How does the period of oscillation depend on the mass of the block attached to the end of a spring?

$A$ 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

$A$ block of mass $2 \ kg$ is attached to one end of a massless spring whose other end is fixed at a wall. The spring-mass system moves on a frictionless horizontal table. The spring's natural length is $2 \ m$ and the spring constant is $200 \ N/m$. The block is pushed such that the length of the spring becomes $1 \ m$ and then released. At a distance $x \ m \ (x < 2)$ from the wall,the speed of the block will be:

One end of a spring of force constant $k$ is fixed to a vertical wall and the other to a block of mass $m$ resting on a smooth horizontal surface. There is another wall at a distance $x_0$ from the block. The spring is then compressed by $2x_0$ and released. The time taken to strike the wall is

$A$ mass $m$ is suspended from two coupled springs connected in series. The force constants for the springs are $K_1$ and $K_2$. The time period of the suspended mass will be:

Three masses $700 \, g, 500 \, g$,and $400 \, g$ are suspended at the end of a spring as shown and are in equilibrium. When the $700 \, g$ mass is removed,the system oscillates with a period of $3 \, s$. When the $500 \, g$ mass is also removed,it will oscillate with a period of ...... $s$.