Why is the time period of a pendulum with a spring of force constant $k$ and a bob of significant mass $m$ the same on the Moon as on the Earth?

Question

A question I came across in class today:

How will the time period of a loaded spring change when it is taken to the Moon?

What I've been told:

The formula for the time period of a loaded spring

$$ T = 2\pi (\frac{m}{k})^\frac{1}{2}$$

$T$ 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.

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?

Answer 1

Please note that mass $m$ and the spring constant $k$ are both the same on earth as they are on the moon. Consequently, the time period of the spring does not depend upon differences due to the acceleration due to gravity. Hence it will not change when it’s taken to the moon.

I think the value of 𝑘 for a spring hung upside down depends on gravitational pull

No. It’s a constant. The mathematical argument you have provided contains many errors, but suffice to say that the period of a loaded spring will not be different.

The only thing that will differ is the equilibrium position of the mass in which case it will be higher on the moon.

Answer 2

To illustrate the independence of the value of the gravitational field one could set up the spring mass system on a horizontal table and with no friction present the period of oscillation would still be $2\pi \sqrt{\frac mk}$ provided the spring could undergo compression as well as extension.

The reason for this independence is that the restoring force, $F$, depends on the displacement of the mass from its equilibrium position, no net force acting on it, which is not a function of gravitational field and the mass $m$, is also not dependent on the gravitational field, thus the acceleration of the mass $a=\frac Fm$ is gravitational field independent.

The $l$ and $g$ relationship which appears to show the dependence of the period on the gravitational field strength does not show this because $l$ and $g$ are not independent of one another being liked by the equation $kl=mg$, so a $g$ increases so does $l$ in the same proportion.

Answer 3

... we have that the restoring force in the spring depends linearly on displacement. $$F(x) = -k(x) \\ k = \frac{-F(x)}{x}$$

This calculation of the force is wrong.

Actually the force ($F$) has two parts:

• The restoring force of the spring ($-kx$) which is proportional to the current displacement ($x$). And the spring constant $k$ is still a constant.
• The gravitational force ($mg$) which is independent of the current displacement ($x$)

So we have the total force $$F=-kx+mg. \tag{1}$$

According to Newton's second law ($m\ddot{x}=F$) we get the equation of motion $$m\ddot{x}=-kx+mg. \tag{2}$$

The most general solution of (2) can be found to be $$x(t)=\frac{mg}{k}+A\sin\left(\sqrt{\frac{k}{m}}\ t+\phi\right) \tag{3}$$ where $A$ (the amplitude) and $\phi$ (the initial phase) are arbitrary constants. You can verify the correctness of this solution by plugging it into differential equation (2).

From solution (3) you see two features.

• The oscillation period $T$ can be determined from $\frac{2\pi}{T}=\sqrt{\frac{k}{m}}$. Thus $T$ is independent of $g$.
• The equilibrium displacement is $\frac{mg}{k}$. Thus it depends on $g$. On the moon it is smaller than on the earth.