# Period of a Pendulum on the Moon

• imccnj
In summary, the equation T=2π√L/g governs the period of a pendulum's swinging, where T is the period, L is the length of the pendulum, and g is a constant representing the strength of Earth's gravity. This equation also applies to other planets and moons, where g may have a different value. On the Moon, where g is 1/6th of its normal value, the period of a pendulum would be longer compared to Earth. This can be calculated using the formula Tm/Te = √(ge/gm), where Tm is the period on the Moon, Te is the period on Earth, ge is the strength of Earth's gravity, and gm is the strength
imccnj
Homework Statement
I need help and clarification about "Period of Pendulum on the moon"
Relevant Equations
T=2π√L/g
The equation that governs the period of a pendulum’s swinging. T=2π√L/g

Where T is the period, L is the length of the pendulum and g is a constant, equal to 9.8 m/s2. The symbol g is a measure of the strength of Earth’s gravity, and has a different value on other planets and moons.

On our Moon, the strength of earth’s gravity is only 1/6th of the normal value. If a pendulum on Earth has a period of 4.9 seconds, what is the period of that same pendulum on the moon?

What specific question do you have?

I am getting 12 as an answer when the right answer is 2

Please show your work then. We can't tell where your mistake lies if we can't see what you did. Also, don't forget the units.

I think you are right. It is clear from the equation that if g is smaller, T must be greater.
Tm/Te = √(ge/gm)
I think "they" have mistakenly multiplied the period by √(gm/ge).

vela
T=2π√L/g

T = 4.9 g = 9.8

g = (T/2 π)2 = L

9.8 (9.8/2 π)2 = 5.96 = L
T = 2 π √L/g

12 = 2 π √5.96/1.64

Other than the lack of units and a few obvious typos, your work looks fine. As @mjc123 points out, you should expect from the formula the period to increase as ##g## decreases. Intuitively, if the weight of the mass decreases, the restoring torque decreases as well, so you would expect it to take longer to oscillate.

Thank you for your help and clarification.

## 1. What is the period of a pendulum on the moon?

The period of a pendulum on the moon is approximately 2.4 seconds.

## 2. How does the period of a pendulum on the moon compare to the period on Earth?

The period of a pendulum on the moon is longer than on Earth because the gravitational pull on the moon is weaker.

## 3. What factors affect the period of a pendulum on the moon?

The period of a pendulum on the moon is affected by the length of the pendulum, the strength of gravity on the moon, and the amplitude of the pendulum's swing.

## 4. Can the period of a pendulum on the moon be calculated using the same formula as on Earth?

Yes, the period of a pendulum on the moon can be calculated using the same formula as on Earth: T = 2π√(L/g), where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity.

## 5. How does the period of a pendulum on the moon affect timekeeping?

The longer period of a pendulum on the moon means that a clock based on a pendulum would run slower on the moon compared to Earth. This would need to be taken into account for accurate timekeeping on the moon.

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