Answer
(a) The period of the pendulum on the moon is greater than $T$.
(b) $\frac{T_m}{T} = \sqrt{6}$
(c) The period of the mass-spring system on the moon is equal to $T$.
(d) $\frac{T_m}{T} = 1$
Work Step by Step
(a) We can write the expression for the period of a pendulum:
$T = 2\pi~\sqrt{\frac{L}{g}}$
We can see that decreasing $g$ will increase the period. Therefore, the period of the pendulum on the moon is greater than $T$.
(b) We can find the ratio $\frac{T_m}{T}$:
$T_m = 2\pi~\sqrt{\frac{L}{(g/6)}}$
$T_m = \sqrt{6}\times 2\pi~\sqrt{\frac{L}{g}}$
$T_m = \sqrt{6}~T$
$\frac{T_m}{T} = \sqrt{6}$
(c) We can write the expression for the period of a mass-spring system:
$T = 2\pi~\sqrt{\frac{m}{k}}$
We can see that decreasing $g$ will have no effect on the period. Therefore, the period of the mass-spring system on the moon is equal to $T$.
(d) Since $T_m = T$, we can see that $\frac{T_m}{T} = 1$
