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

AI Thread Summary
The discussion revolves around calculating the ratio of free-fall acceleration between Tokyo and Cambridge based on the lengths of seconds pendulums at each location. The lengths are 0.9927 m in Tokyo and 0.9942 m in Cambridge, prompting questions about how these lengths relate to gravitational acceleration. Participants highlight the importance of the pendulum's period and its dependence on both length and acceleration, referencing the formula T = 2π√(L/g). The conversation also touches on the need to derive acceleration from the period of oscillation and the relationship between the two locations. Ultimately, the focus is on understanding how pendulum length influences free-fall acceleration.
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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