Two bodies $P$ and $Q$ of equal mass are suspended from two separate massless springs of force constants $k_1$ and $k_2$ respectively. If the maximum velocities of them are equal during their motion,the ratio of the amplitude of $P$ to $Q$ is:

$A$ mass $m_1$ performs $S.H.M.$ with amplitude $A$ while connected to a horizontal spring. While mass $m_1$ is passing through the mean position, another mass $m_2$ $(m_2 < m_1)$ is placed on it so that both masses move together with amplitude $A_1$. The ratio of $A/A_1$ is

$A$ body of mass $1 \,kg$ is attached to the lower end of a vertically suspended spring of force constant $600 \,N \,m^{-1}$. If another body of mass $0.5 \,kg$ moving vertically upward hits the suspended body with a velocity $3 \,m \,s^{-1}$ and gets embedded in it, then the frequency of the oscillation is

Initially,the system is in equilibrium. The time period of $SHM$ of the block in the vertical direction is

$A$ block of mass $M_1$ is hung by a light spring of force constant $k$ from the top bar of a reverse $U$-frame of mass $M_2$ resting on the floor. The block is pulled down from its equilibrium position by a distance $x$ and then released. Find the minimum value of $x$ such that the reverse $U$-frame will leave the floor momentarily.

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 amplitudes of $A$ and $B$ is