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

In summary, the length of the pendulum has no impact on the free-fall acceleration at either location.
  • #1
leighzer
10
0

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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  • #2
Think about what forces are keeping the pendulum from flying off. And why exactly does a pendulum swing in a circular motion?
 
  • #3
So is this question about centripetal force and acceleration then?
 
  • #4
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...
 
  • #5
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.
 
  • #6
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?
 
  • #7
so...

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

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

The ratio of the free-fall acceleration at two locations is a comparison of the acceleration due to gravity at each location. This ratio can be calculated by dividing the acceleration due to gravity at one location by the acceleration due to gravity at the other location.

How is the free-fall acceleration calculated?

The free-fall acceleration is calculated using the equation a = F/m, where a is the acceleration, F is the force of gravity, and m is the mass of the object. This equation is derived from Newton's second law of motion, which states that the acceleration of an object is directly proportional to the net force acting on the object and inversely proportional to its mass.

What factors can affect the free-fall acceleration?

The free-fall acceleration can be affected by factors such as altitude, latitude, and the density of the medium through which the object is falling. Other factors that can affect the free-fall acceleration include the shape and size of the object and the presence of air resistance.

How does the free-fall acceleration vary on Earth?

The free-fall acceleration varies slightly on Earth due to differences in altitude and latitude. At higher altitudes, the force of gravity is slightly weaker, resulting in a slightly lower free-fall acceleration. Similarly, the free-fall acceleration is slightly higher at the poles compared to the equator due to the Earth's rotation.

Why is the free-fall acceleration considered constant?

The free-fall acceleration is considered constant because it is independent of the mass and shape of the object falling, and is only affected by the force of gravity and the medium through which the object is falling. This means that all objects, regardless of their size or mass, will experience the same acceleration when falling in a vacuum.

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