### Video Transcript

In this video, weโre going to be
talking about the period of a pendulum. Weโll see how the motion of a
pendulum is affected by its length and by the strength of gravity. And weโll also be looking closely
at these formulas, which tell us the approximate period and frequency of a
pendulum.

Now, a pendulum consists simply of
a mass known as a bob, suspended from a pivot. But despite how simple it is to
construct a pendulum, describing its motion can actually be quite complex. As a pendulum swings, its position,
velocity, and acceleration all vary constantly, both in terms of their magnitude and
direction. If we think about the energies in
the system, we can see that these vary continuously as well. The kinetic energy of the pendulum
depends on its velocity. And the gravitational potential
energy of the pendulum depends on its height and, therefore, on its position. And both of these quantities vary
over time.

So when weโre considering a
pendulum, these quantities can all be quite difficult to keep track of. However, there are other ways that
we can describe the pendulum motion that are much more straightforward. For example, it turns out that as
long as our pendulum only swings over a small angle, its period, that is, the time
it takes to swing back and forth once, is constant. This means that its frequency, in
other words, the rate at which it swings, is also constant. This property actually means that
pendulums can be used as timekeeping devices, which is the reason why some types of
clock feature a pendulum.

Talking about the frequency or
period of a pendulum is a useful way of describing its motion without trying to keep
track of its position, velocity, and acceleration at any given time. And conveniently, these quantities
are easy to calculate. The time period of a simple
pendulum, denoted by a capital ๐, is approximately equal to two ๐ times the square
root of ๐ฟ over ๐, where ๐ฟ is the length of the pendulum and ๐ is the local
acceleration due to gravity. And we can recall that, near the
surface of the Earth, ๐ takes a value of 9.81 meters per second squared. We can also calculate the frequency
of a pendulum simply by taking the reciprocal of its period. In other words, the frequency of a
simple pendulum is given by one over two ๐ times the square root of ๐ฟ over ๐.

Now, these formulas are very
simple, but this is because theyโre derived using a number of approximations. This means we need to be careful
about when we can use them. For example, weโve already
mentioned that the period of a pendulum is constant for small swings. When we say small swing, we
specifically mean that the initial angle of the pendulum from the vertical position,
which we can call ๐ zero, is much less than one radian. So this is a necessary condition if
we want to use these two formulas.

In addition to this, the two
formulas assume that the string or rod used to construct the pendulum has a
negligible mass. And they ignore the effects of air
resistance. They also ignore any other
resistances that might slow the pendulum down, such as friction of the pivot. So when weโre using these two
formulas, we should try and make sure that the pendulum didnโt have a large initial
angle and that the string or rod used to construct the pendulum has a much smaller
mass than the bob. And we should also just bear in
mind that these formulas donโt take air resistance or friction into account.

Itโs worth noting that, in most
practical applications, air resistance and friction are what cause pendulums to slow
down and eventually stop moving. But here weโre using a simplified
model of a pendulum in which it keeps moving back and forth forever.

One final interesting thing to note
about these formulas is that neither of them depend on mass. This means that if we have two
pendulums that are the same length and subject to the same gravitational
acceleration, then their periods and frequencies of oscillation will be the same,
even if one has a much larger mass than the other one.

So now that weโve seen these two
formulas and looked at the assumptions used in their derivation, letโs try using
them to solve some practice problems.

A pendulum has a length of 0.500
meters. What is the period of the
pendulum? Use a value of 9.81 meters per
second squared for the local acceleration due to gravity. Give your answer to three
significant figures.

So in this question, weโre
considering a pendulum, and weโre being asked to find its period. For any oscillating system, the
period refers to the amount of time taken to complete a full cycle. For a pendulum, that means the time
taken for it to swing from its starting position to the other side and then back to
its starting position. In this case, weโre told that our
pendulum has a length of 0.500 meters. And weโre also told that the local
acceleration due to gravity is 9.81 meters per second squared, which is the standard
gravitational acceleration for an object near the surface of the Earth.

