$A$ 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:

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 $m_2 = 27 \ kg$?

When the mass attached to a spring is increased from $4 \ kg$ to $9 \ kg$,the time period of oscillation increases by $0.2 \pi \ s$. Then the spring constant of the spring is (in $N \ m^{-1}$)

$A$ mass $M$ is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes $S.H.M.$ of time period $T$. If the mass is increased by $m$,the time period becomes $5T/3$. Then the ratio of $m/M$ is

$A$ 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:

Find the maximum amplitude (in $cm$) of the $SHM$ such that block $A$ will not slip on block $B$. Given: spring constant $K = 100 \ N/m$,mass of block $A$ $m_A = 0.25 \ kg$,mass of block $B$ $m_B = 1.25 \ kg$,and coefficient of friction between $A$ and $B$ is $\mu = 0.4$. Take $g = 10 \ m/s^2$.