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Lesson Video: The Period of a Pendulum Physics • 9th Grade

In this video, we will learn how to calculate the period and frequency of a pendulum given its length and the local acceleration due to gravity.

17:05

Video Transcript

In this video, weโ€™re going to be talking about the period of a pendulum. Weโ€™ll see how the motion of a pendulum is affected by its length and by the strength of gravity. And weโ€™ll 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 weโ€™re 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œ‹ times the square root of ๐ฟ over ๐‘”, where ๐ฟ 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œ‹ times the square root of ๐ฟ over ๐‘”.

Now, these formulas are very simple, but this is because theyโ€™re derived using a number of approximations. This means we need to be careful about when we can use them. For example, weโ€™ve 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œƒ 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 weโ€™re using these two formulas, we should try and make sure that the pendulum didnโ€™t 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 donโ€™t take air resistance or friction into account.

Itโ€™s worth noting that, in most practical applications, air resistance and friction are what cause pendulums to slow down and eventually stop moving. But here weโ€™re 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 weโ€™ve seen these two formulas and looked at the assumptions used in their derivation, letโ€™s 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, weโ€™re considering a pendulum, and weโ€™re 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, weโ€™re told that our pendulum has a length of 0.500 meters. And weโ€™re 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 weโ€™ve 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œ‹ times the square root of ๐ฟ over ๐‘”, where ๐‘‡ is the period of the pendulum, ๐ฟ 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 itโ€™s only accurate if the initial angle of the pendulum from the vertical, ๐œƒ 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, weโ€™re assuming that the pendulum only swings over a small angle.

In this question, weโ€™re 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œ‹ times the square root of 0.500 meters divided by 9.81 meters per second squared.

At this point, itโ€™s worth mentioning that this formula only works when we use the magnitude of the acceleration. Itโ€™s fairly common in physics problems to define upward acceleration as positive and downward acceleration as negative. But when weโ€™re using this formula, itโ€™s important that we donโ€™t 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 weโ€™ve used in our calculation, we can see that this is the case. Weโ€™re dividing the length, which is a quantity in meters, by an acceleration, measured in meters per second squared. And weโ€™re 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 letโ€™s 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 ๐ฟ is the length of the pendulum and ๐‘” is the local acceleration due to gravity.

Now, often when weโ€™re dealing with problems in mechanics, such as this one, we treat ๐‘” as if itโ€™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 weโ€™d 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 itโ€™s the gravitational acceleration near the surface of the Earth.

In this question, weโ€™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. Letโ€™s call this ๐‘” m. Now, the question doesnโ€™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 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 weโ€™re not changing the pendulum itself, we know that ๐ฟ 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 wonโ€™t stay the same as when it was on Earth.

However, looking at this formula, itโ€™s not immediately obvious whether decreasing ๐‘” will cause ๐‘“ to increase or decrease. In a situation like this, when weโ€™re trying to determine how a change in one variable will affect another variable but we donโ€™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 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 weโ€™re not actually calculating an answer, letโ€™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 ๐‘“ is proportional to one over the square root of one over ๐‘”. Now, this isnโ€™t much clearer since we still have a fraction within a fraction. However, using algebra, we can clear some of this up.

Firstly, letโ€™s 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, thatโ€™s the square root of ๐‘”. This shows us that the frequency ๐‘“ is proportional to the square root of ๐‘”.

Now that weโ€™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 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, letโ€™s 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 itโ€™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, weโ€™re considering two different scenarios. In one scenario, we have a pendulum on the surface of the Earth. And weโ€™re told that this pendulum has a period, letโ€™s call this ๐‘‡, of 5.60 seconds. And in the other scenario, we have the same pendulum on the Moon. Weโ€™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 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.

Weโ€™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 ๐‘‡ is approximately equal to two ๐œ‹ times the square root of ๐ฟ over ๐‘”, where ๐ฟ is the length of the pendulum and ๐‘” is the acceleration due to gravity.

Now, in many problems, we treat ๐‘” as if itโ€™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 itโ€™s important to realize that, in this question, ๐‘” is a variable.

Conversely, since weโ€™re just considering one pendulum, the length is effectively a constant. So in this question, weโ€™re 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 weโ€™ve been given. And to make a distinction, weโ€™ll call this ๐‘‡ e.

In situations like this, since weโ€™re effectively comparing ๐‘‡ m and ๐‘‡ e, itโ€™s useful to attempt to calculate the ratio between them. In other words, itโ€™s useful to calculate ๐‘‡ m over ๐‘‡ e. Using this formula, we can say that ๐‘‡ m is approximately equal to two ๐œ‹ multiplied by the square root of the length of the pendulum ๐ฟ 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 ๐ฟ over ๐‘” as the square root of ๐ฟ over the square root of ๐‘”. We can now see that the factors of two ๐œ‹ in the numerator and the denominator of this fraction cancel, as do the factors of root ๐ฟ, 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 weโ€™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 ๐‘” e divided by root ๐‘” m.

So weโ€™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.

Letโ€™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 weโ€™ve 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 weโ€™re 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 weโ€™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.

Letโ€™s now recap some of the key points that weโ€™ve 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, weโ€™ve seen that the period of a pendulum ๐‘‡ is approximately equal to two ๐œ‹ times the square root of ๐ฟ over ๐‘”. And the frequency ๐‘“ is approximately equal to one over two ๐œ‹ times the square root of ๐ฟ over ๐‘”, where ๐ฟ is the length of the pendulum and ๐‘” is the local gravitational acceleration. This is a summary of the period of a pendulum.

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