Two masses $m_1$ and $m_2$ are suspended together by a massless spring of constant $k$. When the masses are in equilibrium,$m_1$ is removed without disturbing the system. Then the angular frequency of oscillation of $m_2$ is

$A$ 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:

$A$ block of mass $m$ has two similar rubber ribbons attached to it as shown in the figure. The force constant of each rubber ribbon is $K$ and the surface is frictionless. The block is displaced from the mean position by $x$ and released. At the mean position,the ribbons are undeformed. The vibration period is

$A$ block of mass $m$ attached to a massless spring is performing oscillatory motion of amplitude $A$ on a frictionless horizontal plane. If half of the mass of the block breaks off when it is passing through its equilibrium point,the amplitude of oscillation for the remaining system becomes $fA$. The value of $f$ is

The drawing shows a top view of a frictionless horizontal surface,where there are two identical springs with particles of mass $m_1$ and $m_2$ attached to them. Each spring has a spring constant of $1200 \ N/m$. The particles are pulled to the right and then released from the positions shown in the drawing. How much time passes before the particles are again side by side for the first time if $m_1 = 3.0 \ kg$ and $m_2 = 27 \ kg$?

$A$ spring of unstretched length $l$ has a mass $m$ with one end fixed to a rigid support. Assuming the spring to be made of a uniform wire,the kinetic energy possessed by it if its free end is pulled with uniform velocity $v$ is