Question Video: Comparing the Frequency of a Pendulum on Earth and on the Moon Physics

Video Transcript

A pendulum is taken from Earth to the Moon. Will the pendulum have a higher frequency, a lower frequency, or the same frequency on the surface of the Moon as on the surface of Earth?

In order to answer this question, we first need to recall that the frequency of a pendulum, that is, the rate at which it oscillates from side to side, depends on the gravitational acceleration that the pendulum experiences. Specifically, the frequency of a pendulum can be calculated using this formula, where 𝐿 is the length of the pendulum and 𝑔 is the local acceleration due to gravity.

Now, often when we’re dealing with problems in mechanics, such as this one, we treat 𝑔 as if it’s just a constant. Specifically, we tend to use a value of 9.81 meters per second squared. However, this value is just the standard gravitational acceleration near the surface of Earth. If we were to move away from the surface of the Earth or indeed to the Moon or a different planet, then we’d find that the local gravitational acceleration would be different. So to be precise, we should really call this quantity 𝑔 sub e to signify that it’s the gravitational acceleration near the surface of the Earth.

In this question, we’re asked to think about what would happen if we took this pendulum to the Moon. In this case, it will experience a different gravitational acceleration. Let’s call this 𝑔 m. Now, the question doesn’t actually tell us what the gravitational acceleration is like on the surface of the Moon. However, we know that the Moon has a much lower mass than the Earth. And therefore, the gravitational acceleration on its surface will be lower. So we can say that 𝑔 m is less than 𝑔 e.

We can see that taking our pendulum from the surface of the Earth to the surface of the Moon will decrease the value of 𝑔. And since we’re not changing the pendulum itself, we know that 𝐿 is effectively a constant. Since 𝑔 is changing and everything else on the right-hand side of this formula is constant, this means that 𝑓, the frequency of the pendulum, must also change. So now at least we know that the frequency won’t stay the same as when it was on Earth.

However, looking at this formula, it’s not immediately obvious whether decreasing 𝑔 will cause 𝑓 to increase or decrease. In a situation like this, when we’re trying to determine how a change in one variable will affect another variable but we don’t have any numbers to work with, it can be useful to try to establish what type of proportionality exists between the two variables. For example, is 𝑓 proportional to 𝑔? Could it be proportional to the square of 𝑔? How about inversely proportional to 𝑔? Establishing this type of relationship between the two variables 𝑓 and 𝑔 will enable us to easily see whether decreasing 𝑔 will increase or decrease 𝑓.

Now, for the sake of this question, where we’re not actually calculating an answer, let’s just assume that our approximately equal to sign is an equal sign. Now, to turn an equation like this one into a statement about proportionality, we can effectively remove all the constants in the equation and replace the equal sign with a proportional to sign. In this case, this gives us 𝑓 is proportional to one over the square root of one over 𝑔. Now, this isn’t much clearer since we still have a fraction within a fraction. However, using algebra, we can clear some of this up.

Firstly, let’s deal with the square root sign. The square root of one over 𝑔 is equivalent to the square root of one over the square root of 𝑔. And the square root of one is of course just one. We can now see that we have one over one over the square root of 𝑔. In other words, we have the reciprocal of the reciprocal of the square root of 𝑔. Taking the reciprocal over reciprocal in this way just leaves us with the quantity that we started with. And in this case, that’s the square root of 𝑔. This shows us that the frequency 𝑓 is proportional to the square root of 𝑔.

Now that we’ve obtained a nice, simple proportionality statement like this, our problem becomes much simpler. When we move our pendulum from the Earth to the Moon, we know that the size of 𝑔 is decreased. And if we decrease the size of 𝑔, then we decrease the size of the square root of 𝑔. Since 𝑓 is proportional to the square root of 𝑔, this means that 𝑓 must decrease too. In other words, we find that the pendulum has a lower frequency on the Moon than it did on the Earth. And this is the final answer to our question. If a pendulum is taken from the Earth to the Moon, the pendulum will have a lower frequency on the surface of the Moon as on the surface of the Earth.