Now, it may not seem like weโve
been given much information about this pendulum. For example, we have no idea what
its mass is. However, this is all the
information we need to calculate its period. In fact, we can recall that the
period of a pendulum is given by this formula. ๐ is approximately equal to two ๐
times the square root of ๐ฟ over ๐, where ๐ is the period of the pendulum, ๐ฟ is
the length of the pendulum, and ๐ is the gravitational acceleration experienced by
the pendulum.

Note that in this formula, ๐ is
only approximately equal to the expression on the right-hand side. This is because the formula is
derived using a number of approximations. Importantly, it uses a small-angle
approximation, which means itโs only accurate if the initial angle of the pendulum
from the vertical, ๐ zero, is much smaller than one radian. The bigger the initial angle is,
the more ๐ will vary from the expression given on the right. So in our answer to this question,
weโre assuming that the pendulum only swings over a small angle.

In this question, weโre told that
the length of the pendulum is 0.500 meters and the local acceleration due to gravity
is 9.81 meters per second squared. So all we need to do to calculate
the period of the pendulum is substitute these values into this formula. Substituting them in gives us two
๐ times the square root of 0.500 meters divided by 9.81 meters per second
squared.

At this point, itโs worth
mentioning that this formula only works when we use the magnitude of the
acceleration. Itโs fairly common in physics
problems to define upward acceleration as positive and downward acceleration as
negative. But when weโre using this formula,
itโs important that we donโt use a negative value for acceleration.

If we type all of this into our
calculator, we obtain a value of 1.418 and so on, which, rounded to three
significant figures, is 1.42. Since the period is a measure of
time, we would expect our answer to have units of time. And if we look at the units weโve
used in our calculation, we can see that this is the case. Weโre dividing the length, which is
a quantity in meters, by an acceleration, measured in meters per second squared. And weโre then taking the square
root of this quantity. We can see that in this fraction,
the factor of meters on the top and the bottom will cancel, leaving us with the
square root of one over one over second squared.

Now, when we have one over one over
a quantity, in other words, the reciprocal of the reciprocal of a quantity, this
actually just leaves us with whatever quantity it was that we started with, in this
case second squared. And the square root of second
squared is of course seconds, which shows us that our answer is 1.42 seconds. And this is the answer to the
question. If a pendulum has a length of 0.500
meters and the local acceleration due to gravity is 9.81 meters per second squared,
then the period of the pendulum is 1.42 seconds.

Now letโs take a look at another
question.

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 ๐ฟ is the length of the
pendulum and ๐ is the local acceleration due to gravity.

Now, often when weโre dealing with
problems in mechanics, such as this one, we treat ๐ as if itโs 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 weโd find
that the local gravitational acceleration would be different. So to be precise, we should really
call this quantity ๐ sub e to signify that itโs the gravitational acceleration near
the surface of the Earth.

In this question, weโre 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. Letโs call this ๐ m. Now, the question doesnโt 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 ๐ m is less
than ๐ 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
๐. And since weโre not changing the
pendulum itself, we know that ๐ฟ is effectively a constant. Since ๐ is changing and everything
else on the right-hand side of this formula is constant, this means that ๐, the
frequency of the pendulum, must also change. So now at least we know that the
frequency wonโt stay the same as when it was on Earth.

However, looking at this formula,
itโs not immediately obvious whether decreasing ๐ will cause ๐ to increase or
decrease. In a situation like this, when
weโre trying to determine how a change in one variable will affect another variable
but we donโt 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 ๐ proportional to
๐? Could it be proportional to the
square of ๐? How about inversely proportional to
๐? Establishing this type of
relationship between the two variables ๐ and ๐ will enable us to easily see
whether decreasing ๐ will increase or decrease ๐.

Now, for the sake of this question,
where weโre not actually calculating an answer, letโs 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 ๐ is
proportional to one over the square root of one over ๐. Now, this isnโt much clearer since
we still have a fraction within a fraction. However, using algebra, we can
clear some of this up.

