A question I came across in class today:

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> How will the time period of a loaded spring change when it is taken to the Moon?
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What I've been told:

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> The formula for the time period of a loaded spring
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> $$ T = 2\pi (\frac{m}{k})^\frac{1}{2}$$
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> $T$ in this expression is independent of $g$. As such, the time period of the loaded spring on the Moon will be the same as its time period on the Earth.
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**My thoughts**

I think the value of $k$ for a spring hung upside down depends on gravitational pull. From Hooke's law, we have that the restoring force in the spring depends linearly on displacement.

$$ F(x) = -k(x) \\ k = \frac{-F(x)}{x} $$

When the mass is hung vertically, $x$ depends on gravitational pull $g$.

$$ x^2 \propto g $$

The same goes for the downward force, $F(x)$. When the spring-bob system is in equilibrium,

$$ F(x) \propto g $$

Therefore,

$$ k = \frac{-F(x)}{x} \propto \sqrt{g} $$

From the formula for a loaded spring,

$$ T \propto (\frac{1}{g})^\frac{1}{4} $$

So the time period will be a little greater on the Moon than on the Earth. Could someone tell me if I've worked this out right?