Period of a Pendulum on the Moon

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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?
 

Answers and Replies

  • #2
vela
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What specific question do you have?
 
  • #3
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I am getting 12 as an answer when the right answer is 2
 
  • #4
vela
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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
mjc123
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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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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
vela
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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.
 
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Thank you for your help and clarification.
 

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