If a vertical mass-spring system is taken to the moon,what will happen to its time period?

Three blocks of masses $700 \,g$,$500 \,g$,and $400 \,g$ are suspended at the end of a spring as shown in the figure and are in equilibrium. When the $700 \,g$ block is removed,the system has a period of oscillation of $3 \,s$. If both $700 \,g$ and $500 \,g$ blocks are removed,the period of oscillation becomes

$A$ block of mass $m$,attached to a spring of spring constant $k$,oscillates on a smooth horizontal table. The other end of the spring is fixed to a wall. The block has a speed $v$ when the spring is at its natural length. Before coming to an instantaneous rest,if the block moves a distance $x$ from the mean position,then

$A$ small block of mass $m$,having charge $q$,is placed on a frictionless inclined plane making an angle $\theta$ with the horizontal. There exists a uniform magnetic field $B$ parallel to the inclined plane but perpendicular to the length of the spring. If $m$ is slightly pulled on the inclined plane in the downward direction and released,the time period of oscillation will be (assume that the block does not leave contact with the plane):

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 $2 x_0$ and released. The time taken by the block to strike the other wall is

The total mechanical energy of a spring-mass system in simple harmonic motion is $E = \frac{1}{2}m\omega^2 A^2$. Suppose the oscillating particle is replaced by another particle of double the mass while the amplitude $A$ remains the same. The new mechanical energy will: