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

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

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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.

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