that it oscillates with a period of $\left( {\frac{\pi }{4}} \right)sec$. is  ... $kg$

Block $A$ is hanging from a vertical spring and it is at rest. Block $'B'$ strikes the block $'A'$ with velocity $v$ and stick to it. Then the velocity $v$ for which the spring just attains natural length is:

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 $S.H.M.$ of time period $T$. If the mass is increased by m, the time period becomes $5T/3$. Then the ratio of $m/M$ is

A $2\, Kg$ block moving with $10\, m/s$ strikes a spring of constant $\pi ^2 N/m$ attached to $2\, Kg$ block at rest kept on a smooth floor, the velocity of the rear $2\, kg$ block after it separates from the spring will be ..... $m/s$

A mass $M$ is suspended from a light spring. An additional mass m added displaces the spring further by a distance $x$. Now the combined mass will oscillate on the spring with period

A particle at the end of a spring executes simple harmonic motion with a period ${t_1}$, while the corresponding period for another spring is ${t_2}$. If the period of oscillation with the two springs in series is $T$, then