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

$A$ particle of mass $200\,g$ executes a simple harmonic motion. The restoring force is provided by a spring of spring constant $80\,N/m$. The time period is .... $\sec$

$A$ body of mass $1 \,kg$ is suspended from a spring of negligible mass. Another body of mass $500 \,g$ moving vertically upwards hits the suspended body with a velocity of $3 \,ms^{-1}$ and gets embedded in it. If the frequency of oscillation of the system of the two bodies after collision is $\frac{10}{\pi} \,Hz$,the amplitude of the motion and the spring constant are respectively,

Two springs,of force constants $k_1$ and $k_2$,are connected to a mass $m$ as shown. The frequency of oscillation of the mass is $f$. If both $k_1$ and $k_2$ are made four times their original values,the frequency of oscillation becomes

$A$ block $A$ of mass $1\, kg$ is connected to two identical springs of spring constant $800\, N/m$ and is placed on a smooth horizontal surface as shown in the figure. Initially,the springs are relaxed. Now,the block $A$ is slightly displaced to the left and released. The time period of oscillation of the system is

The time period of a mass suspended from a spring is $T$. If the spring is cut into four equal parts and the same mass is suspended from one of the parts,then the new time period will be