$A$ simple harmonic oscillator consists of a particle of mass $m$ and an ideal spring with spring constant $k$. The particle oscillates with a time period $T$. The spring is cut into two equal parts. If one part oscillates with the same particle, the new time period will be

$A$ uniform cylinder of length $L$ and mass $M$ having cross-sectional area $A$ is suspended,with its length vertical,from a fixed point by a massless spring,such that it is half submerged in a liquid of density $\sigma$ at equilibrium position. When the cylinder is given a downward push and released,it starts oscillating vertically with a small amplitude. The time period $T$ of the oscillations of the cylinder will be

Two bodies $A$ and $B$ of equal masses are suspended from two separate springs of force constants $k_1$ and $k_2$ respectively. If the two bodies oscillate such that their maximum velocities are equal,the ratio of the amplitudes of oscillation of $A$ and $B$ will be

$A$ spring of length $l$ and force constant $k$ is attached to a mass $m$ and oscillates with frequency $f_1$. If the spring is cut into two equal parts and one part is attached to the same mass $m$ to oscillate,its frequency becomes $f_2$. Find the relation between $f_1$ and $f_2$.

$A$ weightless spring which has a force constant $k$ oscillates with frequency $n$ when a mass $m$ is suspended from it. The spring is cut into two equal halves and a mass $2m$ is suspended from one of the halves. The frequency of oscillation will now become:

$A$ weightless spring of length $60\, cm$ and force constant $200\, N/m$ is kept straight and unstretched on a smooth horizontal table and its ends are rigidly fixed. $A$ mass of $0.25\, kg$ is attached at the middle of the spring and is slightly displaced along the length. The time period of the oscillation of the mass is