# Simple Harmonic Motion of a pendulum

1. Nov 28, 2006

### leighzer

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
All equations for Simple Harmonic Motion

3. 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?

2. Nov 28, 2006

### KoGs

Think about what forces are keeping the pendulum from flying off. And why exactly does a pendulum swing in a circular motion?

3. Nov 28, 2006

### leighzer

So is this question about centripetal force and acceleration then?

4. Nov 28, 2006

### KoGs

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. Nov 28, 2006

### OlderDan

Look up the equation for the period of oscillation of a simple pendulum.

6. Nov 18, 2008

### fanie1031

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. Nov 4, 2009

### vegetto34

so...

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