Question

Calculate the period and frequency of a 3.500 m long pendulum at the following locations:

a. the North Pole, where ag= 9.832 m/s^2
b. Chicago, where ag= 9.803 m/s^2
c. Jakarta, Indonesia, where ag= 9.782 m/s^2

Answer 1

1)

The period of a simple pendulum is given by the formula:

`T=2\pi \sqrt{(L)/(g)}`

where

L is the length of the pendulum

g is the acceleration due to gravity

In this problem we have:

L = 3.50 m is the length of the pendulum

At the North Pole, we have

`g=9.832 m/s^2`

So the period is

`T=2\pi \sqrt{(3.50)/(9.832)}=3.749 s`

At Chicago, we have

`g=9.803 m/s^2`

So the period is

`T=2\pi \sqrt{(3.50)/(9.803)}=3.754 s`

At Jakarta, we have

`g=9.782 m/s^2`

So the period is

`T=2\pi \sqrt{(3.50)/(9.782)}=3.758 s`

2)

The frequency of an object in simple harmonic motion is equal to the reciprocal of the period:

`f=(1)/(T)`

At the North Pole, we have

T = 3.749 s

So the frequency is

`f=(1)/(3.479)=0.267 Hz`

At Chicago, we have

T = 3.754 s

So the frequency is

`f=(1)/(3.754)=0.266 Hz`

At Jakarta, we have

T = 3.758 s

So the frequency is

`f=(1)/(3.758)=0.266 Hz`