$A$ mass $m$ performs oscillations of period $T$ when suspended by a spring of force constant $K$. If the spring is cut into two equal parts and arranged in parallel,and the same mass $m$ is oscillated by them,then the new time period will be:

$A$ mass $M$ is suspended by two springs of force constants $K_1$ and $K_2$ respectively,as shown in the diagram. The total elongation (stretch) of the two springs is

$A$ spring has a certain mass suspended from it and its period for vertical oscillations is $T_1$. The spring is now cut into two equal halves and the same mass is suspended from one of the halves. The period of vertical oscillations is now $T_2$. The ratio $T_1 / T_2$ 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

$A$ spring of force constant $k$ is cut into two pieces such that one piece is double the length of the other. Then the long piece will have a force constant of

$A$ block is placed on a frictionless horizontal table. The mass of the block is $m$ and springs are attached on either side with force constants $K_1$ and $K_2$. If the block is displaced a little and left to oscillate,then the angular frequency of oscillation will be