What is the ratio of the free-fall acceleration at these two locations?

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Homework Help Overview

The discussion revolves around the relationship between the length of a seconds pendulum and the free-fall acceleration at two different locations: Tokyo and Cambridge. Participants are exploring how the physical properties of the pendulum relate to gravitational acceleration.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants question whether free-fall acceleration is constant everywhere and how the lengths of the pendulums influence this acceleration. Some suggest examining the forces acting on the pendulum and the nature of its motion.

Discussion Status

The discussion is active, with participants offering insights into the relationship between pendulum length and acceleration. Some have proposed looking into the equations governing simple harmonic motion, while others are considering the implications of centripetal force.

Contextual Notes

There is a recurring theme of uncertainty regarding the assumptions about gravitational acceleration being uniform and how it relates to the pendulum's period and length. Participants are encouraged to reference the relevant equations for further clarity.

leighzer
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Homework Statement


A "seconds" pendulum is one that moves through its equilibrium position once each second (period = 2.000 s). The length of a seconds pendulum in Tokyo is 0.9927 m and at Cambridge is 0.9942 m. What is the ratio of the free-fall acceleration at these two locations?


Homework Equations


All equations for Simple Harmonic Motion


The Attempt at a Solution


Wouldn't the free-fall acceleration be equal everywhere? If not then can someone please tell me what the length of the pendulums have to do with acceleration?
 
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Think about what forces are keeping the pendulum from flying off. And why exactly does a pendulum swing in a circular motion?
 
So is this question about centripetal force and acceleration then?
 
If you want it to be :). Look at your equation for simple harmonic motion. You should have a term for acceleration in it. How does acceleration relate to the length?

Acceleration is the time derivative of velocity. And velocity is the time derivative of...
 
leighzer said:

Homework Statement


A "seconds" pendulum is one that moves through its equilibrium position once each second (period = 2.000 s). The length of a seconds pendulum in Tokyo is 0.9927 m and at Cambridge is 0.9942 m. What is the ratio of the free-fall acceleration at these two locations?


Homework Equations


All equations for Simple Harmonic Motion


The Attempt at a Solution


Wouldn't the free-fall acceleration be equal everywhere? If not then can someone please tell me what the length of the pendulums have to do with acceleration?

Look up the equation for the period of oscillation of a simple pendulum.
 
I have this question also

The equation for the period of oscillation:
T=2pi* square root (L/g)

Once you find T
Solve for w [w=2pi/T]
Then I thought of using the velocity formula: v=Aw
But... I'm not sure how to find the amplitude.

Can you guide me in the right direction?
 
so...

T2/ T1 = 1 = sqrt(L1/g1) / sqrt (L2/g2)?
 

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