Period of a Pendulum on the Moon

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Homework Help Overview

The discussion revolves around the calculation of the period of a pendulum on the Moon, utilizing the formula T=2π√(L/g), where T is the period, L is the length of the pendulum, and g represents the gravitational acceleration. Participants explore how the value of g differs between Earth and the Moon, affecting the pendulum's period.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the implications of using different values for g in the formula, question the calculations presented, and seek clarification on the relationship between gravitational strength and the period of the pendulum.

Discussion Status

The discussion is active, with participants providing feedback on calculations and questioning assumptions. Some guidance has been offered regarding the expected relationship between gravitational strength and period, but no consensus has been reached on the correct answer.

Contextual Notes

Participants note the importance of including units in calculations and highlight potential errors in the application of the formula. There are indications of confusion regarding the expected outcomes based on the gravitational differences.

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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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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).
 
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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
 
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.
 

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