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

In the given arrangement,the spring constant $k$ has a value of $2\,N\,m^{-1}$,mass $M = 3\,kg$,and mass $m = 1\,kg$. Mass $M$ is in contact with a smooth surface. The coefficient of friction between the two blocks is $0.1$. The time period of $SHM$ executed by the system is

$A$ mass $m$ is suspended separately by two different springs of spring constant $K_1$ and $K_2$ giving the time periods $t_1$ and $t_2$ respectively. If the same mass $m$ is connected by both springs as shown in the figure,then the time period $t$ is given by the relation:

The scale of a spring balance which can measure from $0$ to $15 \ kg$ is $0.25 \ m$ long. If a body suspended from this balance oscillates with a time period $\frac{2 \pi}{5} \ s$,neglecting the mass of the spring,find the mass of the body suspended. (in $kg$)

The potential energy of a particle of mass $m$ situated in a unidimensional potential field varies as $U(x) = U_0(1 - \cos ax)$,where $U_0$ and $a$ are constants. The time period of small oscillations of the particle about the mean position is

$A$ force of $6.4 \ N$ stretches a vertical spring by $0.1 \ m$. The mass that must be suspended from the spring so that it oscillates with a time period of $\pi/4 \ s$ is .... $kg$.