Firstly, letโs deal with the square
root sign. The square root of one over ๐ is
equivalent to the square root of one over the square root of ๐. 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 ๐. In other words, we have the
reciprocal of the reciprocal of the square root of ๐. Taking the reciprocal over
reciprocal in this way just leaves us with the quantity that we started with. And in this case, thatโs the square
root of ๐. This shows us that the frequency ๐
is proportional to the square root of ๐.

Now that weโve 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 ๐ is decreased. And if we decrease the size of ๐,
then we decrease the size of the square root of ๐. Since ๐ is proportional to the
square root of ๐, this means that ๐ 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.

So with that answered, letโs now
take a look at one more practice question.

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 itโs on the surface of the Earth, what would its period be if it were
on the surface of the Moon?

So in this question, weโre
considering two different scenarios. In one scenario, we have a pendulum
on the surface of the Earth. And weโre told that this pendulum
has a period, letโs call this ๐, of 5.60 seconds. And in the other scenario, we have
the same pendulum on the Moon. Weโre 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 ๐ e and acceleration due to gravity on the
surface of the Moon ๐ m, then we could say that ๐ m is 0.165 times ๐ e.

Weโre 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 ๐ is approximately equal to two ๐ times the square root of ๐ฟ
over ๐, where ๐ฟ is the length of the pendulum and ๐ is the acceleration due to
gravity.

Now, in many problems, we treat ๐
as if itโs 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 itโs important to realize that,
in this question, ๐ is a variable.

Conversely, since weโre just
considering one pendulum, the length is effectively a constant. So in this question, weโre being
asked to calculate the period of the pendulum on the moon, which we can call ๐ m,
based on the period of the pendulum on Earth, which weโve been given. And to make a distinction, weโll
call this ๐ e.

In situations like this, since
weโre effectively comparing ๐ m and ๐ e, itโs useful to attempt to calculate the
ratio between them. In other words, itโs useful to
calculate ๐ m over ๐ e. Using this formula, we can say that
๐ m is approximately equal to two ๐ multiplied by the square root of the length of
the pendulum ๐ฟ divided by the acceleration due to gravity on the surface of the
Moon ๐ m. And ๐ e is equal to the same
expression, but this time with the gravitational acceleration on Earth ๐ 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 ๐ฟ over ๐ as the square root of ๐ฟ over the square
root of ๐. We can now see that the factors of
two ๐ in the numerator and the denominator of this fraction cancel, as do the
factors of root ๐ฟ, which leaves us with one over the square root of ๐ m divided by
one over the square root of ๐ e.

Now, we can simplify this
expression by recognizing that weโre 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 ๐
e divided by root ๐ m.

So weโve 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.

Letโs 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 weโve summarized this statement
in the expression ๐ m equals 0.165 ๐ e. And we can substitute this into our
expression in order to eliminate a variable. Specifically, we can substitute
0.165 ๐ e in place of ๐ m. We can then rewrite the square root
of 0.165 ๐ e as the square root of 0.165 multiplied by the square root of ๐ e. Doing this, we can see that we can
now cancel a factor of the square root of ๐ 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 ๐ m. We can do this by multiplying both
sides of the expression by ๐ e, leaving us with ๐ m, that is, the period of the
pendulum on the Moon, is approximately equal to ๐ e, the period of the pendulum on
Earth, divided by the square root of 0.165.

Since weโre told in the question
that the period of the pendulum on Earth ๐ e is 5.60 seconds, we can substitute
this into our expression to give us a value of 13.786 and so on. Because weโre 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.

Letโs now recap some of the key
points that weโve learned in this video. For a pendulum with a small initial
angle measured from the downward vertical position, the frequency and period depend
only on the length of the pendulum and the local gravitational acceleration. Specifically, weโve seen that the
period of a pendulum ๐ is approximately equal to two ๐ times the square root of ๐ฟ
over ๐. And the frequency ๐ is
approximately equal to one over two ๐ times the square root of ๐ฟ over ๐, where ๐ฟ
is the length of the pendulum and ๐ is the local gravitational acceleration. This is a summary of the period of
a pendulum.