Video Transcript
The acceleration due to gravity on
the surface of the moon is 0.165 times that on the surface of the Earth. If a pendulum has a period of 5.60
seconds when it’s on the surface of the Earth, what would its period be if it were
on the surface of the Moon?
So in this question, we’re
considering two different scenarios. In one scenario, we have a pendulum
on the surface of the Earth. And we’re told that this pendulum
has a period, let’s call this 𝑇, of 5.60 seconds. And in the other scenario, we have
the same pendulum on the Moon. We’re also told that the
acceleration due to gravity on the surface of the Moon is 0.165 times that on the
surface of the Earth. So if we call acceleration due to
gravity on the surface of the Earth 𝑔 e and acceleration due to gravity on the
surface of the Moon 𝑔 m, then we could say that 𝑔 m is 0.165 times 𝑔 e.
We’re being asked to determine what
the period of the pendulum is on the Moon. In order to answer this question,
we need to recall that the period of a pendulum depends on the acceleration due to
gravity that it experiences. Specifically, we can say that the
period of a pendulum 𝑇 is approximately equal to two 𝜋 times the square root of 𝐿
over 𝑔, where 𝐿 is the length of the pendulum and 𝑔 is the acceleration due to
gravity.
Now, in many problems, we treat 𝑔
as if it’s just a constant. Generally, we say it has a value of
9.81 meters per second squared. However, this is just the
acceleration due to gravity near the surface of the Earth. And if we travel to the Moon as in
this question, then the acceleration due to gravity is different. So it’s important to realize that,
in this question, 𝑔 is a variable.
Conversely, since we’re just
considering one pendulum, the length is effectively a constant. So in this question, we’re being
asked to calculate the period of the pendulum on the moon, which we can call 𝑇 m,
based on the period of the pendulum on Earth, which we’ve been given. And to make a distinction, we’ll
call this 𝑇 e.
In situations like this, since
we’re effectively comparing 𝑇 m and 𝑇 e, it’s useful to attempt to calculate the
ratio between them. In other words, it’s useful to
calculate 𝑇 m over 𝑇 e. Using this formula, we can say that
𝑇 m is approximately equal to two 𝜋 multiplied by the square root of the length of
the pendulum 𝐿 divided by the acceleration due to gravity on the surface of the
Moon 𝑔 m. And 𝑇 e is equal to the same
expression, but this time with the gravitational acceleration on Earth 𝑔 e.
Looking at this expression, we can
see that many of the factors in the numerator and the denominator cancel. And we can make this even clearer
if we express the square root of 𝐿 over 𝑔 as the square root of 𝐿 over the square
root of 𝑔. We can now see that the factors of
two 𝜋 in the numerator and the denominator of this fraction cancel, as do the
factors of root 𝐿, which leaves us with one over the square root of 𝑔 m divided by
one over the square root of 𝑔 e.
Now, we can simplify this
expression by recognizing that we’re dividing a fraction by another fraction. This is equivalent to multiplying
the top fraction by the reciprocal of the bottom fraction. And this then simplifies to root 𝑔
e divided by root 𝑔 m.
So we’ve now shown that the ratio
between the period of the pendulum on the Moon and the period of the pendulum on
Earth is approximately equal to the ratio between the square root of the
gravitational acceleration on Earth and the square root of the gravitational
acceleration on the Moon. We can now use this expression to
calculate the period of the pendulum on the Moon.
Let’s recall that the question
tells us the acceleration due to gravity on the surface of the Moon is 0.165 times
that on the surface of the Earth. And we’ve summarized this statement
in the expression 𝑔 m equals 0.165 𝑔 e. And we can substitute this into our
expression in order to eliminate a variable. Specifically, we can substitute
0.165 𝑔 e in place of 𝑔 m. We can then rewrite the square root
of 0.165 𝑔 e as the square root of 0.165 multiplied by the square root of 𝑔 e. Doing this, we can see that we can
now cancel a factor of the square root of 𝑔 e from the denominator and numerator of
this expression, which leaves us with one over the square root of 0.165.
All we need to do now is rearrange
this expression slightly, and we can use it to calculate 𝑇 m. We can do this by multiplying both
sides of the expression by 𝑇 e, leaving us with 𝑇 m, that is, the period of the
pendulum on the Moon, is approximately equal to 𝑇 e, the period of the pendulum on
Earth, divided by the square root of 0.165.
Since we’re told in the question
that the period of the pendulum on Earth 𝑇 e is 5.60 seconds, we can substitute
this into our expression to give us a value of 13.786 and so on. Because we’re dividing a time in
seconds by a dimensionless number, this means that our answer is also in
seconds. And rounding our answer to three
significant figures, we obtain a final answer of 13.8 seconds. Given that the acceleration due to
gravity on the surface of the Moon is 0.165 times that on the surface of the Earth,
then a pendulum with a period of 5.60 seconds on Earth would have a period of 13.8
seconds on the Moon.
