Period of a Pendulum on the Moon

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
  • #1
imccnj
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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?
 
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  • #2
What specific question do you have?
 
  • #3
I am getting 12 as an answer when the right answer is 2
 
  • #4
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.
 
  • #5
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).
 
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  • #6
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
 
  • #7
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.
 
  • #8
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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