### 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 πΏ is the length of the
pendulum and π is the local acceleration due to gravity.

Now, often when weβre dealing with
problems in mechanics, such as this one, we treat π as if itβs 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 weβd find
that the local gravitational acceleration would be different. So to be precise, we should really
call this quantity π sub e to signify that itβs the gravitational acceleration near
the surface of the Earth.

In this question, weβre 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. Letβs call this π m. Now, the question doesnβt 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 π m is less
than π 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
π. And since weβre not changing the
pendulum itself, we know that πΏ is effectively a constant. Since π is changing and everything
else on the right-hand side of this formula is constant, this means that π, the
frequency of the pendulum, must also change. So now at least we know that the
frequency wonβt stay the same as when it was on Earth.

However, looking at this formula,
itβs not immediately obvious whether decreasing π will cause π to increase or
decrease. In a situation like this, when
weβre trying to determine how a change in one variable will affect another variable
but we donβt 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 π proportional to
π? Could it be proportional to the
square of π? How about inversely proportional to
π? Establishing this type of
relationship between the two variables π and π will enable us to easily see
whether decreasing π will increase or decrease π.

Now, for the sake of this question,
where weβre not actually calculating an answer, letβs 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 π is
proportional to one over the square root of one over π. Now, this isnβt much clearer since
we still have a fraction within a fraction. However, using algebra, we can
clear some of this up.

Firstly, letβs deal with the square
root sign. The square root of one over π is
equivalent to the square root of one over the square root of π. 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 π. In other words, we have the
reciprocal of the reciprocal of the square root of π. Taking the reciprocal over
reciprocal in this way just leaves us with the quantity that we started with. And in this case, thatβs the square
root of π. This shows us that the frequency π
is proportional to the square root of π.

Now that weβve 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 π is decreased. And if we decrease the size of π,
then we decrease the size of the square root of π. Since π is proportional to the
square root of π, this means that π 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.